# Stability in quadratic variation, with applications

Philip Kennerberg∗ and Magnus Wiktorsson111Centre for Mathematical Sciences,
Lund University, Box 118 SE-22100, Lund, Sweden,
<EMAIL_ADDRESS>
###### Abstract

We show that non continuous Dirichlet processes, defined as in Coquet, Mémin
and Słomiński (2003) are closed under a wide family of locally Lipschitz
continuous maps (similar to the time-homogeneous variants of the maps
considered in Lowther (2010)) thus extending Theorem 2.1. from that paper. We
provide an Itô formula for these transforms and apply it to study of how
$[f(X^{n})-f(X)]\to 0$ when $X^{n}\to X$ (in some appropriate sense) for
certain Dirichlet processes $\\{X^{n}\\}_{n}$, $X$ and certain locally
Lipschitz continuous maps. We also consider how $[f_{n}(X^{n})-f(X)]\to 0$ for
$C^{1}$ maps $\\{f_{n}\\}_{n}$, $f$ when $f_{n}^{\prime}\to f^{\prime}$
uniformly on compacts. For applications we give examples of jump removal and
stability of integrators.

##  1 Introduction

We will study stability of process limits in quadratic variation when the
processes are transformed by (a wide class of) locally Lipschitz continuous or
$C^{1}$ maps. More precisely, if $\\{f_{n}\\}_{n}$ and $f$ are some functions,
$\\{X^{n}\\}_{n}$ and $X$ are cadlag processes such that $X^{n}\to X$ (in a
sense specified below) then we investigate when $[f_{n}(X^{n})-f(X)]\to 0$ in
a corresponding sense. We will mostly consider Dirichlet processes in this
paper but we will also show some results for the more general class of
processes that admit to a quadratic variation. We will study stability in
quadratic variation both in probability, $L^{p}$ and in the almost sure sense.
We will assume that our processes live on some compact time interval say
$[0,t]$ for some $t\in{\mathbb{R}}^{+}$. The term refining sequence will refer
to a sequence of partitions of $[0,t]$, $\\{D_{k}\\}_{k}$ say, such that
$D_{k}\subseteq D_{k+1}$ and $\lim_{k\to\infty}\max_{t_{i}\in
D_{k}}|t_{i+1}-t_{i}|=0$. We say that a cadlag process $X$ admits to a
quadratic variation $[X]$ if there exists an increasing continuous process
$[X]^{c}$ such that

$\displaystyle[X]_{s}=[X]^{c}_{s}+\sum_{u\leq s}(\Delta X_{u})^{2},$ (1.1)

for every $0<s\leq t$, there exists at least one refining sequence
$\\{D_{k}\\}_{k}$ such that if we let

$(S_{n}(X))_{s}:=\sum_{t_{i}\in D_{n},t_{i}\leq
s}\left(X_{t_{i+1}}-X_{t_{i}}\right)^{2},$

for every $0<s\leq t$ (where we use the convention that $t_{i+1}=s$ if
$t_{i}=s$) then

$\displaystyle(S_{n}(X))_{s}\xrightarrow{{\mathbb{P}}}[X]_{s}\textit{ as
}n\to\infty.$ (1.2)

We say that two cadlag processes admit to a covariation $[X,Y]$ along
$\\{D_{n}\\}_{n}$ if $[X,Y]^{c}$ is a continuous finite variation process such
that

$[X,Y]_{s}=[X,Y]^{c}_{s}+\sum_{u\leq s}\Delta X_{u}\Delta Y_{u},$

for every $0<s\leq t$ and if we let

$S_{n}(X,Y)_{s}:=\sum_{t_{i}\in D_{n},t_{i}\leq
s}\left(X_{t_{i+1}}-X_{t_{i}}\right)\left(Y_{t_{i+1}}-Y_{t_{i}}\right),$

for every $0<s\leq t$ then $S_{n}(X,Y)_{s}\xrightarrow{{\mathbb{P}}}[X,Y]_{s}$
as $n\to\infty$.
As in Coquet, Mémin and Słomiński (2003) we define a (non-continuous)
Dirichlet process as a sum of a semimartingale and an adapted continuous
process with zero quadratic variation along some refining sequence. By this
definition we see that Dirichlet processes are a subclass of the processes
admitting to quadratic variation.

##  2 Preliminaries

We assume that all of our processes are defined on a common filtered
probability space $(\Omega,\mathcal{F},\\{\mathcal{F}_{t}\\}_{t\geq
0},{\mathbb{P}})$ and that all the defined processes are adapted to the same
filtration $\\{\mathcal{F}_{t}\\}_{t\geq 0}$ fulfilling the usual hypothesis.

Given a cadlag process $X_{t}$ with $t\geq 0$ and a stopping time $T$ we
defined $X$ stopped at $T$ as $X^{T}_{t}=X_{t\wedge T}$, we also define the
supremum process of $X$ as $X^{*}_{t}=\sup_{s\leq t}|X_{s}|$.

######  Definition 2.1.

A property of a stochastic process is said to hold locally (pre-locally) if
there exist a sequence of stopping times $T_{k}$ increasing to infinity such
that the property holds for the stopped process $X^{T_{k}}$ ($X^{T_{k}-}$) for
each $k$.

######  Lemma 2.2.

Suppose $X^{1},...,X^{n}$ are processes with quadratic variations along the
same refining sequence and assume that $\sum_{k=1}^{m}X^{k}$ also has
quadratic variation along this refining sequence for all $m\leq n$ then

$\left[\sum_{k=1}^{n}X^{k}\right]_{t}\leq\left(\sum_{k=1}^{n}[X^{k}]_{t}^{\frac{1}{2}}\right)^{2}.$

###### Proof.

First of all we note that since $\sum_{k=1}^{m}X^{k}$ has a quadratic
variation along the same refining sequence for each $m\leq n$ it follows that
the covariation $\left[\sum_{k=1}^{m}X^{k},X^{m+1}\right]$ also exists along
the same refining sequence. We will now prove the stated inequality by
induction, the case $n=1$ is trivial. Assume the statement is true for $n=m$
then for $n=m+1$,

$\displaystyle\left[\sum_{k=1}^{m+1}X^{k}\right]_{t}=\left[\sum_{k=1}^{m}X^{k}\right]_{t}+2\left[\sum_{k=1}^{m}X^{k},X^{m+1}\right]_{t}+\left[X^{m+1}\right]_{t}\leq\left[\sum_{k=1}^{m}X^{k}\right]_{t}+$
$\displaystyle
2\left[\sum_{k=1}^{m}X^{k}\right]_{t}^{\frac{1}{2}}\left[X^{m+1}\right]^{\frac{1}{2}}_{t}+\left[X^{m+1}\right]_{t}=\left(\left[\sum_{k=1}^{m}X^{k}\right]^{\frac{1}{2}}_{t}+\left[X^{m+1}\right]^{\frac{1}{2}}_{t}\right)^{2}\leq\left(\sum_{k=1}^{m}[X_{k}]_{t}^{\frac{1}{2}}+\left[X^{m+1}\right]^{\frac{1}{2}}_{t}\right)^{2}$
$\displaystyle=\left(\sum_{k=1}^{m+1}[X_{k}]_{t}^{\frac{1}{2}}\right)^{2},$

where we used the Kunita-Watanabe inequality in the second step. ∎

######  Lemma 2.3.

Suppose $X^{1},...,X^{n}$ are processes with quadratic variations along the
same refining sequence then

$\left[\sum_{k=1}^{n}X^{k}\right]_{t}\leq 2^{n-1}\sum_{k=1}^{n}[X^{k}]_{t}.$

###### Proof.

We again prove the stated inequality by induction, the case $n=1$ is trivial
and the case $n=2$ follows from Lemma 2.2 combined with the inequality
$(a+b)^{2}\leq 2a^{2}+2b^{2}$. Assume the statement is true for $n=m$ then for
$n=m+1$,

$\displaystyle\left[\sum_{k=1}^{m+1}X^{k}\right]_{t}\leq
2\left[\sum_{k=1}^{m}X^{k}\right]_{t}+2\left[X^{m+1}\right]_{t}\leq
2^{(m-1)+1}\sum_{k=1}^{m}[X^{k}]_{t}+2\left[X^{m+1}\right]_{t}\leq
2^{m}\sum_{k=1}^{m+1}[X^{k}]_{t},$

where we used the case $n=2$ in the first step and the induction hypothesis in
the second step. ∎

The following definition is taken from Coquet, Mémin and Słomiński (2003)

######  Definition 2.4.

Given two cadlag processes $X,Y$ we say that the integral
$U_{.}=\int_{0}^{.}Y_{s-}dX_{s}$ is well defined if there exists a refining
sequence $\\{D_{k}\\}_{k}$, $U$ is a cadlag process such that for every $s\leq
t$, $\Delta U_{s}=Y_{s-}\Delta X_{s}$ and $I^{k}(X,Y)\to U,$ as $k\to\infty$
in probability under the $J1$ topology, where $I^{k}(X,Y)_{t}=\sum_{t_{i}\in
D_{k},t_{i}\leq t}Y_{t_{i}}(X_{t_{i+1}}-X_{t_{i}})$.

The next lemma shows that all processes that admit to quadratic variations are
admissible integrators in the sense of Defnition 2.4 for semimartingale
integrands.

######  Lemma 2.5.

Let $Y$ be a semimartingale and $X$ a cadlag process such that both $X$ and
$X+Y$ have quadratic variations along $\\{D_{k}\\}_{k}$ then the integral
$\int_{0}^{.}Y_{s-}dX_{s}$ exists in the sense of Definition 2.4. Moreover
this integral satisfies the integration by parts formula (along
$\\{D_{k}\\}_{k}$).

###### Proof.

For any $k$ and $s\leq t$

$\displaystyle I^{k}(X,Y)=Y_{.}X_{.}-Y_{0}X_{0}-I^{k}(Y,X)-\sum_{t_{i}\in
D_{k},t_{i}\leq.}(Y_{t_{i+1}}-Y_{t_{i}})(X_{t_{i+1}}-X_{t_{i}}).$ (2.3)

Since $Y$ is a semimartingale, $I^{k}(Y,X)$ converges in ucp (and therefore in
probability under the $J1$ topology). We now expand the last term,

$\displaystyle\sum_{t_{i}\in
D_{k},t_{i}\leq.}(Y_{t_{i+1}}-Y_{t_{i}})(X_{t_{i+1}}-X_{t_{i}})=$
$\displaystyle\frac{1}{2}\left(\sum_{t_{i}\in
D_{k},t_{i}\leq.}(Y_{t_{i+1}}+X_{t_{i+1}}-Y_{t_{i}}-X_{t_{i}})^{2}-\sum_{t_{i}\in
D_{k},t_{i}\leq.}(Y_{t_{i+1}}-Y_{t_{i}})^{2}-\sum_{t_{i}\in
D_{k},t_{i}\leq.}(X_{t_{i+1}}-X_{t_{i}})^{2}\right),$ (2.4)

all three terms on the right side of (2) converge in the $J1$ topology due to
Lemma 1.3 in Coquet, Mémin and Słomiński (2003) and therefore the final term
in (2.3) converges in the $J1$ topology. Furthermore, for any $s\leq t$

$(\Delta[X,Y])_{s}=\frac{1}{2}\left(\Delta[X+Y]-\Delta[X]-\Delta[Y]\right)_{s}=\Delta
X_{s}\Delta Y_{s},$

so $[X,Y]$ exists and is the J1 limit of (2) as $k\to\infty$. Taking limits on
the right-hand side of (2.3) and evaluating the jump at time $u$ gives us

$\displaystyle\Delta\left(\int_{0}^{.}Y_{s-}dX_{s}\right)_{u}=\Delta(XY)_{u}-\Delta\left(\int_{0}^{.}X_{s-}dY_{s}\right)_{u}-\Delta
X_{u}\Delta Y_{u}=$ $\displaystyle\Delta(XY)_{u}-\left(X_{u-}+\Delta
X_{u}\right)\Delta Y_{u}=Y_{u-}\Delta X_{u},$

and hence the requirements of Definition 2.4 are met. Evaluating (2.3) at time
$s\leq t$ and taking limits (as $k\to\infty$) in probability gives us the
integration by parts formula. ∎

Combining Remark 2 on page 148 of Föllmer (1981) (which states that if one
transforms a process with quadratic variation by a $C^{1}$ map this new
process will also have a quadratic variation) and Lemma 2.5 gives us the
following.

######  Corollary 2.6.

Suppose $X$ has a quadratic variation along the refining sequence
$\\{D_{k}\\}_{k}$, $f\in C^{1}$ and $Y$ is a semimartingale such that $X+Y$
also has a quadratic variation along $\\{D_{k}\\}_{k}$, then the integral
$\int Y_{-}df(X)$ is well defined in the sense of Definition 2.4

For general a semimartingale $X$ we define (see e.g. Protter, 1992, p. 245.)
the ${\stackrel{{\scriptstyle H}}{{=}}}^{p}$-norm, we here however work on a
finite interval $[0,t]$.

######  Definition 2.7.

For $1\leq p\leq\infty$ define

$j_{p}(M,A)=\lVert\leavevmode\nobreak\
|X_{0}|+[M]_{t}^{1/2}+\int_{0}^{t}|dA_{s}|\leavevmode\nobreak\
\rVert_{\text{${L^{p}}$}},$

and then

$\lVert X\rVert_{{\stackrel{{\scriptstyle
H}}{{=}}}^{p}}=\inf_{X=X_{0}+M+A}j_{p}(M,A),$

for all possible decompositions $X=X_{0}+M+A$ where $M$ is a local martingale
with $M_{0}=0$ and $A$ is a process of finite variation with $A_{0}=0$.

Recall that ${\stackrel{{\scriptstyle H}}{{=}}}^{p}$ is the Banach space of
all semi martingales with finite ${\stackrel{{\scriptstyle H}}{{=}}}^{p}$-norm
and note that all semimartingales with $X_{0}\in L^{p}$ are pre-locally in
${\stackrel{{\scriptstyle H}}{{=}}}^{p}$ (see e.g. Protter, 1992, p. 247.).

##  3 Main results and lemmas

We will work with the following assumptions with regards to transforms

######  Assumption 3.1.

Suppose that $f$ is a locally Lipschitz continuous function and $X$ is a
dirichlet process such that

$\displaystyle\int_{0}^{t}1_{\\{s:X_{s}\not\in\mathsf{diff}(f)\\}}d[X]^{c}_{s}=0\hskip
5.69054pta.s.,$ (3.5)

where $\mathsf{diff}(f)$ is the set of points where $f$ is differentiable

######  Remark 3.2.

Note that if the set of points where $f$ is not differentiable is at most
countable then (3.5) is always fulfilled (see the proof of Theorem 3.16 for
details). Another case when Assumption 3.1 is always satisfied is when
$d[X]^{c}_{s}<<ds$, i.e. when $d[X]^{c}$ is absolutely continuous with respect
to Lebesgue measure.

######  Assumption 3.3.

$\\{f_{n}\\}_{n}$ and $f$ are assumed to be $C^{1}$, $f_{n}(x_{0})\to
f(x_{0})$ for some fixed point $x_{0}$ and $f_{n}^{\prime}\to f^{\prime}$
uniformly on compacts.

Throughout the remainder of this paper, whenever we write $f^{\prime}(x)$ we
will be referring to the Dini derivative $f^{\prime}(x)=\limsup_{h\to
0}\frac{f(x+h)-f(x)}{h}$, if $f$ is in-fact differentiable at $x$ then this
becomes, as we all know, the usual derivative. The following Lemma is an
application of Theorem 2.1 in Lowther (2010).

######  Lemma 3.4.

Suppose that $f$ is a locally Lipschitz continuous function and $X$ is a
Dirichlet process such that they satisfy Assumption 3.1 then

$[f(X)]_{s}=\int_{0}^{s}f^{\prime}(X_{u-})^{2}d[X]^{c}_{s}+\sum_{u\leq
s}\Delta f(X_{u})^{2}.$

###### Proof.

According to Theorem 2.1. in Lowther (2010) we have that if $X$ is sum of a
semimartingale and a process admitting a quadratic variation, with zero
continuous quadratic variation then

$\displaystyle f(X_{s})=\int_{0}^{s}f^{\prime}(X_{u-})dZ_{u}+V_{s},$ (3.6)

where $V_{s}$ is a cadlag process that has a quadratic variation and zero
continuous quadratic variation. Since $X=Z+C$ and $[C]^{c}=[C]=0$, $X$ does
indeed satisfy this assumption. Note that $f(X)$ has a quadratic variation so
we must have that $[f(X)]_{s}=[f(X)]^{c}_{s}+\sum_{u\leq s}\Delta
f(X_{u})^{2}$ while if we compute the quadratic variation of 3.6 we get

$\displaystyle[f(X)]_{s}=\int_{0}^{s}f^{\prime}(X_{u-})^{2}d[Z]^{c}_{s}+\sum_{u\leq
s}f^{\prime}(X_{u-})^{2}(\Delta Z_{u})^{2}+2\sum_{u\leq
s}f^{\prime}(X_{u-})^{2}\Delta Z_{u}\Delta V_{u}+\sum_{u\leq s}\Delta
V_{u}^{2}$ (3.7)

and therefore we may conclude that
$[f(X)]^{c}_{s}=\int_{0}^{s}f^{\prime}(X_{u-})^{2}d[Z]^{c}_{s}=\int_{0}^{s}f^{\prime}(X_{u-})^{2}d[X]^{c}_{s}$
from which the result follows. ∎

Just as in Coquet, Mémin and Słomiński (2003) we will make use of the
following variant of the Itô formula from Föllmer (1981).

######  Theorem 3.5.

Suppose $X$ is a process admitting to a quadratic variation, $f$ a $C^{2}$
function then

$\displaystyle f(X_{t})=f(X_{0})+\sum_{s\leq t}\left(\Delta f(X_{s})-\Delta
X_{s}f^{\prime}(X_{s_{-}})\right)+\int_{0}^{t}f^{\prime}(X_{s_{-}})dX_{s}+\frac{1}{2}\int_{0}^{t}f^{\prime\prime}(X_{s_{-}})d[X]^{c}_{s}.$

######  Theorem 3.6.

Suppose $X$ is a Dirichlet process with fixed time jumps of finite variation,
that $f$ (and $X$) fulfils the assumption of Lemma 3.4, that $\limsup_{h\to
0}\frac{f(x+h)-f(x)}{h}$ is a Baire category 1 function and $a>0$. Then $f(X)$
is a Dirichlet process (with quadratic variation along the same refining
sequence as $X$) that admits to the following decomposition,
$f(X)=Y^{a}+\Gamma^{a}$ with

$\displaystyle Y^{a}_{t}$ $\displaystyle=f(X_{0})+\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}+\int_{0}^{t}f^{\prime}(X_{s-})dZ_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\nu(\\{s\\},dx),$
(3.8)

where $\mu$ is the jump measure of $X$ with $\nu$ as its compensator,
$f^{\prime}$ is as in Lemma 3.4 and $\Gamma^{a}$ is an adapted continuous
process with zero quadratic variation.

This theorem generalizes the results of Theorem 2.1 in Coquet, Mémin and
Słomiński (2003) and the decomposition is exactly of the same type. One should
compare this result with Theorem 2.1. in Lowther (2010) which, in the time
homogeneous case shows that given a map $f$ satisfying Assumption 3.1 and a
process $X$, which is the sum of a semimartingale and process of zero
continuous quadratic variation then $f(X)$ is also a sum of a semimartingale
and process of zero continuous quadratic variation so this results differs
from ours.

######  Remark 3.7.

Instead of assuming fixed time jumps of finite variation one may instead
assume (as is done in Coquet, Mémin and Słomiński (2003)) the equivalent
condition, that $\sum_{s\leq t}\int_{|x|\leq a}|x|\nu(\\{s\\},dx)<\infty$
a.s., but we feel this is a slightly less natural assumption.

###### Proof.

Define $\Gamma=f(X)-Y^{a}$. Note that the jumps of $f(X)$ and $Y^{a}$
coincide. Indeed, the $\Delta X_{s}f^{\prime}(X_{s-})$ jumps in the first sum
are cancelled by the jumps exceeding $a$ in the $dZ$ integral, the smaller
jumps in the $dZ$ integral are cancelled from the $(\mu-\nu)$ and $\nu$
integrals. Therefore $\Gamma$ is continuous and adapted so the result is true
if we can show that $[\Gamma]_{t}=0$ and that $Y^{a}$ is a semimartingale.

If we let $T_{m}:=\inf\\{s:(\Delta X)^{*}_{s}\vee X^{*}_{s}\vee Z^{*}_{s}\geq
m\\}$ then $\lim_{m\to\infty}{\mathbb{P}}(T_{m}>t)=1$. We will also denote
$S_{m}=\sup_{x\in[-m,m]}|f^{\prime}(x)|$. For any given $\epsilon>0$ we may
chose $m$ such that ${\mathbb{P}}(T_{m}>t)>1-\epsilon$ we can therefore
restrict our attention to $X^{T_{m}-}$ and for our purpose we can without loss
of generality assume that $X=X^{T_{m}-}$ for some large $m$. Since
$f^{\prime}$ is a Baire category 1 function we may approximate it point-wise
by a sequence of continuous functions, $\\{h_{n}\\}_{n}$. By considering
$h_{n}\wedge S_{m}$ (which is still continuous) we may assume that $h_{n}\leq
S_{m}$ on $[-m,m]$. Due to Weierstrass Theorem we can take a polynomial
$p_{n}$ such that $|p_{n}(x)-h_{n}(x)|<1/n$ for every $x\in[-m,m]$. It follows
that $p_{n}\to f^{\prime}$ point-wise on $[-m,m]$ and that
$\sup_{x\in[-m,m]}|p_{n}(x)|\leq S_{m}+1<\infty$, for all $n$. Moreover note
that since $X_{-}\in\mathbb{L}$ then $p_{n}(X_{-})\in\mathbb{L}$ and so
$f^{\prime}(X_{-})=\lim_{n\to\infty}p_{n}(X_{-})\in\mathcal{P}$
($f^{\prime}(X_{-})$ is predictable).

We now verify that $Y^{a}$ is a semimartingale for each $a>0$. First note that
by assumption $\sum_{s\leq t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)<\infty$ a.s.
so we can consider the localizing sequence $U_{0}=0$,
$U_{m}=\inf\\{t>U_{m-1}:\sum_{s\leq t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)\geq
m\\}$ for $m\geq 1$. Since ${\mathbb{E}}\left[\int_{|x|\leq a}|x|1_{s\leq
U_{m}}\mu(\\{s\\},dx)\right]={\mathbb{E}}\left[\int_{|x|\leq a}|x|1_{s\leq
U_{m}}\nu(\\{s\\},dx)\ \right]$, it then follows from monotone convergence
that

$\displaystyle{\mathbb{E}}\left[\sum_{s\leq t}\int_{|x|\leq
a}|x|\nu(\\{s\\},dx)1_{t\leq
U_{m}}\right]\leq{\mathbb{E}}\left[\left(\sum_{s\leq.}\int_{|x|\leq
a}|x|\nu(\\{s\\},dx)\right)_{t\wedge U_{m}}\right]$ $\displaystyle=$
$\displaystyle{\mathbb{E}}\left[\sum_{s\leq t}\int_{|x|\leq a}|x|1_{s\leq
U_{m}}\nu(\\{s\\},dx)\right]={\mathbb{E}}\left[\sum_{s\leq t}\int_{|x|\leq
a}|x|1_{s\leq U_{m}}\mu(\\{s\\},dx)\right]\leq m+a$

so $\sum_{s\leq t}\int_{|x|\leq a}|x|\nu(\\{s\\},dx)<\infty$ on $\\{t\leq
U_{m}\\}$ and since ${\mathbb{P}}\left(\bigcup_{m\geq
1}\\{U_{m}>t\\}\right)=1$ we conclude that
$\sum_{s\leq t}\int_{|x|\leq a}|x|\nu(\\{s\\},dx)<\infty$ a.s.. Therefore one
may also deduce that

$\displaystyle\left|\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\nu(\\{s\\},dx)\right|\leq$
$\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}|x|\int_{0}^{1}\left|f^{\prime}(X_{s-}+\theta
x)-f^{\prime}(X_{s-})\right|d\theta\nu(\\{s\\},dx)\leq 2S_{m}\sum_{s\leq
t}\int_{|x|\leq a}|x|\nu(\\{s\\},dx)<\infty,$

a.s.. This allows us to expand

$\displaystyle\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(ds,dx)=$
$\displaystyle\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)+$
$\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(\\{s\\},dx),$
(3.9)

where $\tilde{\mu}$ and $\nu_{c}$ denotes the measures $\mu$ and $\nu$ with
all fixed time jumps removed respectively. Let us now consider Equation (3.6)
and re-write it using (3)

$\displaystyle Y^{a}_{t}$ $\displaystyle=f(X_{0})+\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}+\int_{0}^{t}f^{\prime}(X_{s-})dZ_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\mu(\\{s\\},dx).$
(3.10)

Our task is now to verify that the right-hand side of (3) is a semimartingale:

  * •

The jump sum $\sum_{s\leq t}\left(f(X_{s})-f(X_{s-})-\Delta
X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta X_{s}|>a}$ only contains a finite
number of jumps for each path and is trivially of finite variation

  * •

For the term $\int_{0}^{.}f^{\prime}(X_{s-})dZ_{s}$, it suffices to show that
$(\int_{0}^{.}f^{\prime}(X_{s-})dZ_{s})^{T_{m}}$ is a semimartingale for each
$m$ (so that it is a total semimartingale). Note that $f^{\prime}(X_{s-})$ is
predictable and bounded, so clearly the integral is a semimartingale.

  * •

The integrand in the $(\tilde{\mu}-\nu_{c})$\- integral may be re-written as

$W(s,x)=\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)1_{|x|\leq
a}=x\left(\int_{0}^{1}f^{\prime}(X_{s-}+\theta
x)d\theta-f^{\prime}(X_{s-})\right)1_{|x|\leq a}.$

Using the notation from Jacod and Shiryaev (2003), 1.24 chapter 2, page 72, we
let

$\widehat{W}_{t}=\int_{{\mathbb{R}}}W(t,x)\nu_{c}(\\{t\\}\times
dx)1_{\int_{{\mathbb{R}}}|W(t,x)|\nu_{c}(\\{t\\}\times dx)<\infty}$

then it follows that $\widehat{W}_{t}=a_{c}(t)=0$ (where
$a_{c}(t)=\nu_{c}(\\{t\\}\times{\mathbb{R}})$) since $\nu_{c}$ has no fixed
time jumps by construction. Therefore, if we now also use the notation from
page 73 chapter 2 in Jacod and Shiryaev (2003) and denote

$C(W)_{t}=\int_{{\mathbb{R}}}\int_{0}^{t}\left(W(s,x)-\widehat{W}_{s}\right)^{2}\nu_{c}(ds,dx)+\sum_{s\leq
t}(1-a_{c}(s))\widehat{W}_{s}^{2}$

then

$\displaystyle C(W)^{T_{m}}_{t}=\left(\int_{0}^{t}\int_{|x|\leq
a}(W(s,x)-\widehat{W}_{s})^{2}\nu_{c}(ds,dx)\right)^{T_{m}}+0$
$\displaystyle=\left(\int_{0}^{t}\int_{|x|\leq
a}W(s,x)^{2}\nu_{c}(ds,dx)\right)^{T_{m}}=\left(\int_{0}^{t}\int_{|x|\leq
a}x^{2}\left(\int_{0}^{1}f^{\prime}(X_{s-}+\theta
x)d\theta-f^{\prime}(X_{s-})\right)^{2}\nu_{c}(ds,dx)\right)^{T_{m}}$
$\displaystyle\leq\int_{0}^{t\wedge T_{m}}\int_{|x|\leq
a}x^{2}4S_{m}^{2}\nu(ds,dx),$

which is a process with jumps smaller than $4S_{m}^{2}a^{2}$, so after further
localization we see that this is locally integrable. This implies that (by
Theorem 1.33 a) of Chapter II in Jacod and Shiryaev (2003))
$\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)$
is a well-defined local martingale.

  * •

For the final term we have the following estimate

$\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}\left|f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right|\nu(\\{s\\},dx)\leq\sum_{s\leq
t}\int_{|x|\leq a}|x|2S_{m}\nu(\\{s\\},dx),$

which shows that the series is summable, so it is a pure jump semimartingale.

Define $f_{n}(x)=f(-m)+\int_{-m}^{x}p_{n}(u)du$ for $x\in[-m,m]$ so that both
$f_{n}(x)\to f(x)$ and $p_{n}(x)\to f^{\prime}(x)$ on $[-m,m]$. By Theorem 3.5
it follows that

$f_{n}(X_{t})=f_{n}(X_{0})+\sum_{s\leq t}\left(\Delta f_{n}(X_{s})-\Delta
X_{s}p_{n}(X_{s_{-}})\right)+\int_{0}^{t}p_{n}(X_{s_{-}})dX_{s}+\frac{1}{2}\int_{0}^{t}p_{n}^{\prime}(X_{s_{-}})d[X,X]^{c}_{s}.$

We now let

$\displaystyle Y^{n,a}_{t}$ $\displaystyle=f_{n}(X_{0})+\sum_{s\leq
t}\left(f_{n}(X_{s})-f_{n}(X_{s-})-\Delta X_{s}p_{n}(X_{s-})\right)I_{|\Delta
X_{s}|>a}+\int_{0}^{t}p_{n}(X_{s-})dZ_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f_{n}(X_{s-})-xp_{n}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f_{n}(X_{s-})-xp_{n}(X_{s-})\right)\mu(\\{s\\},dx).$
(3.11)

Since each term on the right-hand side of (3) is a semimartingale, so is the
sum.
Define
$\Gamma^{n}_{t}=\int_{0}^{t}p_{n}(X_{s_{-}})dC_{s}+\int_{0}^{t}p_{n}^{\prime}(X_{s_{-}})d[X,X]^{c}_{s}$
($\Gamma^{n}_{t}$ does in fact depend on $a$ but we suppress this dependence
for notational convenience since $a$ is fixed at this point), so that
$f_{n}(X)=Y^{n,a}+\Gamma^{n}$. Clearly
$\int_{0}^{t}p_{n}^{\prime}(X_{s_{-}})d[X,X]^{c}_{s}$ has zero quadratic
variation and by part (2) of the proof of Lemma 2.5 in Coquet, Mémin and
Słomiński (2003), $\int_{0}^{t}p_{n}(X_{s_{-}})dC_{s}$ has zero quadratic
variation as well which therefore must also be true for $\Gamma^{n}$. We will
now show that $[\Gamma]_{t}=0$. A-priori we do not even know that
$[\Gamma]_{t}$ exists but it will follow from our argument that it both exists
and is zero. We make the following estimate for a fixed $n$

$\displaystyle\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left(\Gamma_{t_{i}}-\Gamma_{t_{i-1}}\right)^{2}}$
$\displaystyle\leq\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left((\Gamma_{t_{i}}-\Gamma^{n}_{t_{i}})-(\Gamma_{t_{i-1}}-\Gamma^{n}_{t_{i-1}})\right)^{2}}+\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left(\Gamma^{n}_{t_{i}}-\Gamma^{n}_{t_{i-1}}\right)^{2}}$
$\displaystyle\leq\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left((f(X)_{t_{i}}-f_{n}(X)_{t_{i}})-(f(X)_{t_{i-1}}-f_{n}(X)_{t_{i-1}})\right)^{2}}$
$\displaystyle+\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left((Y^{a}_{t_{i}}-Y^{n,a}_{t_{i}})-(Y^{a}_{t_{i-1}}-Y^{n,a}_{t_{i-1}})\right)^{2}}+\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left(\Gamma^{n}_{t_{i}}-\Gamma^{n}_{t_{i-1}}\right)^{2}}$

and this shows that

$\displaystyle\limsup_{k\to\infty}{\mathbb{P}}\left(\sqrt{\sum_{t_{i}\in
D_{k}^{t}}\left(\Gamma_{t_{i}}-\Gamma_{t_{i-1}}\right)^{2}}\geq\epsilon\right)\leq\limsup_{k\to\infty}{\mathbb{P}}\left(\sum_{t_{i}\in
D_{k}^{t}}\left((f(X)_{t_{i}}-f_{n}(X)_{t_{i}})-(f(X)_{t_{i-1}}-f_{n}(X)_{t_{i-1}})\right)^{2}\geq\frac{\epsilon^{2}}{3}\right)$
$\displaystyle+\limsup_{k\to\infty}{\mathbb{P}}\left(\sum_{t_{i}\in
D_{k}^{t}}\left((Y^{a}_{t_{i}}-Y^{n,a}_{t_{i}})-(Y^{a}_{t_{i-1}}-Y^{n,a}_{t_{i-1}})\right)^{2}\geq\frac{\epsilon^{2}}{3}\right)+\limsup_{k\to\infty}{\mathbb{P}}\left(\sum_{t_{i}\in
D_{k}^{t}}\left(\Gamma^{n}_{t_{i}}-\Gamma^{n}_{t_{i-1}}\right)^{2}\geq\frac{\epsilon^{2}}{3}\right)$
$\displaystyle={\mathbb{P}}\left(\left[f(X)-f_{n}(X)\right]_{t}\geq\frac{\epsilon^{2}}{3}\right)+{\mathbb{P}}\left(\left[Y^{a}-Y^{n,a}\right]_{t}\geq\frac{\epsilon^{2}}{3}\right)+{\mathbb{P}}\left(\left[\Gamma^{n}\right]_{t}\geq\frac{\epsilon^{2}}{3}\right)$
$\displaystyle={\mathbb{P}}\left(\left[f(X)-f_{n}(X)\right]_{t}\geq\frac{\epsilon^{2}}{3}\right)+{\mathbb{P}}\left(\left[Y^{a}-Y^{n,a}\right]_{t}\geq\frac{\epsilon^{2}}{3}\right),$
(3.12)

where we used the fact that the map $f-f_{n}$ fulfils the hypothesis (since
$f_{n}$ is a polynomial) of Lemma 3.4, so it has a quadratic variation,
$Y^{a}-Y^{n,a}$ is a semimartingale so it obviously has a quadratic variation
as well and finally we also used the fact that $\Gamma^{n}$ has zero quadratic
variation. For the first term on the right-most side of (3) we have by Lemma
3.4 that

$\displaystyle\left[f(X)-f_{n}(X)\right]_{t}=\int_{0}^{t}\left(f^{\prime}(X_{s-})-f^{\prime}_{n}(X_{s-})\right)^{2}d[X]^{c}_{s}+\sum_{s\leq
t}\Delta\left(f(X_{s})-f_{n}(X_{s})\right)^{2}.$ (3.13)

The second term on the right-hand side of (3.13) will vanish a.s. as
$n\to\infty$ due to the uniform convergence of $f_{n}\to f$. The first term
will vanish a.s. by dominated convergence. We now tackle the second term on
the right-most side of (3). Since $\tilde{\mu}$ and it’s compensator measure
$\nu_{c}$ are void of any fixed time jumps it follows from Theorem 1 of
chapter 3, section 5 in Liptser and Shiryaev (1986), that if $g(s,x,\omega)$
is locally integrable then

$\left[\int_{[0,.]}\int_{|x|\leq
a}g(s,x,\omega)(\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}=\int_{[0,t]}\int_{|x|\leq
a}g(s,x,\omega)^{2}\tilde{\mu}(ds,dx).$

Using this fact for the second term as well as applying Lemma 2.2, we make the
following estimates,

$\displaystyle[Y^{a}-Y^{n,a}]_{t}^{\frac{1}{2}}\leq\left[\sum_{s\leq
t}\left(f_{n}(X_{s})-f(X_{s})-(f_{n}(X_{s-})-f(X_{s-}))-\Delta
X_{s}(p_{n}(X_{s-})-f^{\prime}(X_{s-}))\right)I_{|\Delta
X_{s}|>a}\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}(p_{n}(X_{s-})-f^{\prime}(X_{s-}))dZ_{s}\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f(X_{s-}+x)-(f_{n}(X_{s-})-f(X_{s-}))-x(f^{\prime}(X_{s-})-p_{n}(X_{s-}))\right)(\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\sum_{s\leq.}\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f(X_{s-}+x)-(f_{n}(X_{s-})-f(X_{s-}))-x(f^{\prime}(X_{s-})-p_{n}(X_{s-})\right)\mu(\\{s\\},dx)\right]^{\frac{1}{2}}_{t}$
$\displaystyle\leq\sqrt{\sum_{s\leq t}\left((f_{n}(X_{s})-f(X_{s}))I_{|\Delta
X_{s}|>a})\right)^{2}}+\sqrt{\sum_{s\leq
t}\left((f_{n}(X_{s_{-}})-f(X_{s_{-}}))I_{|\Delta X_{s}|>a})\right)^{2}}$
$\displaystyle+\sqrt{\sum_{s\leq t}\left(\Delta
X_{s}(p_{n}(X_{s-})-f^{\prime}(X_{s-}))I_{|\Delta
X_{s}|>a}\right)^{2}}+\sqrt{\int_{0}^{t}(p_{n}(X_{s-})-f^{\prime}(X_{s-}))^{2}d[Z]_{s}}$
$\displaystyle+\left(\int_{0}^{t}\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f(X_{s-}+x)-(f_{n}(X_{s-})-f(X_{s-}))-x(f^{\prime}(X_{s-})-p_{n}(X_{s-}))\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}$
$\displaystyle+\left(\sum_{s\leq t}\left(\int_{|x|\leq
a}\left(f_{n}(X_{s-}+x)-f(X_{s-}+x)-(f_{n}(X_{s-})-f(X_{s-}))-x(f^{\prime}(X_{s-})-p_{n}(X_{s-})\right)\mu(\\{s\\},dx)\right)^{2}\right)^{\frac{1}{2}},$

Since the first three sums contain only finitely many jumps a.s., the
pointwise convergence ensures pathwise convergence to zero. For the fourth
term we use the fact that
$(p_{n}(X_{s-})-f^{\prime}(X_{s-}))^{2}\leq(2S_{m}+1)^{2}$, so we can employ
the stochastic version of the dominated convergence theorem to conclude that
this term vanishes. Now we consider term five (term six is dealt with
afterwards). We have the following dominating bound for the integrand,

$\displaystyle\left(f_{n}(X_{s-}+x)-f(X_{s-}+x)-(f_{n}(X_{s-})-f(X_{s-}))-x(f^{\prime}(X_{s-})-p_{n}(X_{s-}))\right)^{2}=$
$\displaystyle\left(x\int_{0}^{1}(p_{n}(X_{s_{-}}+\theta
x)-f^{\prime}(X_{s_{-}}+\theta
x))d\theta-x(f^{\prime}(X_{s-})-p_{n}(X_{s-}))\right)^{2}\leq$ $\displaystyle
x^{2}\left(\int_{0}^{1}(|p_{n}(X_{s_{-}}+\theta
x)|+S_{m})d\theta+(S_{m}+|p_{n}(X_{s-})|)\right)^{2}\leq
4x^{2}\left(2S_{m}+1\right)^{2},$ (3.14)

which is $\tilde{\mu}(ds,dx)$-integrable (since $\int_{0}^{t}\int_{|x|\leq
a}x^{2}\tilde{\mu}(ds,dx)\leq\int_{0}^{t}\int_{|x|\leq
a}x^{2}\mu(ds,dx)\leq[X]_{t}$), so we may apply the dominated convergence
theorem (pathwise). For term six we can analogously dominate the integrand by
$2|x|(2S+1)$ and apply dominated convergence again. ∎

Let us recall the following definition from Coquet, Mémin and Słomiński (2003)

######  Definition 3.8.

A Dirichlet process $X=Z+C$ along the refining sequence $\\{D_{k}\\}_{k}$ is
said to be of class $\mathcal{D}^{p}$, for $p\in[1,2]$ if

$\left\\{\sum_{i=1}^{m}|C_{s_{i}}-C_{s_{i-1}}|^{p},0=s_{0},\leq...\leq
s_{m}=t,s_{i}\in D_{k},m,k\in\mathbb{N}\right\\}$

is bounded in probability.

What follows is a corollary which is an exact analogue of Corollary 3.8 in
Coquet, Mémin and Słomiński (2003).

######  Corollary 3.9.

Suppose $X$ is a Dirichlet process of class $\mathcal{D}^{2}$ and $f$
satisfies the assumptions of Lemma 3.6 then $f(X)\in\mathcal{D}^{2}$ as well.

###### Proof.

We argue, partially at least, as in the proof of Corollary 3.7 in Coquet,
Mémin and Słomiński (2003). Let $X=Z+C$ be a decomposition of $X$ into a
semimartingale $Z$ and $C$ a continuous process of zero quadratic variation.
Let also $Z=M+A$ be a decomposition of $Z$, where $M$ is a local martingale
and $A$ is a process of finite variation. By localization, as in Theorem 3.6,
we may assume $[M]_{t}$ is bounded and $|X|\leq m$ on $[0,t]$, we will let
$S=\sup_{x\in[-m,m]}|f^{\prime}(x)|$. Due to Theorem 3.6, (3) and the
inequality $\left(\sum_{k=1}^{m}a_{k}\right)^{2}\leq
m^{2}\sum_{k=1}^{m}a_{k}^{2}$, which we apply to
$\left(\Gamma_{s_{i}}-\Gamma_{s_{i-1}}\right)^{2}$, we see that

$\displaystyle\sum_{i}\left(\Gamma_{s_{i}}-\Gamma_{s_{i-1}}\right)^{2}$
$\displaystyle\leq
6^{2}\left(\sum_{i}\left(f(X_{s_{i}})-f(X_{s_{i-1}})\right)^{2}+\sum_{i}\left(\sum_{s_{i-1}<s\leq
s_{i}}\left(\Delta f(X_{s})-f^{\prime}(X_{s-})\Delta X_{s}\right)I_{|\Delta
X_{s}|>1}\right)^{2}\right.$
$\displaystyle\left.+\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}f^{\prime}(X_{s-})dM_{s}\right)^{2}+\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}f^{\prime}(X_{s-})dA_{s}\right)^{2}\right.$
$\displaystyle\left.+\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}\int_{|x|\leq
1}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)\right)^{2}\right.$
$\displaystyle\left.+\sum_{i}\left(\sum_{s_{i-1}<s\leq s_{i}}\int_{|x|\leq
1}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\mu(\\{s\\},dx)\right)^{2}\right),$
(3.15)

where we used the fact that $\Gamma=f-Y^{1}$. For the first term in (3)

$\sum_{i}\left(f(X_{s_{i}})-f(X_{s_{i-1}})\right)^{2}\leq
S^{2}\sum_{i}\left(X_{s_{i}}-X_{s_{i-1}}\right)^{2}\leq
2S^{2}\left(\sum_{i}\left(Z_{s_{i}}-Z_{s_{i-1}}\right)^{2}+\sum_{i}\left(C_{s_{i}}-C_{s_{i-1}}\right)^{2}\right),$

which is bounded uniformly in probability over all choices of
$\\{s_{i}\\}_{i}$ since $Z$ is a semimartingale and $X\in\mathcal{D}^{2}$. We
argue as in Coquet, Mémin and Słomiński (2003) and note that the second,
fourth and sixth terms of the right-hand side of (3) are of the form
$\sum_{i}(U_{s_{i}}-U_{s_{i-1}})^{2}$ where $U_{s}$ is a finite variation
process so these terms are trivially bounded. The Burkholder-Davis-Gundy
inequality implies,

${\mathbb{E}}\left[\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}f^{\prime}(X_{s-})dM_{s}\right)^{2}\right]\leq\sum_{i}L{\mathbb{E}}\left[\int_{s_{i-1}}^{s_{i}}f^{\prime}(X_{s-})^{2}d[M]_{s}\right]\leq
LK{\mathbb{E}}\left[[M]_{t}\right],$

for some $L$ and it therefore it follows from the Markov inequality that
$\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}f^{\prime}(X_{s-})dM_{s}\right)^{2}$ is
bounded in probability regardless of how we choose $\\{s_{i}\\}_{i}$ and this
takes care of the third term in (3). Similarly

$\displaystyle{\mathbb{E}}\left[\sum_{i}\left(\int_{s_{i-1}}^{s_{i}}\int_{|x|\leq
1}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)\right)^{2}\right]\leq$
$\displaystyle\sum_{i}L{\mathbb{E}}\left[\int_{s_{i-1}}^{s_{i}}\int_{|x|\leq
1}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)^{2}\tilde{\mu}(ds,dx)\right]$
$\displaystyle\leq\sum_{i}L{\mathbb{E}}\left[\int_{s_{i-1}}^{s_{i}}\int_{|x|\leq
1}4x^{2}S^{2}\tilde{\mu}(ds,dx)\right]\leq
LS^{2}{\mathbb{E}}\left[[X]_{t}\right],$

so the fifth term is bounded. ∎

We now first study stability in quadratic variation in the more general
context of processes admitting to quadratic variation. This result is not very
deep and is arrived at mostly from first principles. We will then later see
that due to the nature of the decomposition in Theorem 3.6 that this result
can be substantially improved for Dirichlet processes.

######  Proposition 3.10.

Given $0\leq u\leq t$, suppose that $(\\{f_{n}\\}_{n},f)$ fulfil Assumption
3.3, for each $n$ $X$, $X^{n}$, $f_{n}(X^{n})-f(X)$, $f_{n}(X)-f(X)$ and
$f_{n}(X^{n})-f_{n}(X)$ all have quadratic variations along the same refining
sequence $\\{D^{n}_{k}\\}_{k}$, that
$[X^{n}-X]_{u}\xrightarrow{{\mathbb{P}}}0$ and
$(X^{n}-X)^{*}_{u}\xrightarrow{{\mathbb{P}}}0$ then

$[f_{n}(X^{n})-f(X)]_{u}\xrightarrow{{\mathbb{P}}}0.$

###### Proof.

See Appendix A.1.1. ∎

A straightforward application of Proposition 3.10 is the following result
concerning (ucp) stability of integrators for semimartingale integrands.

######  Corollary 3.11.

Suppose that $(\\{f_{n}\\}_{n},f)$ fulfil Assumptions 3.3, for each $n$ $X$
and $X^{n}$ are Dirichlet processes along the same refining sequence
$\\{D^{n}_{k}\\}_{k}$, that $[X^{n}-X]_{t}\xrightarrow{{\mathbb{P}}}0$ and
$(X^{n}-X)^{*}_{t}\xrightarrow{{\mathbb{P}}}0$ then for any semimartingale $Y$

$\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})\xrightarrow{ucp}\int_{0}^{.}Y_{s-}df(X_{s}).$

######  Remark 3.12.

An immediate consequence of this is that

$\displaystyle\lim_{n\to\infty}\lim_{m\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{m}_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq
c\right)=\lim_{m\to\infty}\lim_{n\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{m}_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq
c\right)$
$\displaystyle=\lim_{n\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq
c\right)=0,$

for all $c>0$.

###### Proof.

We first note that since $X,X^{n},Y$ and $f(X)$ (by Theorem 3.6) are all
Dirichlet processes, they all have mutual brackets due to Lemma 1.5 in Lowther
(2010). Fix $t\in{\mathbb{R}}^{+}$ and $\epsilon,c>0$ arbitrary. By Corollary
2.6 the integrals $\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})$ and
$\int_{0}^{.}Y_{s-}df(X_{s})$ both exist and we also have

$\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})=f_{n}(X^{n}_{s})Y_{s}-\int_{0}^{.}f_{n}(X^{n}_{s-})dY_{s}-[f_{n}(X^{n}),Y],$

as well as

$\int_{0}^{.}Y_{s-}df(X_{s})=f(X_{s})Y_{s}-\int_{0}^{.}f(X_{s-})dY_{s}-[f(X),Y],$

so

$\displaystyle\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\leq$
$\displaystyle\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}+\left(\int_{0}^{.}f_{n}(X^{n}_{s-})dY_{s}-\int_{0}^{.}f(X_{s-})dY_{s}-\right)^{*}_{t}+\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t},$
(3.16)

and it therefore suffices to show that each term in the right hand side of (3)
converges to zero in probability. Since $f_{n}\to f$ uniformly on compacts and
since $\\{f_{n}\\}_{n}$ and $f$ are continuous this implies that
$\\{f_{n}\\}_{n}$ are equicontinuous. Now take take $R$ so large that
${\mathbb{P}}\left(\sup_{n\geq 1}(X^{n})^{*}_{t}\geq R\right)<\epsilon$ and
$\delta>0$ so small that $|f_{n}(x)-f_{n}(y)|<c/2$ when $|x-y|<\delta$,
$x,y\in[-R,R]$ and all $n$. Pick $n_{1}$ so large that $n\geq n_{1}$ implies
$|f_{n}(x)-f(x)|<c$ for $x\in[-R,R]$ and
${\mathbb{P}}\left((X^{n}-X)^{*}_{t}\geq c/2\right)<\epsilon$ then for $n\geq
n_{1}$

$\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}\leq\left(f_{n}(X^{n})-f_{n}(X)\right)^{*}_{t}+\left(f_{n}(X)-f(X)\right)^{*}_{t}\leq
c,$

on the set $\\{(X^{n}-X)^{*}_{t}<c\\}\cap\\{\sup_{n\geq
1}(X^{n})^{*}_{t}<R\\}$ which has measure no less than $1-2\epsilon$ and this
shows that $f_{n}(X^{n})\xrightarrow{ucp}f(X)$ so by the continuity of the
$dY$-integral, the second term of (3) converges to zero in probability.
Moreover for any $L\in{\mathbb{R}}^{+}$

$\displaystyle{\mathbb{P}}\left(\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}\geq
c\right)\leq{\mathbb{P}}\left(\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}\geq
c,Y^{*}_{t}\leq L\right)+{\mathbb{P}}\left(Y^{*}_{t}>L\right)$
$\displaystyle\leq{\mathbb{P}}\left(\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}\geq\frac{c}{L}\right)+{\mathbb{P}}\left(Y^{*}_{t}>L\right),$

by first letting $n\to\infty$ the first term vanishes and then by letting
$L\to\infty$ the second term vanishes by continuity of probability. For the
last term in (3) we note that by the Kunita-Watanabe inequality

$\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t}\leq\left([f_{n}(X^{n})-f(X)]^{\frac{1}{2}}[Y]^{\frac{1}{2}}\right)^{*}_{t}=[f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}}[Y]_{t}^{\frac{1}{2}},$

and therefore, for any $L\in{\mathbb{R}}^{+}$

$\displaystyle{\mathbb{P}}\left(\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t}\geq
c\right)\leq{\mathbb{P}}\left([f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}}[Y]_{t}^{\frac{1}{2}}\geq
c\right)\leq$
$\displaystyle{\mathbb{P}}\left([f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}}[Y]_{t}^{\frac{1}{2}}\geq
c,[Y]_{t}^{\frac{1}{2}}\leq
L\right)+{\mathbb{P}}\left([Y]_{t}^{\frac{1}{2}}>L\right)\leq{\mathbb{P}}\left([f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}}\geq\frac{c}{L}\right)+{\mathbb{P}}\left([Y]_{t}^{\frac{1}{2}}>L\right),$

where the first term vanishes as $n\to\infty$ by Proposition 3.10 and the
second term vanishes as $L\to\infty$, hence all four terms in (3) vanishes as
$n\to\infty$ which completes the proof. ∎

We present an application of Corollary 3.11 for continuous semimartingales in
the following corollary.

######  Corollary 3.13.

Suppose that $\\{f_{n}\\}_{n}$ and $f$ fulfil Assumption 3.3. Assume that
$\\{X^{n}\\}_{n}$ are continuous semimartingales such that
$X^{n}\xrightarrow{ucp}X$, where $X$ is some semimartingale. Let $M$ and
$M_{n}$ denote the (unique) local martingale parts of $X$ and $X^{n}$
respectively. Suppose also that $M^{n}\xrightarrow{ucp}M$ and $|M^{n}|\leq|Z|$
for every $n$, for some semimartingale $Z$ with $Z_{0}\in L^{1}$, then $\int
Y_{-}df_{n}(X^{n})\xrightarrow{ucp}\int Y_{-}df(X)$, for any semimartingale
$Y$.

###### Proof.

Let $Z^{\prime}=Z-Z_{0}$, by Theorem V.4 in Protter (1992) $Z^{\prime}$ is
pre-locally in ${{\stackrel{{\scriptstyle H}}{{=}}}^{1}}$ (See Definitions 2.1
and 2.7). So let $\\{T(m)\\}_{m}$ be stopping times such that
$T(m)\xrightarrow{a.s.}\infty$ and $Z^{T(m)-}\in{\stackrel{{\scriptstyle
H}}{{=}}}^{1}$. Take $\epsilon>0$, $c>0$ arbitrary and pick $m$ so large that
${\mathbb{P}}(T(m)\leq t)<\epsilon$. Let $X^{n}=M^{n}+A^{n}$, $X=M+A$ be the
canonical decompositions of $X^{n}$ and $X$ respectively. Since
$\\{X^{n}\\}_{n}$ are continuous so are $X$, $A$, $M$, $\\{M^{n}\\}_{n}$ and
$\\{A^{n}\\}_{n}$. By the continuity of $M^{n}-M$,

$|M^{n}-M|^{T(m)}=|M^{n}-M|^{T(m)-}\leq 2|Z|^{T(m)-}.$

Since $A^{n}-A$ is both FV and continuous it follows that

$[M^{n}-M,A^{n}-A]_{t}=[A^{n}-A]_{t}=0,$

and therefore $[X^{n}-X]_{t}=[M^{n}-M]_{t}.$ Due to the Burkholder Davis-Gundy
inequality

$\|[(X^{n}-X)^{T(m)})]^{\frac{1}{2}}_{t}\|_{L^{1}}=\|[(M^{n}-M)^{T(m)})]^{\frac{1}{2}}_{t}\|_{L^{1}}\leq
C_{1}\|((M^{n}-M)^{T(m)})^{*}_{t}\|_{L^{1}},$

For some positive constant $C_{1}$. Meanwhile $((M^{n}-M)^{T(m)})^{*}_{t}\leq
2(Z^{T(m)-})^{*}_{t}$, however by Theorem V.2 in Protter (1992)

$\|(Z^{T(m)-})^{*}_{t}\|_{L^{1}}\leq\left(\|(Z^{\prime
T(m)-})^{*}_{t}\|_{L^{1}}+\|Z_{0}\|_{L^{1}}\right)\leq C_{2}\|Z^{\prime
T(m)-}\|_{{\stackrel{{\scriptstyle H}}{{=}}}^{1}}+\|Z_{0}\|_{L^{1}}<\infty,$

for some positive constant $C_{2}$. Since $(M^{n}-M)^{T(m)}\xrightarrow{ucp}0$
and $|M^{n}-M|^{T(m)}\leq 2|Z|^{T(m)-}\in L^{1}$, it follows from a variant of
the dominated convergence theorem that
$\lim_{n}\|((M^{n}-M)^{T(m)})^{*}_{t}\|_{L^{1}}=0$ and therefore
$[(X^{n}-X)^{T(m)})]^{\frac{1}{2}}_{t}\xrightarrow{L^{1}}0$. By the Markov
inequality

$\displaystyle\lim_{n\to\infty}{\mathbb{P}}\left([(X^{n}-X)^{T(m)})]_{t}\geq
c\right)=\lim_{n\to\infty}{\mathbb{P}}\left([(X^{n}-X)^{T(m)})]_{t}^{\frac{1}{2}}\geq\sqrt{c}\right)\leq$
$\displaystyle\lim_{n\to\infty}\frac{1}{\sqrt{c}}\|[(X^{n}-X)^{T(m)})]_{t}^{\frac{1}{2}}\|_{L^{1}}=0,$

for any $c>0$ so in other words
$[(X^{n}-X)^{T(m)})]_{t}\xrightarrow{{\mathbb{P}}}0$. Hence,

$\displaystyle{\mathbb{P}}([(X^{n}-X)]_{t}\geq
c)\leq{\mathbb{P}}([(X^{n}-X)^{T(m)})]_{t}\geq c)+{\mathbb{P}}(T(m)\leq
t)\leq$ $\displaystyle{\mathbb{P}}([(X^{n}-X)^{T(m)})]_{t}\geq c)+\epsilon,$

which converges to $\epsilon$ which was arbitrary and therefore the hypothesis
in Corollary 3.11 is fulfilled and the result follows. ∎

We now shift focus entirely to Dirichlet processes. The next Theorem concerns
(uniform) convergence in the $L^{p}$ sense, this will obviously require moment
conditions of some kind as well as some kind of tail growth conditions on
$\\{f_{n}\\}_{n}$. The proof is rather tedious and has been deferred to the
Appendix.

######  Theorem 3.14.

Let $(\\{f_{n}\\}_{n},f)$ fulfill Assumption 3.3, $\\{X^{n}\\}_{n}$ and $X$ be
Dirichlet processes such that for each $n$, $X^{n}$ and $X$ have quadratic
variations along the same refining sequence and such that
$X^{n}\xrightarrow{ucp}X$. Assume also that $X$ and $X^{n}$ have fixed time
jumps of finite variation. Letting $\mathcal{A}=\\{s\leq
t:{\mathbb{P}}\left(\Delta X_{s}\not=0\right)>0\\}$ and
$\mathcal{A}_{n}=\\{s\leq t:{\mathbb{P}}\left(\Delta
X^{n}_{s}\not=0\right)>0\\}$ we also assume $\sum_{s\leq
t,s\in\mathcal{A}_{n}}|\Delta X^{n}_{s}|\xrightarrow{{\mathbb{P}}}\sum_{s\leq
t,s\in\mathcal{A}}|\Delta X_{s}|$ (i.e. the total variation of the fixed time
jumps of $X^{n}$ converge in probability to the total variation of the fixed
time jumps of $X$).
a) Suppose $|f_{n}^{\prime}|\leq U$ for some $U\in{\mathbb{R}}^{+}$,
$[X]_{t}^{p}\in L^{1}$, ${\mathbb{E}}\left[[X^{n}-X]_{t}^{p}\right]\to 0$ with
$p\in\\{\frac{1}{2},1\\}$ and
$\left\\{\left(\sum_{s\leq t,s\in\mathcal{A}_{n}}|\Delta
X^{n}_{s}|\right)^{2p}\right\\}_{n}$ is u.i. then ,
$[f_{n}(X^{n})-f(X)]^{p}_{t}\xrightarrow{L^{1}}0$ as $n\to\infty$.
b) Suppose $[X]_{t}^{q}\in L^{1}$,
${\mathbb{E}}\left[[X^{n}-X]_{t}^{q}\right]\to 0$ with $q\in\\{1,2\\}$,
$\\{((X^{n})^{*}_{t})^{2q}\\}_{n}$ is u.i.,
$\lim_{|x|\to\infty}\sup_{n}\frac{|f_{n}^{\prime}(x)|}{|x|}\leq C$, for some
$C\in{\mathbb{R}}^{+}$ and $\left\\{\left(\sum_{s\leq
t,s\in\mathcal{A}_{n}}|\Delta X^{n}_{s}|\right)^{2q}\right\\}_{n}$ is u.i.
then
$[f_{n}(X^{n})-f(X)]^{\frac{q}{2}}\xrightarrow{L^{1}}0$ as $n\to\infty$.

###### Proof.

See Appendix A.1.2. ∎

We now study an analogous result in the almost sure convergence setting.
Thanks to the decomposition (3.6) we are able to remove the requirement of
uniform convergence for $f_{n}^{\prime}\to f^{\prime}$ and $X^{n}\to X$.

######  Theorem 3.15.

Let $(\\{f_{n}\\}_{n},f)$ fulfill Assumption 3.3, $\\{X^{n}\\}_{n}$ and $X$ be
Dirichlet processes such that for each $n$, $X^{n}$ and $X$ have quadratic
variations along the same refining sequence and such that
$X^{n}_{s}\xrightarrow{a.s.}X_{s}$ for each $s\leq t$,
$[X^{n}-X]_{t}\xrightarrow{a.s.}0$ and that $\\{(X^{n})^{*}_{t}\\}_{n}$ is
bounded in probability. Assume also that $X$ and $X^{n}$ have fixed time jumps
of finite variation. We also assume $\sum_{s\leq t,s\in\mathcal{A}_{n}}|\Delta
X^{n}_{s}|\xrightarrow{a.s.}\sum_{s\leq t,s\in\mathcal{A}}|\Delta X_{s}|$
(i.e. the total variation of the fixed time jumps of $X^{n}$ converge a.s. to
the total variation of the fixed time jumps of $X$). Then
$[f_{n}(X^{n})-f(X)]_{t}\xrightarrow{a.s.}0$ as $n\to\infty$.

###### Proof.

See Appendix A.1.3. ∎

######  Theorem 3.16.

Let $f$ be a Lipschitz continuous function such that
$f^{\prime}(x)=\limsup_{h\to 0}\frac{f(x+h)-f(x)}{h}$ has at most a countable
set of discontinuities, assume that $X$ has independent increments and that
$X$ is nowhere constant on $[0,t]$. Let $\\{X^{n}\\}_{n}$ be Dirichlet
processes such that for each $n$, $X^{n}$ and $X$ have quadratic variations
along the same refining sequence and such that
$X^{n}_{s}\xrightarrow{a.s.}X_{s}$ for each $s\leq t$,
$[X^{n}-X]_{t}\xrightarrow{a.s.}0$ and that $\\{(X^{n})^{*}_{t}\\}_{n}$ is
bounded in probability. Assume also that $X$ and $X^{n}$ have fixed time jumps
of finite variation and finally that $\sum_{s\leq
t,s\in\mathcal{A}_{n}}|\Delta X^{n}_{s}|\xrightarrow{a.s.}\sum_{s\leq
t,s\in\mathcal{A}}|\Delta X_{s}|$ (i.e. the total variation of the fixed time
jumps of $X^{n}$ converge a.s. to the total variation of the fixed time jumps
of $X$). Then
$[f(X^{n})-f(X)]_{t}\xrightarrow{a.s.}0$ as $n\to\infty$.

###### Proof.

See Appendix A.1.4. ∎

Now we consider integrator stability in the $L^{p}$ setting. This will
naturally require us to make some type of moment assumptions on the integrand
$Y$.

######  Corollary 3.17.

Assume $Y\in\underline{\underline{H}}^{2}([0,t])$, as well as either
1) Hypothesis a) of Theorem 3.14 and $\\{((X^{n})^{*}_{t})^{2}\\}_{n}$ is u.i.
or
2) Hypothesis b) of Theorem 3.14 and $Y^{*}_{t}\in L^{\infty}$,
or
3) Hypothesis b) of Theorem 3.14 $\\{((X^{n})^{*}_{t})^{4}\\}_{n}$ is u.i.,
then

${\mathbb{E}}\left[\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\right]\to
0,$

as $n\to\infty$.

###### Proof.

It suffices to show that all for terms on the right-hand side of (3) converges
to zero in expectation. Assume 1). We have that

$\displaystyle|f_{n}(x)|\leq|f_{n}(x_{0})|+\left|\int_{x_{0}}^{x}|f_{n}^{\prime}(y)|dy\right|\leq|f_{n}(x_{0})|+U(|x_{0}|+|x|)$
(3.17)

and similarly $|f(x)|\leq|f(x_{0})|+U(|x_{0}|+|x|)$. Let $Y=M+A$ be a
decomposition of $Y$ such that
$j_{2}(M,A)<2\|Y\|_{{}_{{\stackrel{{\scriptstyle H}}{{=}}}^{2}}([0,t])}$. Note
that

${\mathbb{E}}\left[(Y^{*}_{t})^{2}\right]\leq{\mathbb{E}}\left[2(M^{*}_{t})^{2}+2(A^{*}_{t})^{2}\right]\leq
2D{\mathbb{E}}\left[[M]_{t}\right]+2{\mathbb{E}}\left[\left(\int_{0}^{t}|dA_{s}|\right)^{2}\right]\leq
4(D+1)\|Y\|_{{}_{{\stackrel{{\scriptstyle H}}{{=}}}^{2}}([0,t])},$

for some constant $D$ by the Burkholder Davis-Gundy inequality. Therefore

$\displaystyle{\mathbb{E}}\left[\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}1_{E}\right]\leq{\mathbb{E}}\left[\left(|f_{n}(x_{0})|+U|x_{0}|+U(X^{*}_{t}+(X^{n})^{*}_{t})\right)Y^{*}_{t}1_{E}\right]\leq$
$\displaystyle\left(|f_{n}(x_{0})|+U|x_{0}|\right){\mathbb{E}}\left[Y^{*}_{t}1_{E}\right]+U{\mathbb{E}}\left[(X^{*}_{t}+(X^{n})^{*}_{t}))Y^{*}_{t}1_{E}\right]\leq$
$\displaystyle\left(|f_{n}(x_{0})|+U|x_{0}|\right){\mathbb{E}}\left[Y^{*}_{t}1_{E}\right]+U\sqrt{E\left[\left(2(X^{*}_{t})^{2}+2((X^{n})^{*}_{t})^{2})1_{E}\right)\right]}\sqrt{{\mathbb{E}}\left[(Y^{*}_{t})^{2}1_{E}\right]},$

for any measurable set $E$, which shows that
$\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}$ is u.i. and since this term
converges to zero in probability as established by the proof of Theorem 3.11
it follows that it converges to zero in expectation as well. Since
$\int_{0}^{.}\left(f_{n}(X^{n})-f(X)\right)dM_{s}$ is a local martingale (by
the martingale preservation property),

$\displaystyle{\mathbb{E}}\left[\left(\int_{0}^{.}\left(f_{n}(X^{n})-f(X)\right)dY_{s}\right)^{*}_{t}\right]\leq{\mathbb{E}}\left[\left(\int_{0}^{.}\left(f_{n}(X^{n})-f(X)\right)dM_{s}\right)^{*}_{t}\right]+{\mathbb{E}}\left[\left(\int_{0}^{.}\left(f_{n}(X^{n})-f(X)\right)dA_{s}\right)^{*}_{t}\right]$
$\displaystyle\leq D$
$\displaystyle{\mathbb{E}}\left[\left[\int_{0}^{t}\left(f_{n}(X^{n})-f(X)\right)dM_{s}\right]^{\frac{1}{2}}\right]+{\mathbb{E}}\left[\int_{0}^{t}\left|f_{n}(X^{n}_{s-})-f(X_{s-})\right||dA_{s}|\right]$
$\displaystyle=D$
$\displaystyle{\mathbb{E}}\left[\sqrt{\int_{0}^{t}\left(f_{n}(X^{n})-f(X)\right)^{2}d[M]_{s}}\right]+{\mathbb{E}}\left[\int_{0}^{t}\left|f_{n}(X^{n}_{s-})-f(X_{s-})\right||dA_{s}|\right]$
$\displaystyle\leq D$
$\displaystyle{\mathbb{E}}\left[(f_{n}(X^{n})-f(X))^{*}_{t}\sqrt{[M]_{t}}\right]+{\mathbb{E}}\left[\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}\int_{0}^{t}|dA_{s}|\right]\leq
D\sqrt{{\mathbb{E}}\left[\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}\right]}\sqrt{{\mathbb{E}}\left[[M]_{t}\right]}+$
$\displaystyle\sqrt{{\mathbb{E}}\left[\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}\right]}\sqrt{{\mathbb{E}}\left[\left(\int_{0}^{t}|dA_{s}|\right)^{2}\right]}\leq(D+1)\sqrt{{\mathbb{E}}\left[\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}\right]}2\|Y\|_{{}_{{\stackrel{{\scriptstyle
H}}{{=}}}^{2}}([0,t])},$

where we applied both the Burkholder Davis-Gundy and the Cauchy-Schwarz
inequalities. From (3.17) it follows that
$\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}\leq
A+B((X^{*}_{t})^{2}+((X^{n})^{*}_{t})^{2})$ for some positive constants $A,B$
and since $\\{((X^{n})^{*}_{t})^{2}\\}_{n}$ is u.i. so is
$(\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}$ which also converges to zero
in probability which then implies
${\mathbb{E}}\left[\left((f_{n}(X^{n})-f(X))^{*}_{t}\right)^{2}\right]\to 0$
and hence
${\mathbb{E}}\left[\left(\int_{0}^{.}\left(f_{n}(X^{n})-f(X)\right)dY_{s}\right)^{*}_{t}\right]\to
0$. For the final term we proceed as in the proof Theorem 3.11 and note that
$\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t}\leq[f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}}[Y]_{t}^{\frac{1}{2}}$.
By Lemma 2.3, the Cauchy-Schwarz and Kunita-Watanabe inequalities

$\displaystyle{\mathbb{E}}\left[\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t}\right]\leq\sqrt{{\mathbb{E}}\left[[f_{n}(X^{n})-f(X)]_{t}\right]}\sqrt{{\mathbb{E}}\left[[Y]_{t}\right]}\leq$
$\displaystyle\sqrt{{\mathbb{E}}\left[[f_{n}(X^{n})-f(X)]_{t}\right]}\sqrt{{\mathbb{E}}\left[2[M]_{t}+2[A]_{t}\right]}\leq\sqrt{{\mathbb{E}}\left[[f_{n}(X^{n})-f(X)]_{t}\right]}\sqrt{2}\sqrt{{\mathbb{E}}\left[[M]_{t}+\left(\int_{0}^{t}|dA_{s}|\right)^{2}\right]}$
$\displaystyle\leq\sqrt{{\mathbb{E}}\left[[f_{n}(X^{n})-f(X)]_{t}\right]}2^{\frac{3}{2}}\|Y\|_{{}_{{\stackrel{{\scriptstyle
H}}{{=}}}^{2}}([0,t])},$

which converges to zero by Theorem 3.14 a) and this concludes the proof when
assuming 1).

Assuming either 2) or 3) then we may choose $R>0$ such that
$|f_{n}^{\prime}(x)|\leq 2C|x|$ for $|x|\geq R$ and by letting
$M=\sup_{n}\sup_{y\in[-R,R]}|f_{n}^{\prime}(y)|$ we have that for $|x|\leq R$

$|f_{n}(x)|\leq|f_{n}(x_{0})|+M(|x_{0}|+|x|),$

while for $|x|\geq R$

$|f_{n}(x)|\leq|f_{n}(x_{0})|+2C\left|\int_{x_{0}}^{x}|y|dy\right|\leq|f_{n}(x_{0})|+C(x_{0}^{2}+x^{2}),$

which implies

$\displaystyle|f_{n}(x)|\leq|f_{n}(x_{0})|+C(x_{0}^{2}+x^{2})+M(|x_{0}|+|x|)$
(3.18)

and similarly $|f(x)|\leq|f(x_{0})|+C(x_{0}^{2}+x^{2})+M(|x_{0}|+|x|)$.

Assuming 2) and letting $L=\|Y_{t}^{*}\|_{L^{\infty}}$ then for the first term
on the right-hand side of (3) we have that for any measurable subset $E$,

$\displaystyle{\mathbb{E}}\left[\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}1_{E}\right]\leq$
$\displaystyle\left(|f_{n}(x_{0})|+M|x_{0}|\right)L{\mathbb{P}}(E)+ML{\mathbb{E}}\left[(X^{*}_{t}+(X^{n})^{*}_{t})1_{E}\right]+CL{\mathbb{E}}\left[((X^{*}_{t})^{2}+(X^{n})^{*}_{t})^{2})1_{E}\right],$

which shows that $\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}$ is also
u.i., so the first term converges to zero in expectation. For the second term
we proceed as in the case when assuming 1) but use the bound (3.18) instead of
(3.17) and use the $L^{\infty}$ bound on $Y^{*}_{t}$ instead of the Cauchy-
Schwarz inequality. For the final term,

${\mathbb{E}}\left[\left([f_{n}(X^{n})-f(X),Y]\right)^{*}_{t}\right]\leq\sqrt{L}{\mathbb{E}}\left[[f_{n}(X^{n})-f(X)]^{\frac{1}{2}}\right],$

which converges to zero by Theorem 3.14 b), this concludes the proof for the
case 2).
Assuming 3) then for the first term on the right-hand side of (3) we apply the
bound in (3.18) and find that

$\displaystyle{\mathbb{E}}\left[\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}1_{E}\right]\leq$
$\displaystyle{\mathbb{E}}\left[\left(|f_{n}(x_{0})|+|f(x_{0})|+M(X^{*}_{t}+(X^{n})^{*}_{t}+|x_{0}|)+C(2x_{0}^{2}+(X^{*}_{t})^{2}+(X^{n})^{*}_{t})^{2})\right)Y^{*}_{t}1_{E}\right]\leq$
$\displaystyle\left(|f_{n}(x_{0})|+|f(x_{0})|+M|x_{0}|+2Cx_{0}^{2}\right){\mathbb{E}}\left[Y^{*}_{t}1_{E}\right]+M{\mathbb{E}}\left[(X^{*}_{t}+(X^{n})^{*}_{t})Y^{*}_{t}1_{E}\right]+C{\mathbb{E}}\left[((X^{*}_{t})^{2}+(X^{n})^{*}_{t})^{2})Y^{*}_{t}1_{E}\right]$
$\displaystyle\leq\left(|f_{n}(x_{0})|+|f(x_{0})|+M|x_{0}|+2Cx_{0}^{2}\right){\mathbb{E}}\left[Y^{*}_{t}1_{E}\right]+M\sqrt{{\mathbb{E}}\left[(2(X^{*}_{t})^{2}+2((X^{n})^{*}_{t})^{2}1_{E}\right]}\sqrt{{\mathbb{E}}\left[(Y^{*}_{t})^{2}1_{E}\right]}+$
$\displaystyle\sqrt{E\left[\left(2(X^{*}_{t})^{4}+2((X^{n})^{*}_{t})^{4})1_{E}\right)\right]}\sqrt{{\mathbb{E}}\left[(Y^{*}_{t})^{2}1_{E}\right]},$

for any measurable set $E$, which shows that
$\left(f_{n}(X^{n})-f(X)\right)^{*}_{t}Y^{*}_{t}$ is u.i.. For the second term
we proceed as in the case when assuming 1) with the only difference that we
use the bound (3.18) instead of (3.17) together with the assumption that
$\\{((X^{n})^{*}_{t})^{4}\\}_{n}$ is u.i..

The third term is dealt with analogously to the case when assuming 2) but we
instead invoke Theorem 3.14 b). ∎

##  4 Jump truncation

By a jump truncation we mean a modification of some given process $X$ where
jumps below some level are not present. Jump truncations serve as important
applications for our results. For processes with jumps of finite variation one
may simply discard all jumps of $X$ with modulus less than say $a>0$ and let
$a\to 0^{+}$ and in the limit we retain our original process. Given a cadlag
process $X$ with jumps of finite variation we define

$X(a)_{t}=X_{t}^{c}+\sum_{s\leq t}\Delta X_{s}I_{|\Delta X_{s}|\geq a}.$

In the general case however, $X(a)$ will not have a well-defined limit as
$a\to 0^{+}$ since $X^{d}=\sum_{s\leq.}\Delta X_{s}$ is not well-defined. We
will denote by $\mu$ the jump measure corresponding to a given cadlag process
$X$ and we will denote by $\nu$ the compensator measure of $\mu$ w.r.t. the
filtration $\\{\mathcal{F}_{t}\\}_{t\geq 0}$ (see Jacod and Shiryaev (2003)
ch. II for details on this). Suppose now that we are given a cadlag process
$X=Z+C$ which is the sum of a semimartingale $Z$ and a continuous adapted
process $C$. This means that all jumps of $X$ belong to the semimartingale
part. We may represent $X$ by

$X=M^{c}+b+\sum_{s\leq.}\Delta X_{s}I_{|\Delta
X_{s}|>1}+\int_{0}^{.}\int_{|x|\leq 1}xd(\mu-\nu)(s,x)+C,$

where $M^{c}$ is the continuous part of the local martingale part of $Z$ and
$b$ is a process of finite variation. Given a process $X=Z+C$ of the form
above we define for $a<1$,

$\widehat{X}(a)_{t}=M_{t}^{c}+b(a)_{t}+\sum_{s\leq t}\Delta X_{s}I_{|\Delta
X_{s}|>1}+\int_{0}^{t}\int_{a\leq|x|\leq 1}xd(\mu-\nu)(s,x)+C_{t}.$

Notice that $\widehat{X}(a)$ only has jumps greater or equal to $a$.

We shall now show that if has jumps at fixed times of finite variation, then
$\widehat{X}(a)\xrightarrow{ucp}X$ as well as
$[\widehat{X}(a)-X]_{t}\xrightarrow{{\mathbb{P}}}0$ as $a\rightarrow 0^{+}$.

Let
$\tilde{X}(a):=X_{t}-\widehat{X}(a)_{t}=\int_{0}^{t}\int_{0<|x|<a}xd(\mu-\nu)(s,x)+b-b(a)$.
Since $b$ is of finite variation (and therefore has jumps of finite variation)
we trivially have that $[b-b(a)]_{t}\xrightarrow{a.s.}0$ as well as
$b-b(a)\xrightarrow{ucp}0$ as $a\to 0^{+}$. Now we rewrite

$\int_{0}^{t}\int_{0<|x|<a}x(\mu-\nu)(ds,dx)=\int_{0}^{t}\int_{0<|x|<a}x(\tilde{\mu}-\nu_{c})(ds,dx)+\sum_{s\leq
t}\int_{0<|x|<a}x(\mu+\nu)(\\{s\\},dx),$

where $\tilde{\mu}$ is the measure $\mu$ with all fixed time jumps removed and
$\nu_{c}$ is the compensator of $\tilde{\mu}$ (which is the time-continuous
part of the compensator of $\mu$). By Lemma 2.2,

$\displaystyle[\tilde{X}(a)]_{t}^{\frac{1}{2}}\leq[b-b(a)]_{t}^{\frac{1}{2}}+\left[\int_{0}^{.}\int_{|x|<a}x^{2}(\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}+\left[\sum_{s\leq.}\int_{|x|<a}|x|(\mu+\nu)(\\{s\\},dx)\right]_{t}=$
$\displaystyle[b-b(a)]_{t}^{\frac{1}{2}}+\int_{0}^{t}\int_{0<|x|<a}x^{2}\tilde{\mu}(ds,dx)+\sum_{s\leq
t}\left(\int_{|x|<a}|x|(\mu+\nu)(\\{s\\},dx)\right)^{2},$

since $\int_{0}^{t}\int_{|x|\leq 1}x^{2}\mu(ds,dx)\leq[X]_{t}<\infty$ a.s. and
$\sum_{s\leq t}\int_{|x|\leq 1}|x|(\mu+\nu)(\\{s\\},dx)<\infty$ a.s., it
follows that $[\widehat{X}(a)-X]_{t}$ converges to zero a.s. (and hence in
probability) as $a\to 0$. If we let
$N(a)_{t}=\int_{0}^{t}\int_{|x|<a}xd(\tilde{\mu}-\nu_{c})(ds,dx)$ then $N(a)$
is a local martingale with $N(a)_{0}=0$ for every $a>0$, by Theorem 2.21 in
Jacod and Shiryaev (2003) (since $\nu_{c}$ is the compensator measure of
$\tilde{\mu}$). Let $T_{k}:=\inf\\{t>0:\int_{0}^{t}\int_{|x|\leq
1}x^{2}d\tilde{\mu}\geq k\\}$. For a given $t$ and $\epsilon>0$ we may take
$k$ such that ${\mathbb{P}}\left(T_{k}\leq t\right)<\epsilon$. By the
Burkholder Davis-Gundy and Jensen inequalities

$\displaystyle{\mathbb{E}}\left[(N(a)^{T_{k}})^{*}_{t}\right]\leq
C{\mathbb{E}}\left[\sqrt{\left[N(a)^{T_{k}}\right]_{t}}\right]\leq
C\sqrt{{\mathbb{E}}\left[\left[N(a)^{T_{k}}\right]_{t}\right]}=C\sqrt{{\mathbb{E}}\left[\int_{0}^{t\wedge
T_{k}}\int_{|x|<a}x^{2}d\tilde{\mu}(ds,dx)\right]},$

for some $C>0$. Since $\int_{0}^{t\wedge
T_{k}}\int_{|x|<a}x^{2}d\tilde{\mu}(ds,dx)\leq k+1$ ,for all $a\leq 1$, it
follows from the dominated convergence theorem that $\lim_{a\to
0^{+}}{\mathbb{E}}\left[(N(a)^{T_{k}})^{*}_{t}\right]=0$ and therefore
$(N(a)^{T_{k}})^{*}_{t}\xrightarrow{{\mathbb{P}}}0$ as $a\to 0^{+}$. We
conclude that for any $c>0,$

$\displaystyle\lim_{a\to
0^{+}}{\mathbb{P}}\left((\widehat{X}(a)-X)^{*}_{t}\geq c\right)\leq\lim_{a\to
0^{+}}{\mathbb{P}}\left((N(a))^{*}_{t}\geq c\right)+\lim_{a\to
0^{+}}{\mathbb{P}}\left((b-b(a))^{*}_{t}\right)$ $\displaystyle\leq\lim_{a\to
0^{+}}{\mathbb{P}}\left((N(a))^{*}_{t\wedge T_{k}}\geq
c\right)+{\mathbb{P}}\left(T_{k}\leq t\right)={\mathbb{P}}\left(T_{k}\leq
t\right)<\epsilon,$

hence $\widehat{X}(a)\xrightarrow{ucp}X$.

To apply this, let $X$ be as in the above proposition and $f$ be as in Theorem
3.6 then $[f(\widehat{X}(a))-f(X)]_{t}\xrightarrow{{\mathbb{P}}}0$ by Theorems
3.6, 3.10 and the fact Dirichlet processes along the same refining sequence
form a vector space. Similarly if $\\{f_{n}\\}_{n}$ is a sequence of smooth
approximating functions that fulfill Assumption 3.3 then
$[f_{n}(\widehat{X}(a_{n}))-f(X)]_{t}\xrightarrow{{\mathbb{P}}}0$ for any
sequence of positive real numbers $\\{a_{n}\\}_{n}$ such that $a_{n}\to
0^{+}$.

######  Definition 4.1.

Given a cadlag process $X$ we define the following set of reals

$\mathbb{A}_{X}:=\left\\{a\in{\mathbb{R}}^{+}:{\mathbb{P}}\left(\exists s\leq
t:|\Delta X_{s}|=a\right)>0\right\\}.$

######  Remark 4.2.

Since $X$ is cadlag it follows that $\mathbb{A}_{X}$ is countable

###  4.1 The case with finite variation of jumps

We now take a closer look at jump truncation for processes with jumps of
finite variation. Assuming that $X$ has jumps of finite variation then
$(X-X(a))^{*}_{t}\leq\sum_{s\leq t}|\Delta X_{s}|I_{|\Delta X_{s}|\leq a}$ so
it follows that $(X-X(a))^{*}_{t}\to 0$ a.s. when $a\to 0^{+}$ and therefore,
obviously, $X(a)\xrightarrow{ucp}X$ as $a\to 0^{+}$. Also
$[X-X(a)]_{t}=\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq a}$ which
also converges a.s. to zero since the $l^{1}$ norm dominates the $l^{2}$ norm.
Moreover this kind of truncation actually has some slightly peculiar
properties, one may wonder whether $X^{n}\xrightarrow{ucp}X$ implies
$X^{n}(a)\xrightarrow{ucp}X(a)$ for all $a>0$ and if
$[X^{n}-X]_{t}\xrightarrow{{\mathbb{P}}}0$ implies
$[X^{n}(a)-X(a)]_{t}\xrightarrow{{\mathbb{P}}}0$. In general neither is true,
as the following example shows.

######  Example 4.3.

Let $X$ be a Poisson process with intensity 1 and let $X^{n}=(1-\frac{1}{n})X$
then $X^{n}\xrightarrow{ucp}X$ as well as
$[X^{n}-X]_{t}=\frac{1}{n^{2}}[X]_{t}\to 0$ as $n\to\infty$, but $X^{n}(1)=0$
for all $n$ so $X^{n}(1)\xrightarrow{ucp}0\neq X(1)=X$ and
$[X^{n}(1)-X(1)]_{t}\not\xrightarrow{{\mathbb{P}}}0$. Notice that in this
example there is only a single jump size which is troublesome but we may
generalize this example to have countably many troublesome points as follows.
Let $X_{t}=\sum_{k=1}^{\infty}\frac{1}{k^{2}}N_{k}(t)$, where
$\\{N_{k}(t)\\}_{k}$ are i.i.d. Poisson processes with intensity 1. Setting
$X^{n}=(1-\frac{1}{n})X$ just as before shows despite
$X^{n}\xrightarrow{ucp}X$ as well as $[X^{n}-X]_{t}=\frac{1}{n^{2}}[X]_{t}\to
0$ as $n\to\infty$ we get
$X^{n}(\frac{1}{k^{2}})\not\xrightarrow{ucp}X(\frac{1}{k^{2}})$ and
$[X^{n}(\frac{1}{k^{2}})-X(\frac{1}{k^{2}})]_{t}\not\xrightarrow{{\mathbb{P}}}0$
for any $k\in{\mathbb{N}}$.

The above example illustrates the difficulty of truncation when the
distribution of the jump size of $X$ is non-continuous. There are however some
reasonable assumptions under which these problems do not arise.

######  Proposition 4.4.

Suppose $\\{X^{n}\\}_{n}$ and $X$ are cadlag processes withs jumps of finite
variation. If $X^{n}\xrightarrow{ucp}X$ then the following three statements
are equivalent,
1) $(X^{n})^{c}\xrightarrow{ucp}X^{c}$
2) $\sum_{s\leq t}\Delta X^{n}_{s}\xrightarrow{ucp}\sum_{s\leq t}\Delta X_{s}$
3) $\lim_{a\to
0^{+},a\not\in\mathbb{A}_{X}}\lim_{n\to\infty}{\mathbb{P}}\left(\left(\sum_{s\leq.}\Delta
X^{n}_{s}I_{|\Delta X^{n}_{s}|\leq a}\right)^{*}_{t}\geq c\right)=0$ for all
$c>0$ and $t\in{\mathbb{R}}^{+}$.
Moreover if either one of the three above conditions hold then
$X^{n}(a)\xrightarrow{ucp}X(a)$ for any $a\not\in\mathbb{A}_{X}$. Finally, if
assume we the slightly stronger condition,
4) $\lim_{a\to 0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left(\sum_{s\leq t}|\Delta
X^{n}_{s}|I_{|\Delta X^{n}_{s}|\leq a}\geq c\right)=0$ for all $c>0$ and
$t\in{\mathbb{R}}^{+}$ then

$\lim_{a\to
0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left((X^{n}(a)-X(a))^{*}_{t}\geq
c\right)=0,$

for all $c>0$.

######  Remark 4.5.

Notice that if $X^{n}\xrightarrow{ucp}X$ and we assume one of 1)-3) then
(obviously)

$\lim_{a\to
0^{+},a\not\in\mathbb{A}_{X}}\lim_{n\to\infty}{\mathbb{P}}\left((X^{n}(a)-X(a))^{*}_{t}\geq
c\right)=0,$

in comparison to the final statement of the Proposition.

###### Proof.

See Appendix A.1.5. ∎

######  Definition 4.6.

For $X(a)$ we define the following properties,
(V1): $X$ and $\\{X^{n}\\}_{n}$ all have jumps of finite variation and either
one of 1)-3) in Lemma 4.4 holds true.
(V2): $X$ and $\\{X^{n}\\}_{n}$ all have jumps of finite variation and 4) in
Lemma 4.4 holds true.

######  Proposition 4.7.

Suppose $a\not\in\mathbb{A}_{X}$, $[X^{n}-X]_{t}\xrightarrow{{\mathbb{P}}}0$
and $X^{n}\xrightarrow{ucp}X$ then
$[X^{n}(a)-X(a)]_{t}\xrightarrow{{\mathbb{P}}}0$ and if we assume assumption
(V2) then

$\lim_{a\to 0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left([X^{n}(a)-X(a)]_{t}\geq
c\right)=0,$

for all $c>0$.

###### Proof.

Since

$[(X^{n})^{c}-X^{c},J^{n}-J]_{t}=\sum_{s\leq
t}\Delta((X^{n})^{c}-X^{c})_{s}\Delta(J^{n}-J)_{s}=0,$

it follows that $[X^{n}-X]_{t}=[(X^{n})^{c}-X^{c}]_{t}+[J^{n}-J]_{t}$, since
the left hand side converges to zero and both terms on the right are non-
negative we conclude that
$[(X^{n})^{c}-X^{c}]_{t}\xrightarrow{{\mathbb{P}}}0$. Similarly
$[X^{n}(a)-X(a)]_{t}=[(X^{n})^{c}-X^{c}]_{t}+[J^{n}(a)-J(a)]_{t}$ so it
suffices to show that $[J^{n}(a)-J(a)]_{t}\xrightarrow{{\mathbb{P}}}0$. A
similar approach to the proof of the implication 3) $\rightarrow$ 2) in
Proposition 4.4 shows that also
$[J^{n}(a)-J(a)]_{t}\xrightarrow{{\mathbb{P}}}0$ if $a\not\in\mathbb{A}_{X}$
(indeed as in that proof on the set $B_{(k,n)}$ we have
$[J^{n}(a)-J(a)]_{t}=\sum_{l=1}^{k}(\Delta(X^{n}-X))^{2}_{T_{l}}$). Let us now
assume assumption 4) of Proposition 4.4 then, since the $l^{1}$-norm dominates
the $l^{2}$-norm, $\lim_{a\to
0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left(\sum_{s\leq t}|\Delta
X^{n}_{s}|^{2}I_{|\Delta X^{n}_{s}|\leq a}\geq c\right)=0$ and therefore we
may take $a^{\prime}\not\in\mathbb{A}_{X}$ so small and $n_{1}$ so large that
$a\leq a^{\prime}$ and $n\geq n_{1}$ implies ${\mathbb{P}}\left(\sum_{s\leq
t}|\Delta X^{n}_{s}|^{2}I_{|\Delta X^{n}_{s}|\leq a}\geq c\right)<\epsilon$ as
well as ${\mathbb{P}}\left(\sum_{s\leq t}|\Delta X_{s}|^{2}I_{|\Delta
X_{s}|\leq a}\geq c\right)<\epsilon$. Letting

$\displaystyle B_{n}$ $\displaystyle=\left\\{\sum_{s\leq t}|\Delta
X^{n}_{s}|^{2}I_{|\Delta X^{n}_{s}|\leq a}<c/9\right\\}\cap\left\\{\sum_{s\leq
t}|\Delta X_{s}|^{2}I_{|\Delta X_{s}|\leq a}\geq c/9\right\\}$
$\displaystyle\cap\left\\{[J^{n}(a^{\prime})-J(a^{\prime})]_{t}<c/9]\right\\},$

so on $B_{n}$,

$\displaystyle[J^{n}(a)-J(a)]_{t}\leq
3\left([J^{n}(a^{\prime})-J^{n}(a)]_{t}+[J^{n}(a^{\prime})-J(a^{\prime})]_{t}+[J(a^{\prime})-J(a)]_{t}\right)$
$\displaystyle<\sum_{s\leq t}|\Delta X^{n}_{s}|^{2}I_{|\Delta X^{n}_{s}|\leq
a}+c/3+\sum_{s\leq t}|\Delta X_{s}|^{2}I_{|\Delta X_{s}|\leq
a}<3(c/9+c/9+c/9)=c,$

where we used Lemma 2.2 together with the inequality $(x+y+z)^{2}\leq
3(x^{2}+y^{2}+z^{2})$ in the first step. Hence,

${\mathbb{P}}\left([J^{n}(a)-J(a)]_{t}\geq
c\right)\leq{\mathbb{P}}\left(\left\\{[J^{n}(a)-J(a)]_{t}\geq c\right\\}\cap
B_{n}\right)+{\mathbb{P}}(B_{n}^{c})={\mathbb{P}}(B_{n}^{c})<3\epsilon,$

when $a\leq a^{\prime}$ and $n\geq n_{1}$, which shows the final statement.

$[J^{n}(a)-J(a)]_{t}\leq\sum_{s\leq
t}\Delta(X^{n}-X)_{s}^{2}I_{\Delta|X^{n}-X|_{s}\leq a},$

converges to zero a.s. for each $t\in{\mathbb{R}}^{+}$ it follows that
$\lim_{a\to 0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left([J^{n}(a)-J(a)]_{t}\geq
c\right)=0$. If we again apply Lemma 4.4,

$[X^{n}(a)-X(a)]_{t}=[(X^{n})^{c}-X^{c}+J^{n}(a)-J(a)]_{t}\leq\sqrt{[(X^{n})^{c}-X^{c}]_{t}}+\sqrt{[J^{n}(a)-J(a)]_{t}},$

therefore $[X^{n}(a)-X(a)]_{t}\xrightarrow{{\mathbb{P}}}0$ for
$a\not\in\mathbb{A}_{X}$ and
$\lim_{a\to 0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left([X^{n}(a)-X(a)]_{t}\geq
c\right)=0$. ∎

The following example serves as a reminder that $X^{n}\xrightarrow{ucp}X$ does
in general not imply $[X^{n}-X]_{t}\xrightarrow{{\mathbb{P}}}0$ or vice versa.

######  Example 4.8.

Take $X\equiv 0$ and let

$X^{n}=2^{-(n+1)/2}\sum_{k=1}^{2^{n-1}}I_{[\frac{2(k-1)}{2^{n}},\frac{2k-1}{2^{n}})}(\cdot)-I_{[\frac{2k-1}{2^{n}},\frac{2k}{2^{n}})}(\cdot).$

We then have that $(X^{n})_{\infty}^{\ast}=2^{-(n+1)/2}\rightarrow 0$ but
$[X^{n}]_{1}=(2^{-(n+1)/2})^{2}(2(2^{n}-1)+1)=1-2^{-(n+1)}\rightarrow 1.$
Which shows that
$[X^{n}-X]_{t}\stackrel{{\scriptstyle{\mathbb{P}}}}{{\not{\rightarrow}}}0$ so
that

$X^{n}\xrightarrow{ucp}X\not{\Rightarrow}[X^{n}-X]_{t}\xrightarrow{{\mathbb{P}}}0.$

On the other hand take any semimartingale $Z$ plus a sequence
$\\{V^{n}\\}_{n}$ of continuous processes of zero quadratic variation not
converging to zero, then if we let $X^{n}:=V^{n}+Z$ then $[X^{n}-Z]_{t}=0$ for
all $t\in{\mathbb{R}}^{+}$.

We now show that when we are dealing with jumps finite variation then under
$(V1)$ or $(V2)$ conditions, taking limits over $n$ in Theorem 3.11 commutes
in a sense with taking limits over the truncation level $a$.

######  Proposition 4.9.

Let $(X,\\{X^{n}\\}_{n})$, $(f,\\{f_{n}\\}_{n})$ fulfil hypothesis of
Proposition 3.10.
1) If $\\{X^{n}\\}_{n}$ has the (V1) property from Definition 4.6 then

$\displaystyle\lim_{a\to
0^{+},a\not\in\mathbb{A}_{X}}\lim_{n\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq\epsilon\right)=$
$\displaystyle\lim_{n\to\infty}\lim_{a\to
0^{+},a\not\in\mathbb{A}_{X}}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq\epsilon\right)=0,$

for every $\varepsilon>0$.
2) If $\\{X^{n}\\}_{n}$ has the (V2) property from Definition 4.6 then

$\displaystyle\lim_{a\to
0^{+}}\lim_{n\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq\epsilon\right)=$
$\displaystyle\lim_{n\to\infty}\lim_{a\to
0^{+}}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq\epsilon\right)=0,$

for every $\varepsilon>0$.

###### Proof.

We start with 1). Assume that $a\not\in\mathbb{A}_{X}$. From the integration
by parts formula we have

$\displaystyle\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})=f_{n}(X^{n}(a))Y-f_{n}(X_{0})Y_{0}-\int_{0}^{.}f_{n}(X^{n}(a)_{s})dY_{s}-[f_{n}(X^{n}(a)),Y],$
(4.19)

so to prove the first claim we will first show that the right-hand side above
converges in ucp to $\int Y_{-}df(X(a))$ as $n\to\infty$ an then let $a\to
0^{+}$. Since $X^{n}(a)\xrightarrow{ucp}X(a)$ and since $f_{n}\to f$ uniformly
as $n\to\infty$, $f_{n}(X^{n}(a))\xrightarrow{ucp}f(X(a))$ so the first (by
continuity of probability) and second (by continuity of the stochastic
integral) term in (4.19) converges in ucp to $f(X(a))Y$ and $\int
f(X(a))\\_dY$ respectively as $n\to\infty$.
We now study the last term in 4.19, $[f_{n}(X^{n}(a)),Y]$. Since
$[X^{n}(a)-X(a)]_{t}\xrightarrow{{\mathbb{P}}}0$ by Lemma 4.7 and
$X^{n}(a)\xrightarrow{ucp}X(a)$ by Proposition 4.4, as $n\to\infty$, it then
follows from Theorem 3.14 that
$[f_{n}(X(a)^{n})-f(X(a))]_{t}\xrightarrow{{\mathbb{P}}}0$ and therefore we
may argue as in the proof of Proposition 3.10 to show that
$[f_{n}(X^{n}(a)),Y]\xrightarrow{ucp}[f(X(a)),Y]$. So we have shown that

$\displaystyle\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})\xrightarrow{ucp}f(X(a))Y-f(X_{0})Y_{0}-\int_{0}^{.}f(X(a)_{s})dY_{s}-[f(X(a)),Y].$
(4.20)

Since $f$ is continuous and $X(a)\xrightarrow{ucp}X$, as $a\to 0^{+}$, it
follows that $f(X(a))\xrightarrow{ucp}f(X)$ so the first and second term
converges in ucp to $f(X)Y$ and $\int f(X)_{-}dY$ respectively as $a\to
0^{+}$. For the last term in (4.20) we have
$[f(X(a)),Y]_{t}\leq[f(X(a))]_{t}^{\frac{1}{2}}[Y]_{t}^{\frac{1}{2}},$ but by
Proposition 3.10 (the fact that we are are working with a continuous parameter
will pose no problem, as is evident by the proof)
$[f(X(a))-f(X)]_{t}\xrightarrow{{\mathbb{P}}}0$ which implies
$[f(X(a))]_{t}\xrightarrow{{\mathbb{P}}}[f(X)]_{t}$ so
$[f(X(a)),Y]\xrightarrow{ucp}[f(X),Y]$ as $a\to 0^{+}$, so the right-hand side
of 4.20 converges in ucp to

$f(X)Y-f(X_{0})Y_{0}-\int_{0}^{.}f(X_{s})dY_{s}-[f(X),Y]=\int_{0}^{.}Y_{s-}df(X_{s})$

as $a\to 0^{+}$, this shows

$\lim_{a\to
0^{+},a\not\in\mathbb{A}_{X}}\lim_{n\to\infty}{\mathbb{P}}\left(\left(\int_{0}^{.}Y_{s-}df_{n}(X^{n}(a)_{s})-\int_{0}^{.}Y_{s-}df(X_{s})\right)^{*}_{t}\geq\epsilon\right)=0.$

To show the second equality of 1) note that by Lemma 2.2,

$[f_{n}(X^{n}(a))-f(X)]_{t}^{\frac{1}{2}}\leq[f_{n}(X^{n}(a))-f_{n}(X^{n})]_{t}^{\frac{1}{2}}+[f_{n}(X^{n})-f(X)]_{t}^{\frac{1}{2}},$

by first letting $a\to 0^{+}$ the first term on the right-hand side above
converges to zero in probability. Letting $n\to\infty$ makes the second term
converge to zero, i.e.
$[f_{n}(X^{n}(a))-f(X)]_{t}^{\frac{1}{2}}\xrightarrow{{\mathbb{P}}}0$ which
implies $[f_{n}(X^{n}(a))-f(X),Y]_{t}\xrightarrow{ucp}0$. By the fact that
$f_{n}$ is $C^{1}$, $f_{n}(X^{n}(a))\xrightarrow{ucp}f_{n}(X^{n})$ as $a\to
0^{+}$ and using the integration by parts formula concludes the proof of 1).
Showing 2) is completely analogous. ∎

##  Appendix A.1 Additional proofs

###  A.1.1 Proof of Proposition 3.10

###### Proof.

By Lemma 2.3 we have that

$[f_{n}(X^{n})-f(X)]_{u}\leq
2[f_{n}(X)-f(X)]_{u}+2[f_{n}(X^{n})-f_{n}(X)]_{u},$

so therefore it will suffice to show,
1) $[f_{n}(X)-f(X)]_{u}\xrightarrow{{\mathbb{P}}}0$,
and then
2) $[f_{n}(X^{n})-f_{n}(X)]_{u}\xrightarrow{{\mathbb{P}}}0$. We proceed with
1). By localization we may assume $X^{*}_{t}$ is bounded. Due to Remark 2) on
page 148 of Föllmer (1981),

$[f_{n}(X)-f(X)]_{t}=\int\left(f^{\prime}_{n}(X_{s-})-f^{\prime}(X_{s-})\right)^{2}d[X]^{c}_{s}+\sum_{s\leq
t}\Delta\left(f_{n}(X_{s})-f(X_{s})\right)^{2}.$

The first term on the right above is defined as a path-wise integral and due
to the assumption that $X^{*}_{t}$ is bounded, $f_{n}^{\prime}(X_{s-})\to
f^{\prime}(X_{s-})$ uniformly, therefore this term vanishes. The second term
also vanishes by uniform convergence. We now prove 2). Since
$[X^{n}-X]_{u}\xrightarrow{{\mathbb{P}}}0$ implies that
$[X^{n}]_{u}\xrightarrow{{\mathbb{P}}}[X]_{u}$ and since
$X^{n}\xrightarrow{ucp}X$ there exists $R^{\prime}$ such that

${\mathbb{P}}\left(\max\left([X^{n}]_{u},[X]_{u},(X^{n})^{*}_{u},X^{*}_{u}\right)\geq
R^{\prime}\right)<\epsilon,$

for all $n$. Since $f^{\prime}_{n}\to f^{\prime}$ uniformly on compacts we
have that $\\{f^{\prime}_{n}\\}_{n}$ is equicontinuous on $[-R-1,R+1]$, so we
may take $\delta>0$ so small that if $x,y\in[-R-1,R+1]$ and $|x-y|\leq\delta$
then
$\max\left(\sup_{n}|f_{n}^{\prime}(x)-f_{n}^{\prime}(y)|,|f^{\prime}(x)-f^{\prime}(y)|\right)<\sqrt{\frac{c}{12(R^{\prime}+1)}}$.
By arguments identical to those in the proof of Theorem 3.14 there exists some
$0<\delta^{\prime}<\delta/3$ such that if we let $J_{n}$ be the event that the
jumps of size greater or equal to $\delta^{\prime}$ coincide for $X_{n}$ and
$X$ then $\lim_{n\to\infty}{\mathbb{P}}\left(J_{n}\right)=1$. Let $T_{0}=0$
and for $k\geq 1$, $T_{k}=u\wedge\inf\\{s>T_{k-1}:|\Delta
X_{s}|\geq\delta^{\prime}\\}$ then if $N(\omega)=\inf\\{k:T_{k}=u\\}$ we must
have that $N<\infty$ a.s. by the cadlag property. Note that

$[f_{n}(X_{n})-f_{n}(X)]_{u}=\sum_{k=1}^{N}[f_{n}(X^{n})-f_{n}(X))1_{(T_{k-1},T_{k})}]_{u}+\sum_{k=1}^{N}\left(\Delta\left(f_{n}(X^{n})-f_{n}(X)\right)_{T_{k}}\right)^{2},$

by the ucp convergence we have that the second sum above vanishes so we only
have to concern ourselves with the first sum. Thus we can without loss of
generality show 2) under the assumption that $X^{n}$ or $X$ has no jumps
exceeding $\delta^{\prime}$ on $[0,u]$. Now we may take $n_{1}$ so large that
if $n\geq n_{1}$ then
${\mathbb{P}}\left([X^{n}-X]_{u}\geq\frac{c}{24M^{2}}\right)<\epsilon$. Given
any $n\geq n_{1}$ we may, since $X$, $X^{n}-X$ and $f_{n}(X^{n})-f_{n}(X)$ all
have quadratic variations along $\\{D^{n}_{k}\\}_{k}$, choose a partition
$D^{n}_{k}=\left\\{t^{n}_{i}\right\\}_{i\geq 1}$ (which depends on $n$) such
that
${\mathbb{P}}\left(\max_{i}|X_{t^{n}_{i+1}}-X_{t^{n}_{i}}|>\delta/3\right)<\epsilon$,

$\displaystyle{\mathbb{P}}\left(\left(\sum_{t_{i}\leq.,\hskip
2.84526ptt_{i}\in
D^{n}_{k}}\left(\left(X^{n}_{t_{i+1}^{n}}-X_{t_{i+1}^{n}}\right)-\left(X^{n}_{t_{i}^{n}}-X_{t_{i}^{n}}\right)\right)^{2}-[X^{n}-X]\right)^{*}_{u}<\frac{c}{24M^{2}}\right)<\epsilon,$

${\mathbb{P}}\left(\left(\sum_{t_{i}\leq.,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(X_{t_{i+1}^{n}}-X_{t_{i}^{n}}\right)^{2}-[X]\right)^{*}_{u}\geq
1\right)<\epsilon,{\mathbb{P}}\left((X^{n}-X)^{*}_{u}\geq\delta^{\prime}\right)<\epsilon,$

and

$\displaystyle{\mathbb{P}}\left(\left([f_{n}(X^{n})-f_{n}(X)]-\sum_{t^{n}_{i}\leq.,\hskip
2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X)_{t_{i+1}^{n}}\right)-\left(f_{n}(X^{n})_{t_{i}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\vphantom{{\mathbb{P}}\left(\left([f_{n}(X^{n})-f_{n}(X)]-\sum_{t_{i}\leq.,\hskip
2.84526ptt_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X)_{t_{i+1}^{n}}\right)-\left(f_{n}(X^{n})_{t_{i}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}\right.}\right)<\epsilon.$

Let

$\displaystyle
A(n)=\left\\{\max\left([X^{n}]_{u},[X]_{u},(X^{n})^{*}_{u},X^{*}_{u}\right)<R^{\prime}\right\\}\cap\left\\{[X^{n}-X]_{u}<\frac{c}{24M^{2}}\right\\}$
$\displaystyle\cap\left\\{\left(\sum_{t_{i}\leq.,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(\left(X^{n}_{t_{i+1}^{n}}-X_{t_{i+1}^{n}}\right)-\left(X^{n}_{t_{i}^{n}}-X_{t_{i}^{n}}\right)\right)^{2}-[X^{n}-X]\right)^{*}_{u}<\frac{c}{24M^{2}}\right\\}$
$\displaystyle\cap\left\\{\left([X]-\sum_{t_{i}\leq.,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(X_{t_{i+1}^{n}}-X_{t_{i}^{n}}\right)^{2}\right)^{*}_{u}<1\right\\}\cap\\{(X^{n}-X)^{*}_{u}<\delta^{\prime}\\}\cap\left\\{\max_{i}|X_{t^{n}_{i+1}}-X_{t^{n}_{i}}|<\delta^{\prime}\right\\},$

then ${\mathbb{P}}(A(n)^{c})<6\epsilon$ for $n\geq n_{1}$. So, for $n\geq
n_{1}$,

$\displaystyle{\mathbb{P}}\left([f_{n}(X^{n})-f_{n}(X)]_{u}>c\right)\leq\ $
$\displaystyle{\mathbb{P}}\left(\left([f_{n}(X^{n})-f_{n}(X)]-\sum_{t^{n}_{i}\leq.,\hskip
2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X)_{t_{i+1}^{n}}\right)-\left(f_{n}(X^{n})_{t_{i}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\right)$
$\displaystyle+$
$\displaystyle{\mathbb{P}}\left(\left(\sum_{t^{n}_{i}\leq.,\hskip
2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X)_{t_{i+1}^{n}}\right)-\left(f_{n}(X^{n})_{t_{i}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\right)\leq\epsilon+$
$\displaystyle{\mathbb{P}}\left(\left(\sum_{t^{n}_{i}\leq.,\hskip
2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X)_{t_{i+1}^{n}}\right)-\left(f_{n}(X^{n})_{t_{i}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\right)\leq$
$\displaystyle{\mathbb{P}}\left(\left\\{\left(\sum_{t^{n}_{i}\leq.,\hskip
2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X^{n})_{t_{i}^{n}}\right)-\left(f_{n}(X)_{t_{i+1}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\right\\}\cap
A(n)\right)+{\mathbb{P}}(A(n)^{c})+\epsilon.$

We shall see that

$B^{\prime}:=\left\\{\left(\sum_{t^{n}_{i}\leq.,\hskip 2.84526ptt^{n}_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X^{n})_{t_{i}^{n}}\right)-\left(f_{n}(X)_{t_{i+1}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}\right)^{*}_{u}>\frac{c}{2}\right\\},$

is disjoint from $A(n)$. Using the the mean-value theorem,
$f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X^{n})_{t_{i}^{n}}=\Delta_{i}^{n}f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})$
and similarly
$f_{n}(X)_{t_{i+1}^{n}}-f_{n}(X)_{t_{i}^{n}}=\Delta_{i}f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})$
where $\Delta_{i}=X_{t_{i+1}^{n}}-X_{t_{i}^{n}}$,
$\Delta^{n}_{i}=X^{n}_{t_{i+1}^{n}}-X^{n}_{t_{i}^{n}}$ and
$\theta^{n}_{i},\theta_{i}\in[0,1]$. This allows us to rewrite

$\displaystyle\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X^{n})_{t_{i}^{n}}\right)-\left(f_{n}(X)_{t_{i+1}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)=\Delta_{i}^{n}f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-\Delta_{i}f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})$
$\displaystyle=f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})\left(\Delta^{n}_{i}-\Delta_{i}\right)+\left(f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})\right)\Delta_{i},$

and furthermore on $A(n)$,

$\left|f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})\right|<\sqrt{\frac{c}{12(R^{\prime}+1)}}$

where we used the fact that

$|X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i}-X_{t_{i}^{n}}+\theta_{i}\Delta_{i}|\leq(X^{n}-X)^{*}_{t}+|\Delta^{n}_{i}|+|\Delta_{i}|\leq
3\delta^{\prime}<\delta.$

Plugging in the above expressions, expanding the squares and then using the
Cauchy-Schwartz inequality (for sums) on the cross terms we see that on $A(n)$
for $s\leq u$,

$\displaystyle\sum_{t_{i}\leq s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(\left(f_{n}(X^{n})_{t_{i+1}^{n}}-f_{n}(X^{n})_{t_{i}^{n}}\right)-\left(f_{n}(X)_{t_{i+1}^{n}}-f_{n}(X)_{t_{i}^{n}}\right)\right)^{2}$
$\displaystyle\leq\sum_{t_{i}\leq s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})^{2}\left(\Delta^{n}_{i}-\Delta_{i}\right)^{2}+\sum_{t_{i}\leq
s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})\right)^{2}\Delta_{i}^{2}$
$\displaystyle+2\left(\sum_{t_{i}\leq s,\hskip 2.84526ptt_{i}\in D^{n}_{k}}\
f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})^{2}\left(\Delta^{n}_{i}-\Delta_{i}\right)^{2}\right)^{\frac{1}{2}}\left(\sum_{t_{i}\leq
s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})\right)^{2}\Delta_{i}^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq 3\sum_{t_{i}\leq s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})^{2}\left(\Delta^{n}_{i}-\Delta_{i}\right)^{2}+3\sum_{t_{i}\leq
s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(f^{\prime}_{n}(X^{n}_{t_{i}^{n}}+\theta^{n}_{i}\Delta^{n}_{i})-f^{\prime}_{n}(X_{t_{i}^{n}}+\theta_{i}\Delta_{i})\right)^{2}\Delta_{i}^{2}$
$\displaystyle\leq 3M^{2}\sum_{t_{i}\leq s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(\left(X^{n}_{t_{i+1}^{n}}-X_{t_{i+1}^{n}}\right)-\left(X^{n}_{t_{i}^{n}}-X_{t_{i}^{n}}\right)\right)^{2}+3\frac{c}{12(R^{\prime}+1)}\sum_{t_{i}\leq
s,\hskip 2.84526ptt_{i}\in
D^{n}_{k}}\left(X_{t_{i+1}^{n}}-X_{t_{i}^{n}}\right)^{2}$ $\displaystyle\leq
3M^{2}\left([X^{n}-X]_{s}+\frac{c}{24M^{2}}\right)+3\frac{c}{12(R^{\prime}+1)}([X]_{s}+1)<3\frac{c}{12}+3\frac{c}{12(R^{\prime}+1)}(R^{\prime}+1)=\frac{c}{2},$

where we applied the AM-GM inequality to the cross term after the second
inequality. We therefore conclude that $A(n)$ and $B^{\prime}$ are disjoint
and therefore we conclude that
$[f_{n}(X^{n})-f_{n}(X)]_{u}\xrightarrow{{\mathbb{P}}}0$. ∎

###  A.1.2 Proof of Theorem 3.14

###### Proof.

We will first show the Theorem for a) with $p=\frac{1}{2}$ and b) with $q=1$,
at the end of this proof we outline how to carry out the other cases which are
very much analogous. Let $\epsilon>0$ and $c>0$ be arbitrary. Let $a>1$ be
some (large) constant to be chosen later. Let $X=Z+C$ where $Z$ is a
semimartingale and $C$ has zero quadratic variation. According to Theorem 3.6
we have that $f(X_{s})=Y^{a}_{s}+\Gamma^{a}_{s}$ where $Y^{a}$ is a
semimartingale, $\Gamma^{a}$ is continuous and $[\Gamma^{a}]_{t}=0$ for all
$t>0$. We recall that the expression for $Y^{a}$ is given by

$\displaystyle Y_{t}$ $\displaystyle=f(X_{0})+\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f(X_{s-})\right)I_{|\Delta
X_{s}|>a}+\int_{0}^{t}f^{\prime}(X_{s-})dZ_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\nu(\\{s\\},dx).$
(A.1.21)

If we assume b) then since $\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu_{n}(\\{s\\},dx)\xrightarrow{{\mathbb{P}}}\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu(\\{s\\},dx)$ and
$\left\\{\left(\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu_{n}(\\{s\\},dx)\right)^{2}\right\\}_{n}$ is u.i. it follows that

$\lim_{n\to\infty}{\mathbb{E}}\left[\left|\left(\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu_{n}(\\{s\\},dx)\right)^{2}-\left(\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu(\\{s\\},dx)\right)^{2}\right|\right]=0,$

so $\left(\sum_{s\leq t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)\right)^{2}\in
L^{1}$ and this implies (by Jensen’s inequality) that $\sum_{s\leq
t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)\in L^{1}$. By a similar argument, if we
instead assume a) then
$\sum_{s\leq t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)\in L^{1}$. By the arguments
in the proof of Theorem 3.6, $\sum_{s\leq t}\int_{|x|\leq
a}|x|\nu(\\{s\\},dx)<\infty$ a.s. and

$\displaystyle\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(ds,dx)=$
$\displaystyle\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)+$
$\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\mu-\nu)(\\{s\\},dx),$

where $\tilde{\mu}$ denotes the jump measure $\mu$ with all fixed time jumps
removed. We may now rewrite as

$\displaystyle f(X_{0})+\sum_{s\leq t}\left(f(X_{s})-f(X_{s-})-\Delta
X_{s}f(X_{s-})\right)I_{|\Delta
X_{s}|>a}+\int_{0}^{t}f^{\prime}(X_{s-})dZ_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f(X_{s-}+x)-f(X_{s-})-xf^{\prime}(X_{s-})\right)\mu(\\{s\\},dx).$

Similarly let $X^{n}=Z^{n}+C^{n}$ where $Z^{n}$ is a semimartingale and
$C^{n}$ has zero quadratic variation. We again apply Theorem 3.6, we have that
$f_{n}(X^{n}_{s})=(Y^{n})^{a}_{s}+(\Gamma^{n})^{a}_{s}$ where $(Y^{n})^{a}$ is
a semimartingale, $(\Gamma^{n})^{a}$ is continuous and
$[(\Gamma^{n})^{a}]_{t}=0$ for all $t>0$. Arguing as above we see that the
expression for $(Y^{n})^{a}$ will be given by

$\displaystyle Y^{n}_{t}=f_{n}(X^{n}_{0})+\sum_{s\leq
t}\left(f_{n}(X^{n}_{s})-f_{n}(X^{n}_{s-})-\Delta
X^{n}_{s}f(X^{n}_{s-})\right)I_{|\Delta
X^{n}_{s}|>a}+\int_{0}^{t}f_{n}^{\prime}(X^{n}_{s-})dZ^{n}_{s}$
$\displaystyle+\int_{0}^{t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-})-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)$
$\displaystyle+\sum_{s\leq t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-})-xf_{n}^{\prime}(X^{n}_{s-})\right)\mu_{n}(\\{s\\},dx).$

Since

$\displaystyle
f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-})-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)$
$\displaystyle=x\left(\int_{0}^{1}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta
x)-f^{\prime}(X_{s-}+\theta x)\right)d\theta-
f_{n}^{\prime}(X^{n}_{s-})+f^{\prime}(X_{s-})\right),$

and the term in the parenthesis is clearly locally bounded we can conclude
that

$\displaystyle\int_{0}^{.}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-})-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right)(\tilde{\mu}-\nu_{c})(ds,dx),$

is well defined (as a local martingale). Furthermore, since $X^{n}$ and $X$
have quadratic variations along the same refining sequence then so does
$f_{n}(X^{n})$ and $f(X)$ by Theorem 3.6 which implies that
$(\Gamma^{n})^{a}-\Gamma^{a}$ has a quadratic variation along this sequence
which is zero (the quadratic variation of the semimartingale parts of
$f_{n}(X^{n})$ and $f(X)$ do not depend on the refining sequence). With this
in mind we make the following estimate,

$\displaystyle[f_{n}(X^{n})-f(X)]^{\frac{1}{2}}_{t}=[(Y^{n})^{a}-Y^{a}]^{\frac{1}{2}}_{t}\leq\left[\int
f^{\prime}(X_{-})d\left(Z-Z^{n}\right)\right]_{t}^{\frac{1}{2}}+\left[\int\left(f_{n}^{\prime}(X^{n}_{-})-f^{\prime}(X_{-})\right)dZ^{n}\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}-\nu_{c})(ds,dx)\right.$
$\displaystyle\left.\left.-\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}+\right.$
$\displaystyle\left[\sum_{s\leq t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)\mu(\\{s\\},dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\sum_{s\leq t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-})-xf_{n}^{\prime}(X^{n}_{s-})\right)(\mu-\mu_{n})(\\{s\\},dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\sum_{s\leq
t}\left(f_{n}(X^{n}_{s})-f_{n}(X^{n}_{s-})-\Delta
X^{n}_{s}f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta X^{n}_{s}|>a}-\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}\right]_{t}^{\frac{1}{2}},$ (A.1.22)

where we substituted the expressions for $Y$, $Y^{a}$, did a bit of
rearrangement and then used Lemma 2.2.
$\tilde{\mu}_{n}$ and $\tilde{\mu}_{n}$ are void of fixed time jumps so the
quadratic variation of the $\tilde{\mu}-\nu_{c}$ and
$\tilde{\mu}_{n}-\nu_{c}$-integrals can be computed in the same manner as in
the proof of Theorem 3.6. We now expand (A.1.2) as

$\displaystyle=\left(\int_{0^{+}}^{t}f^{\prime}(X_{-})^{2}d\left[Z-Z^{n}\right]_{s}\right)^{\frac{1}{2}}+\left(\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\right)^{\frac{1}{2}}+$
$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq
a}\left((f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)+f_{n}(X^{n}_{s-}))-f(X_{s-}))-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)((\tilde{\mu}-\nu_{c})(ds,dx)\right.$
$\displaystyle\left.\left.-\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}\right.$
$\displaystyle+\left(\sum_{s\leq t}\left(\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)\mu(\\{s\\},dx)\right)^{2}\right)^{\frac{1}{2}}$
$\displaystyle+\left(\sum_{s\leq t}\left(\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-})-xf_{n}^{\prime}(X_{s-})\right)(\mu-\mu_{n})(\\{s\\},dx)\right)^{2}\right)^{\frac{1}{2}}$
$\displaystyle+\left[\sum_{s\leq
t}\left(f_{n}(X^{n}_{s})-f_{n}(X^{n}_{s-})-\Delta
X^{n}_{s}f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta X^{n}_{s}|>a}-\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}\right]_{t}^{\frac{1}{2}},$ (A.1.23)

here we used the fact that for term five and six of (A.1.2) (corresponding to
term five and six of (A.1.2)), the expressions inside the quadratic variations
are quadratic pure jump semimartingales so their contributions are just the
square sums of their jumps.

We now collect some preliminary facts that will be used to tackle the terms of
(A.1.2). Note that assuming a),
${\mathbb{E}}\left[\left(\int_{0}^{t}\int_{{\mathbb{R}}}x^{2}\mu(ds,dx)\right)^{\frac{1}{2}}\right]\leq{\mathbb{E}}[[X]_{t}^{\frac{1}{2}}]<\infty$
and

$\int_{0}^{t}\int_{|x|\leq r}x^{2}\mu(ds,dx)=\sum_{s\leq t}(\Delta
X_{s})^{2}1_{|\Delta X_{s}|\leq r}\xrightarrow{a.s.}0,$

as $r\to 0^{+}$. So by dominated convergence it follows that
$\lim_{r\to 0^{+}}{\mathbb{E}}\left[\left(\int_{0}^{t}\int_{|x|\leq
r}x^{2}\mu(ds,dx)\right)^{\frac{1}{2}}\right]=0$, while if we instead assume
b) then
$\lim_{r\to 0^{+}}{\mathbb{E}}\left[\left(\int_{0}^{t}\int_{|x|\leq
r}x^{2}\mu(ds,dx)\right)^{\frac{1}{2}}\right]=0$.

Since $[X^{n}]_{t}\leq 2[X^{n}-X]_{t}+2[X]_{t}$ it follows that
$|[X^{n}]_{t}-[X]_{t}|\leq 2[X^{n}-X]_{t}+3[X]_{t}$ and since
$[X^{n}]_{t}\xrightarrow{{\mathbb{P}}}[X]_{t}$ we may conclude by Pratt’s
lemma that if we assume b) then $[X^{n}]_{t}\xrightarrow{L^{1}}[X]_{t}$ and
therefore both $\\{[X^{n}]_{t}\\}_{n}$ and $\\{[X^{n}-X]_{t}\\}_{n}$ are u.i.,
under this assumption. Similarly we have that
$[X^{n}]_{t}^{\frac{1}{2}}\leq[X^{n}-X]_{t}^{\frac{1}{2}}+[X]_{t}^{\frac{1}{2}}$
implying
$|[X^{n}]_{t}^{\frac{1}{2}}-[X]_{t}^{\frac{1}{2}}|\leq[X^{n}-X]_{t}^{\frac{1}{2}}+3[X]_{t}^{\frac{1}{2}}$
and so if we assume a), both $\\{[X^{n}]_{t}^{\frac{1}{2}}\\}_{n}$ and
$\\{[X^{n}-X]_{t}^{\frac{1}{2}}\\}_{n}$ are u.i. in this case. By assumption
a) $\left(\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu_{n}(\\{s\\},dx)\right)^{\frac{1}{2}}$ is u.i. and
$\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu_{n}(\\{s\\},dx)\xrightarrow{{\mathbb{P}}}\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu(\\{s\\},dx)$ as $n\to\infty$, this implies
$\left(\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu_{n}(\\{s\\},dx)\right)^{\frac{1}{2}}\xrightarrow{L^{1}}\left(\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu(\\{s\\},dx)\right)^{\frac{1}{2}}$. Similarly,
$\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu_{n}(\\{s\\},dx)\xrightarrow{L^{1}}\sum_{s\leq
t}\int_{{\mathbb{R}}}|x|\mu(\\{s\\},dx)$ if we instead assume b).

Furthermore,

$\displaystyle\lim_{r\to 0^{+}}\limsup_{n}{\mathbb{P}}\left(\sum_{s\leq
t}\int_{|x|\leq r}|x|\mu_{n}(\\{s\\},dx)\geq c\right)\leq$
$\displaystyle\lim_{r\to
0^{+}}\limsup_{n}\left({\mathbb{P}}\left(\left|\sum_{s\leq t}\int_{|x|\leq
r}|x|\mu_{n}(\\{s\\},dx)-\sum_{s\leq t}\int_{|x|\leq
r}|x|\mu(\\{s\\},dx)\right|\geq c\right)+{\mathbb{P}}\left(\sum_{s\leq
t}\int_{|x|\leq r}|x|\mu(\\{s\\},dx)\geq c\right)\right)$ $\displaystyle=$
$\displaystyle\lim_{r\to 0^{+}}{\mathbb{P}}\left(\sum_{s\leq t}\int_{|x|\leq
r}|x|\mu(\\{s\\},dx)\geq c\right)=0,$

for every $c>0$.

It remains to show that all seven terms on the right-most side of (A.1.2)
converge to zero in expectation under the assumed hypotheses. Our strategy
will be the following. We will begin by showing that if we assume a), term
seven of (A.1.2) can be made arbitrarily small in $L^{1}$ (or in $L^{2}$ if we
instead assume b)) uniformly over $n$ by choosing $a$ large enough. Once we
have shown that term seven converges to zero in $L^{1}$ (or $L^{2}$) as
$a\to\infty$ we fix $a$ so large that this term is smaller than say
$\epsilon$. After this we show that all other terms converge to zero as
$n\to\infty$ which means that the right-hand side of $\eqref{gris}$ is less
than $\epsilon$ when $n\to\infty$ and then we can finally let $\epsilon$,
which is arbitrary, go to zero. So for term seven, first note that
${\mathbb{P}}\left((\Delta
X^{n})^{*}_{t}>a\right)\leq{\mathbb{P}}\left([X^{n}]^{1/2}_{t}>a\right)$,
which can be made arbitrarily small for large enough $a$ and therefore term
seven converges to zero in probability as $a\to\infty$ for all $n$. so letting
$a\to\infty$ shows that the above expression tends to zero in probability for
all $n$ as $a\to\infty$. Assuming a) we have by the mean value theorem that
for some $\theta_{1}(s,\omega)$ and $\theta_{2}(s,\omega)$,

$\displaystyle\left[\sum_{s\leq
t}\left(f_{n}(X^{n}_{s})-f_{n}(X^{n}_{s-})-\Delta
X^{n}_{s}f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta X^{n}_{s}|>a}-\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}\right]^{\frac{1}{2}}$ $\displaystyle=$
$\displaystyle\left[\sum_{s\leq t}\Delta
X^{n}_{s}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}(s)\Delta
X^{n}_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta
X^{n}_{s}|>a}-\sum_{s\leq t}\Delta
X_{s}\left(f^{\prime}(X_{s-}+\theta_{2}(s)\Delta
X_{s})-f^{\prime}(X_{s-})\right)I_{|\Delta X_{s}|>a}\right]^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle\left[\sum_{s\leq t}\Delta
X^{n}_{s}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}(s)\Delta
X^{n}_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta
X^{n}_{s}|>a}\right]^{\frac{1}{2}}+\left[\sum_{s\leq t}\Delta
X_{s}\left(f^{\prime}(X_{s-}+\theta_{2}(s)\Delta
X_{s})-f^{\prime}(X_{s-})\right)I_{|\Delta X_{s}|>a}\right]^{\frac{1}{2}}$
$\displaystyle=$ $\displaystyle\left(\sum_{s\leq t}(\Delta
X^{n}_{s})^{2}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}(s)\Delta
X^{n}_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}I_{|\Delta
X^{n}_{s}|>a}\right)^{\frac{1}{2}}+$ $\displaystyle\left(\sum_{s\leq t}(\Delta
X_{s})^{2}\left(f^{\prime}(X_{s-}+\theta_{2}(s)\Delta
X_{s})-f^{\prime}(X_{s-})\right)^{2}I_{|\Delta
X_{s}|>a}\right)^{\frac{1}{2}}\leq\left(4U^{2}\sum_{s\leq t}(\Delta
X^{n}_{s})^{2}I_{|\Delta X^{n}_{s}|>a}\right)^{\frac{1}{2}}+$
$\displaystyle\left(4U^{2}\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta
X_{s}|>a}\right)^{\frac{1}{2}}\leq
2U\left([X^{n}]_{t}^{\frac{1}{2}}1_{H_{a}(n)}+[X]_{t}^{\frac{1}{2}}1_{H_{a}}\right),$

where $H_{a}(n)=\\{\exists s\leq t:|\Delta X^{n}_{s}|>a\\}$ and
$H_{a}=\\{\exists s\leq t:|\Delta X_{s}|>a\\}$. We know that
${\mathbb{P}}(H_{a}(n))$ tends to zero uniformly over $n$ as $a\to\infty$, as
well as ${\mathbb{P}}(H_{a})\to 0$ as $a\to\infty$. This implies (due to the
u.i. property of $\\{[X^{n}]_{t}^{\frac{1}{2}}\\}_{n}$) that
$\lim_{a\to\infty}\sup_{n}{\mathbb{E}}\left[[X^{n}]_{t}^{\frac{1}{2}}1_{H_{a}(n)}\right]=0$,
as well as
$\lim_{a\to\infty}{\mathbb{E}}\left[[X]_{t}^{\frac{1}{2}}1_{H_{a}}\right]=0$
which shows the $L^{1}$ convergence when assuming a). If we instead assume b)
then by using the bounds $|X^{n}_{s-}+\theta_{1}(s)\Delta X^{n}_{s}|\leq
3(X^{n})^{*}_{t}$ and $|X_{s-}+\theta_{2}(s)\Delta X_{s}|\leq 3X^{*}_{t}$ we
find that,

$\displaystyle{\mathbb{E}}\left[\left[\sum_{s\leq
t}\left(f_{n}(X^{n}_{s})-f_{n}(X^{n}_{s-})-\Delta
X^{n}_{s}f_{n}^{\prime}(X^{n}_{s-})\right)I_{|\Delta X^{n}_{s}|>a}-\sum_{s\leq
t}\left(f(X_{s})-f(X_{s-})-\Delta X_{s}f^{\prime}(X_{s-})\right)I_{|\Delta
X_{s}|>a}\right]^{\frac{1}{2}}\right]\leq$
$\displaystyle{\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta
X^{n}_{s})^{2}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}(s)\Delta
X^{n}_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}I_{|\Delta
X^{n}_{s}|>a}\right)^{\frac{1}{2}}\right]+$
$\displaystyle{\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta
X_{s})^{2}\left(f^{\prime}(X_{s-}+\theta_{2}(s)\Delta
X_{s})-f^{\prime}(X_{s-})\right)^{2}I_{|\Delta
X_{s}|>a}\right)^{\frac{1}{2}}\right]\leq$
$\displaystyle{\mathbb{E}}\left[2(M\vee 2C(X^{n})^{*}_{t})\left(\sum_{s\leq
t}(\Delta X^{n}_{s})^{2}I_{|\Delta
X^{n}_{s}|>a}\right)^{\frac{1}{2}}\right]+{\mathbb{E}}\left[2(M\vee
2CX^{*}_{t})\left(\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta
X_{s}|>a}\right)^{\frac{1}{2}}\right]$ $\displaystyle\leq
2{\mathbb{E}}\left[[X^{n}]_{t}1_{H_{a}(n)}\right]^{\frac{1}{2}}{\mathbb{E}}\left[(M\vee
2C(X^{n})^{*}_{t})^{2}1_{H_{a}(n)}\right]^{\frac{1}{2}}+2{\mathbb{E}}\left[[X]_{t}1_{H_{a}}\right]^{\frac{1}{2}}{\mathbb{E}}\left[(M\vee
2CX^{*}_{t})^{2}1_{H_{a}}\right]^{\frac{1}{2}},$

where we used the Cauchy-Schwarz inequality in the last step. By assumption b)
and the same kind of argument we made when assuming a) this shows uniform
convergence to zero as $a\to\infty$ uniformly over $n$. Since
$X^{n}\xrightarrow{ucp}X$ it follows that
$\lim_{R\to\infty}\sup_{n}{\mathbb{P}}\left((X^{n})^{*}_{t}\geq R\right)=0$
hence we may choose $R$ so large that
${\mathbb{P}}\left(A_{R}(n)\right)<\epsilon$ for all $n$ while also
$|f_{n}^{\prime}(x)|\leq 2C|x|$ for all $n$ if $|x|\geq R$ (under assumption
b)), where $A_{R}(n)=\left\\{\max\left((X^{n})^{*}_{t},X^{*}_{t}\right)\leq
R\right\\}$. Let us also define
$M=\sup_{n}\sup_{x\in[-R-a,R+a]}|f_{n}^{\prime}(x)|$.

For the first term in (A.1.2), under assumption a)

$\left(\int_{0^{+}}^{t}f^{\prime}(X_{-})^{2}d\left[Z-Z^{n}\right]_{s}\right)^{\frac{1}{2}}\leq
U[Z^{n}-Z]_{t}^{\frac{1}{2}}=U[X^{n}-X]_{t}^{\frac{1}{2}},$

which converges to zero in $L^{1}$ by assumption.

If we instead assume b)

$\displaystyle{\mathbb{E}}\left[\sqrt{\int_{0^{+}}^{t}f^{\prime}(X_{-})^{2}d\left[Z-Z^{n}\right]_{s}}\right]\leq{\mathbb{E}}\left[M\vee
2CX^{*}_{t}\sqrt{\int_{0^{+}}^{t}d\left[X-X^{n}\right]_{s}}\right]$
$\displaystyle\leq{\mathbb{E}}\left[[X^{n}-X]_{t}\right]^{\frac{1}{2}}{\mathbb{E}}\left[2C(X^{*}_{t})^{2}+M\right]^{\frac{1}{2}},$

by the Cauchy-Schwartz inequality and this converges to zero by assumption.

For the second term in (A.1.2), fix $c>0$. Let $M_{1}$ be so large that
${\mathbb{P}}\left(X^{*}_{t}\geq M_{1}\right)<\epsilon$. Since
$[X-X^{n}]_{t}\xrightarrow{{\mathbb{P}}}0$ as $n\to\infty$ and
$[X^{n}]_{t}\leq\left([X^{n}-X]^{\frac{1}{2}}_{t}+[X]^{\frac{1}{2}}_{t}\right)^{2}$
it follows that we may take $M_{2}$ so large that
${\mathbb{P}}\left([X^{n}]_{t}\geq M_{2}\right)<\varepsilon$ for $n\geq
n_{2}$. Let $\delta>0$ be so small that if $x,y\in[-M_{1}-a,M_{1}+a]$ and
$|x-y|<\delta$ then $|f(x)-f(y)|<\frac{\sqrt{c}}{\sqrt{M_{2}}}$. Take
$n_{1}>0$ so large that
${\mathbb{P}}\left((X^{n}-X)^{*}_{t}\geq\delta\right)<\epsilon$ for $n\geq
n_{1}$. Define
$E_{2}(n)=\\{(X^{n}-X)^{*}_{t}\leq\delta\\}\cap\left\\{[X^{n}]_{t}\leq
M_{2}\right\\}$ then
$\left(f^{\prime}(X^{n})-f^{\prime}(X)\right)^{*}_{t}<\frac{\epsilon}{\sqrt{M_{2}}}$
on $E_{2}(n)$ and ${\mathbb{P}}(E_{2}(n)^{c})<2\varepsilon$ if $n\geq
n^{\prime}:=\max(n_{1},n_{2})$ and on $E_{2}(n)$,

$\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}<\frac{c}{M_{2}}[Z^{n}]_{t}=\frac{c}{M_{2}}[X^{n}]_{t}\leq
c.$

So for $n\geq n^{\prime}$

$\displaystyle{\mathbb{P}}\left(\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\geq
c\right)\leq$
$\displaystyle{\mathbb{P}}\left(\left\\{\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\geq
c\right\\}\cap
E_{2}(n)\right)+{\mathbb{P}}(E_{2}(n)^{c})={\mathbb{P}}(E_{2}(n)^{c})\leq
2\varepsilon,$

thus
$\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\xrightarrow{{\mathbb{P}}}0.$
To show convergence in expectation we notice similarly to the first term that
under assumption a) we have

$\left(\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\right)^{\frac{1}{2}}\leq
2U[Z^{n}]_{t}^{\frac{1}{2}}=2U[X^{n}]_{t}^{\frac{1}{2}}\in L^{1},$

and by dominated convergence this implies
${\mathbb{E}}\left[\left(\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}\right)^{\frac{1}{2}}\right]\to
0$. If we instead assume b) then note that, by letting $A_{R}(n)$ and $M$ be
as above

$\displaystyle{\mathbb{E}}\left[\sqrt{\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[Z^{n}]_{s}}\right]$
$\displaystyle=$
$\displaystyle{\mathbb{E}}\left[\sqrt{\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[X^{n}]_{s}1_{A_{R}(n)}}\right]+{\mathbb{E}}\left[\sqrt{\int_{0^{+}}^{t}\left(f_{n}^{\prime}(X^{n}_{s_{-}})-f^{\prime}(X_{s_{-}})\right)^{2}d[X^{n}]_{s}1_{A_{R}(n)^{c}}}\right]$
$\displaystyle\leq$ $\displaystyle
M{\mathbb{E}}\left[[X^{n}]_{t}^{\frac{1}{2}}\right]+2C{\mathbb{E}}\left[\left(X^{*}_{t}+(X^{n})^{*}_{t}\right)\sqrt{[X^{n}-X]_{t}}\right]$
$\displaystyle\leq$ $\displaystyle
M{\mathbb{E}}\left[[X^{n}-X]_{t}^{\frac{1}{2}}\right]+2C\sqrt{{\mathbb{E}}\left[[X^{n}-X]_{t}\right]}\sqrt{{\mathbb{E}}\left[2(X^{*}_{t})^{2}+2((X^{n})^{*}_{t})^{2}\right]},$

which converges to zero since
${\mathbb{E}}\left[2(X^{*}_{t})^{2}+2((X^{n})^{*}_{t})^{2}\right]$ is
uniformly bounded in $n$ by the u.i. assumption (note that
$((X^{n})^{*}_{t})^{2}\xrightarrow{ucp}(X^{*}_{t})^{2}$ implies
${\mathbb{E}}\left[(X^{*}_{t})^{2}\right]<\infty$ by the u.i. property as
well).

The third term of (A.1.2) will be split into two terms, we will show that the
first one will vanish and the second terms is completely analogous. Given any
$\epsilon>0$ and $c>0$, take $r>0$ so small that
${\mathbb{P}}\left(\left\\{\int_{0}^{t}\int_{|x|\leq
r}|x|^{2}\tilde{\mu}(ds,dx)\geq\frac{c^{2}}{48M}\right\\}\cap
A_{R}(n)\right)<\epsilon$ and take $K$ so large that
${\mathbb{P}}\left(\tilde{\mu}([0,t]\times[-a,a]\setminus[-r,r])\geq
K\right)<\epsilon.$ Let $\delta>0$ be so small that if $x,y\in[-R-a,R+a]$ and
$|x-y|<\delta$ then $|f(x)-f(y)|<\frac{c^{2}}{24K}$. Take $n_{1}>0$ so large
that ${\mathbb{P}}\left((X^{n}-X)^{*}_{t}\geq\delta\right)<\epsilon$ for
$n\geq n_{1}$. Since $\tilde{\mu}([0,t]\times[-a,a]\setminus[-r,r])<\infty$
and since $f_{n}^{\prime}\to f^{\prime}$ uniformly on compacts there exists
$n_{2}$ such that if $n\geq n_{2}$ then

$\int_{0}^{t}\int_{r<|x|\leq
a}\left|f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right|^{2}\tilde{\mu}(ds,dx)<\frac{c}{10},$

on $A_{R}(n)$. We define

$\displaystyle E_{3}(n)$
$\displaystyle=A_{R}(n)\cap\left\\{(X^{n}-X)^{*}_{t}\leq\delta\right\\}\cap\left\\{\int_{0}^{t}\int_{|x|\leq
r}x^{2}\tilde{\mu}(ds,dx)<\frac{c^{2}}{64M^{2}}\right\\}\cap$
$\displaystyle\cap\left\\{\tilde{\mu}([0,t]\times[-a,a]\setminus[-r,r])<K\right\\}.$

Now let $n\geq\max(n_{1},n_{2})$. If we now restrict our attention to the set
$E_{3}(n)$ and apply the mean-value theorem then for some
$\theta_{1}(s,\omega),\theta_{2}(s,\omega)\in[0,1]$,

$\displaystyle\int_{0}^{t}\int_{|x|\leq
r}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)^{2}\tilde{\mu}(ds,dx)$
$\displaystyle\leq\int_{0}^{t}\int_{|x|\leq
r}\left(\left|f^{\prime}(X_{s-}+\theta_{1}x)\right||x|+\left|f_{n}^{\prime}(X^{n}_{s-}+\theta_{2}x)\right||x|+|x|\left|f^{\prime}(X_{s-})\right|+|x|\left|f_{n}^{\prime}(X^{n}_{s-})\right|\right)^{2}\tilde{\mu}(ds,dx)$
$\displaystyle\leq\int_{0}^{t}\int_{|x|\leq
r}8x^{2}\left(\left|f_{n}^{\prime}(X^{n}_{s-}+\theta_{2}x)\right|^{2}+\left|f^{\prime}(X_{s-}+\theta_{1}x)\right|^{2}+|f^{\prime}(X_{s-})|^{2}+|f_{n}^{\prime}(X^{n}_{s-})|^{2}\right)\tilde{\mu}(ds,dx)$
$\displaystyle\leq 32M^{2}\int_{0}^{t}\int_{|x|\leq
r}x^{2}\tilde{\mu}(ds,dx)<32M^{2}\cdot\frac{c^{2}}{64M^{2}}=c^{2}/2,$

while

$\displaystyle\int_{0}^{t}\int_{r<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x))-f_{n}(X^{n}_{s-})+f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right)^{2}\tilde{\mu}(ds,dx)$
$\displaystyle\leq\int_{0}^{t}\int_{r<|x|\leq
a}\left(\left|f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)\right|+\left|f(X_{s-})-f_{n}(X^{n}_{s-})\right|+|x|\left|f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right|\right)^{2}\tilde{\mu}(ds,dx)$
$\displaystyle\leq
3\frac{c^{2}}{6K}\tilde{\mu}([0,t]\times[-a,a]\setminus[-r,r])<\frac{c^{2}}{2},$

so therefore

$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-}+x)-f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}$
$\displaystyle<$
$\displaystyle\left(\frac{c^{2}}{2}+\frac{c^{2}}{2}\right)^{\frac{1}{2}}=c.$

Hence

${\mathbb{P}}\left(\left(\int_{0}^{t}\int_{|x|\leq
a}\left(\left(f(X_{s-}+x)-f(X_{s-})\right)-xf^{\prime}(X_{s-})\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\geq
c\right)\leq{\mathbb{P}}(E_{3}(n)^{c})<4\epsilon.$

To show convergence in $L^{1}$ under assumption a) we notice that by applying
the mean value theorem

$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}$
$\displaystyle\leq\sqrt{32}U$ $\displaystyle\left(\int_{0}^{t}\int_{|x|\leq
a}x^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\leq\sqrt{32}U[X]_{t}^{\frac{1}{2}},$

so we have a $L^{1}$ bound for this term. Assuming b) then

$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}$
$\displaystyle=\left(\int_{0}^{t}\int_{|x|\leq
a}x^{2}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}x)-f^{\prime}(X_{s-}+\theta_{2}x)-f_{n}^{\prime}(X^{n}_{s-})+f^{\prime}(X_{s-})\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\leq$
$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq a}x^{2}\left(4M\vee
2C(|X^{n}_{s-}+\theta_{1}x|+|X_{s-}+\theta_{2}x|+|X^{n}_{s-}|+|X_{s-}|)\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\leq$
$\displaystyle\left(\int_{0}^{t}\int_{|x|\leq a}x^{2}\left(4M\vee
2C(2|x|+(X^{n})^{*}_{t}+X^{*}_{t})\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\leq(4M+2C(2|a|+(X^{n})^{*}_{t}+X^{*}_{t}))[X]_{t}^{\frac{1}{2}},$

and since

${\mathbb{E}}\left[(4\tilde{M}+2C(2|a|+(X^{n})^{*}_{t}+X^{*}_{t}))[X]_{t}^{\frac{1}{2}}1_{E}\right]\leq
C_{1}{\mathbb{P}}(E)+C_{2}\sqrt{{\mathbb{E}}\left[((X^{n})^{*}_{t})^{2}+(X^{*}_{t}))^{2})1_{E}\right]}\sqrt{{\mathbb{E}}\left[[X]_{t}\right]},$

for any measurable $E$, some constants $C_{1},C_{2}$ that depend only on
$\tilde{M},a$ and $C$, we see that term three is u.i. but we already
established convergence in probability to zero therefore this implies
convergence in $L^{1}$ to zero of term three.
For term four of (A.1.2) let $r\in{\mathbb{R}}^{+}$ and $\tilde{N}$ denote the
number of jumps of $X$ larger than $r/2$ and take $K$ so large that
${\mathbb{P}}(\tilde{N}\geq K)<\epsilon$. We may now take $L(r)\in[r/2,r]$
such that ${\mathbb{P}}\left(\exists s\leq t:|\Delta
X_{s}|\in[L(r)-\gamma,L(r)+\gamma]\right)<\epsilon/K$ for some $\gamma>0$
(this is possible since there are only a finite number of jumps exceeding
$r/2$ on $[0,t]$ for $X$) and we may assume without loss of generality that
$\gamma<r/2$. Now let $N$ and $N_{n}$ denote the number of jumps larger than
$L(r)$ of $X$ and $X^{n}$ in $[0,t]$ respectively, then clearly
$N\leq\tilde{N}$ since $L(r)\geq r/2$. Let $T_{1},...,T_{k}$ and
$T^{n}_{1},...,T^{n}_{N_{n}}$ denote the jumps of $X$ and $X^{n}$ larger than
$L(r)$ in $[0,t]$. Let $A_{k}:=\\{N=k\\}\cap\left(\bigcap_{l=1}^{N}\\{|\Delta
X_{T_{l}}|\not\in[L(r)-\gamma,L(r)+\gamma)\\}\right)$ then
${\mathbb{P}}\left(\left(\bigcup_{k=1}^{K}A_{k}\right)^{c}\right)<2\epsilon$.
Now on $A_{k}$ we know that if $||\Delta X_{T_{l}}|\geq L(r)$ then $|\Delta
X_{T_{l}}|\geq L(r)+\gamma$ and if $|\Delta X^{n}_{T_{l}}|<L(r)$ then
$|\Delta(X^{n}-X)_{T_{l}}|\geq\gamma$ for any $l\in\\{1,...,k\\}$, similarly
if $|\Delta X_{T^{n}_{l}}|<L(r)$ then in fact $|\Delta
X_{T^{n}_{l}}|<L(r)-\gamma$ implying
$|\Delta(X^{n}-X)_{T^{n}_{l}}|\geq\gamma$, so

$\displaystyle\lim_{n\to\infty}{\mathbb{P}}\left(\left\\{\\{T^{n}_{1},...,T^{n}_{N_{n}}\\}\neq\\{T_{1},...,T_{k}\\}\right\\}\cap
A_{k}\right)\leq\lim_{n\to\infty}{\mathbb{P}}\left(\left\\{(\Delta(X^{n}-X))^{*}_{t}\geq\gamma\right\\}\cap
A_{k}\right)$
$\displaystyle\leq\lim_{n\to\infty}{\mathbb{P}}\left((\Delta(X^{n}-X))^{*}_{t}\geq\gamma\right)\leq\lim_{n\to\infty}{\mathbb{P}}\left([X^{n}-X]^{1/2}_{t}\geq\gamma\right)=0,$

for every $k$. Therefore

$\displaystyle\lim_{n\to\infty}{\mathbb{P}}\left(\\{T^{n}_{1},...,T^{n}_{N_{n}}\\}\neq\\{T_{1},...,T_{N}\\}\right)\leq\lim_{n\to\infty}{\mathbb{P}}\left(\left\\{\\{T^{n}_{1},...,T^{n}_{N_{n}}\\}\neq\\{T_{1},...,T_{k}\\}\right\\}\cap\left(\bigcup_{k=1}^{K}A_{k}\right)\right)$
$\displaystyle+{\mathbb{P}}\left(\left(\bigcup_{k=1}^{K}A_{k}\right)^{c}\right)\leq\sum_{k=1}^{K}\lim_{n\to\infty}{\mathbb{P}}\left(\left\\{\\{T^{n}_{1},...,T^{n}_{N_{n}}\\}\neq\\{T_{1},...,T_{k}\\}\right\\}\cap
A_{k}\right)+{\mathbb{P}}\left(\left(\bigcup_{k=1}^{K}A_{k}\right)^{c}\right)$
$\displaystyle={\mathbb{P}}\left(\left(\bigcup_{k=1}^{K}A_{k}\right)^{c}\right)<2\epsilon,$

letting $\epsilon\to 0^{+}$ shows
$\lim_{n\to\infty}{\mathbb{P}}\left(\\{T^{n}_{1},...,T^{n}_{N_{n}}\\}\neq\\{T_{1},...,T_{N}\\}\right)=0$,
i.e. with probability tending to one the jump times corresponding to jumps of
modulus greater or equal to $L(r)$ for $X$ and and $X^{n}$ on $[0,t]$
coincide, this implies

$\displaystyle\sum_{s\leq t}\left|1_{|\Delta X^{n}_{s}|\geq L(r)}-1_{|\Delta
X_{s}|\geq L(r)}\right|\xrightarrow{{\mathbb{P}}}0,$ (A.1.24)

as $n\to\infty$ and since this sum cannot assume any values between 0 and 1
this means that for large $n$ the sum is exactly zero with large probability.
Let
$B(n,r)=\left\\{\sum_{s\leq t}\left|1_{|\Delta X^{n}_{s}|\geq L(r)}-1_{|\Delta
X_{s}|\geq L(r)}\right|=0\right\\}$ then ${\mathbb{P}}(B(n,r))\to 1$ as
$n\to\infty$, for each $r>0$. On $B(n,r)$, by applying the AM-GM inequality

$\displaystyle\sum_{s\leq t}(\Delta X^{n}_{s})^{2}I_{|\Delta X^{n}_{s}|\leq
L(r)}=\sum_{s\leq t}(\Delta(X^{n}-X)_{s}+\Delta X^{n}_{s})^{2}I_{|\Delta
X^{n}_{s}|\leq L(r)}$ $\displaystyle\leq$ $\displaystyle 2\sum_{s\leq
t}(\Delta(X^{n}-X)_{s})^{2}+2\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta
X^{n}_{s}|\leq L(r)}$ $\displaystyle\leq$ $\displaystyle 2\sum_{s\leq
t}(\Delta(X^{n}-X)_{s})^{2}+2\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta
X_{s}|\leq L(r)}\leq 2[X^{n}-X]_{t}+2\sum_{s\leq t}(\Delta
X_{s})^{2}I_{|\Delta X_{s}|\leq L(r)},$ (A.1.25)

the last inequality follows from the fact that $|\Delta X^{n}_{s}|>L(r)$ if
and only if $|\Delta X_{s}|>L(r)$ which is the same as to say that $|\Delta
X^{n}_{s}|\leq L(r)$ if and only if $|\Delta X_{s}|\leq L(r)$ so the last sum
contains at most all the jumps less or equal to $L(r)$ of $X$ ($(\Delta
X_{s})^{2}I_{|\Delta X_{s}|\leq L(r)}>0$ only if $(\Delta X_{s})\leq L(r)$ on
$B(n,r)$).

Applying Lemma 2.2 and doing a bit of rearrangement of term four of (A.1.2)
gives us

$\displaystyle\left[\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)((\tilde{\mu}-(\nu)_{c})(ds,dx)\right.$
$\displaystyle\left.-\int_{0}^{.}\int_{|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle\leq\left[\int_{0}^{.}\int_{|x|\leq
L(r)}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)((\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{|x|\leq
L(r)}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}-\tilde{\mu}_{n})(ds,dx)\right]_{t}^{\frac{1}{2}}$
$\displaystyle+\left[\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\nu-\nu_{n})_{c}(ds,dx)\right]_{t}^{\frac{1}{2}}.$
(A.1.26)

We will now show convergence of each term in the right-hand side of (A.1.3).
Let $\tilde{\epsilon}>0$ be arbitrary. We will show that if we assume a), the
$L^{1}$-limit ($L^{2}$-limit if we assume b)) in $n$, for some small $r$, of
the right-hand side of (A.1.3) can be bounded by some constant times
$\tilde{\epsilon}$ and since $\tilde{\epsilon}$ is arbitrary this will show
that (A.1.3) converges to zero in $L^{1}$ if we assume a) (in $L^{2}$ if we
assume b)). Assuming a), then by taking $r$ so small that
${\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq
L(r)}\right)^{\frac{1}{2}}\right]<\tilde{\varepsilon}$ and using the fact that
$\lim_{n\to\infty}{\mathbb{E}}\left[[X^{n}-X]_{t}^{\frac{1}{2}}\right]=0$ we
see that

$\displaystyle\lim_{n\to\infty}{\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta
X^{n}_{s})^{2}I_{|\Delta X^{n}_{s}|\leq
L(r)}\right)^{\frac{1}{2}}\right]\leq\lim_{n\to\infty}{\mathbb{E}}\left[\left(\sum_{s\leq
t}(\Delta X^{n}_{s})^{2}I_{|\Delta X^{n}_{s}|\leq
L(r)}\right)^{\frac{1}{2}}1_{B(n,r)^{c}}\right]$ $\displaystyle+$
$\displaystyle\sqrt{2}\left(\lim_{n\to\infty}{\mathbb{E}}\left[[X^{n}-X]_{t}^{\frac{1}{2}}1_{B(n,r)}\right]+\lim_{n\to\infty}{\mathbb{E}}\left[\left(\sum_{s\leq
t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq
L(r)}\right)^{\frac{1}{2}}1_{B(n,r)}\right]\right)$ $\displaystyle\leq$
$\displaystyle\lim_{n\to\infty}{\mathbb{E}}\left[[X^{n}-X]_{t}^{\frac{1}{2}}\right]+\sqrt{2}\lim_{n\to\infty}{\mathbb{E}}\left[\left(\sum_{s\leq
t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq
L(r)}\right)^{\frac{1}{2}}\right]<\sqrt{2}\tilde{\epsilon},$

where we used (A.1.3) and the triangle inequality at the first step. In the
same fashion if we take $r$ so small that ${\mathbb{E}}\left[\sum_{s\leq
t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq L(r)}\right]<\varepsilon$ we see
that if we assume b) then $\lim_{n\to\infty}{\mathbb{E}}\left[\sum_{s\leq
t}(\Delta X^{n}_{s})^{2}I_{|\Delta X^{n}_{s}|\leq
L(r)}\right]<2\tilde{\epsilon}$. For the first term on the right-hand side of
(A.1.3) we have assuming a)

$\displaystyle{\mathbb{E}}\left[\left[\int_{0}^{.}\int_{|x|\leq
L(r)}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)((\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}\right]=$
$\displaystyle{\mathbb{E}}\left[\left(\int_{0}^{t}\int_{|x|\leq
L(r)}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\right]\leq$
$\displaystyle{\mathbb{E}}\left[\left(\int_{0}^{t}\int_{|x|\leq
L(r)}x^{2}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta
x)-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}\right]\leq
2U{\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta X_{s})^{2}I_{|\Delta
X_{s}|\leq L(r)}\right)^{\frac{1}{2}}\right]\leq 2U\tilde{\epsilon},$

Assuming b) then

$\displaystyle{\mathbb{E}}\left[\left[\int_{0}^{.}\int_{|x|\leq
L(r)}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)((\tilde{\mu}-\nu_{c})(ds,dx)\right]_{t}^{\frac{1}{2}}\right]\leq$
$\displaystyle{\mathbb{E}}\left[\left(\int_{0}^{t}\int_{|x|\leq
L(r)}x^{2}\left(2M\vee
2C(2(X^{n})^{*}_{t}+|x|)\right)^{2}\tilde{\mu}(ds,dx)\right)^{\frac{1}{2}}1_{E}\right]\leq$
$\displaystyle{\mathbb{E}}\left[\left(2M+2C(2(X^{n})^{*}_{t}+|a|)\right)\left(\sum_{s\leq
t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq
L(r)}\right)^{\frac{1}{2}}\right]\leq$
$\displaystyle\sqrt{{\mathbb{E}}\left[\left(2M+2C(2(X^{n})^{*}_{t}+|a|)\right)^{2}\right]}\sqrt{{\mathbb{E}}\left[\sum_{s\leq
t}(\Delta X_{s})^{2}I_{|\Delta X_{s}|\leq
L(r)}\right]}\leq\sqrt{{\mathbb{E}}\left[\left(2M+2C(2(X^{n})^{*}_{t}+|a|)\right)^{2}\right]}2\tilde{\epsilon},$

since $\\{((X^{n})^{2})^{*}_{t}\\}_{n}$ is u.i.the factor in front of
$\tilde{\epsilon}$ can be made uniformly bounded over all $n$.

For the second term of (A.1.3) the first steps are analogous to those of the
first term. Assuming a) and skipping ahead,

$\displaystyle{\mathbb{E}}\left[\left(\left[\int_{0}^{.}\int_{|x|\leq
L(r)}(((f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-})))-xf_{n}^{\prime}(X^{n}_{s-}))(\tilde{\mu}_{n}-(\nu_{n})_{c})(ds,dx)\right]_{t}\right)^{\frac{1}{2}}\right]$
$\displaystyle\leq$
$\displaystyle{\mathbb{E}}\left[\left(\int_{0}^{.}\int_{|x|\leq
L(r)}x^{2}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta
x)-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}\tilde{\mu}_{n}(ds,dx)\right)^{\frac{1}{2}}\right]\leq
2U{\mathbb{E}}\left[\left(\sum_{s\leq t}(\Delta X^{n}_{s})^{2}1_{|\Delta
X^{n}_{s}|\leq L(r)}\right)^{\frac{1}{2}}\right]\leq
2U\sqrt{2}\tilde{\epsilon}.$

Assuming b) also gives the analogous bound that we had for term one of
(A.1.3).

Next, for the third term in the right-hand side of (A.1.3) fix some arbitrary
$\epsilon,c>0$. Note that since $f_{n}\to f$ uniformly on $[-R-a,R+a]$ there
exists $\delta>0$ such that if $|x-y|<\delta$ then
$|f_{n}(x)-f_{n}(y)|<\frac{\sqrt{c}}{2K}$ for $x,y\in[-R-a,R+a]$. Let

$\displaystyle E_{4}(n)$
$\displaystyle:=\left(\bigcup_{k=1}^{K}A_{k}\right)\cap\left\\{\sum_{s\leq
t}\left|1_{|\Delta X^{n}_{s}|\geq L(r)}-1_{|\Delta X_{s}|\geq
L(r)}\right|=0\right\\}\cap
A_{R}(n)\cap\left\\{(\Delta(X^{n}-X))^{*}_{t}<\min\left(\delta,\frac{\sqrt{c}}{2MK}\right)\right\\},$

then ${\mathbb{P}}\left(E_{4}(n)^{c}\right)<6\epsilon$ for large enough $n$.
We have on $E_{4}(n)$, since $1_{L(r)<|\Delta X_{s}|\leq a}=1_{L(r)<|\Delta
X^{n}_{s}|\leq a}$ for every $s\leq t$,

$\displaystyle\left[\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}-\tilde{\mu}_{n})(ds,dx)\right]_{t}$
$\displaystyle=\left[\sum_{s\leq.}\left(\left(f_{n}(X^{n}_{s-}+\Delta
X_{s})-f_{n}(X^{n}_{s-}))\right)-\Delta
X_{s}f_{n}^{\prime}(X^{n}_{s-})\right)1_{L(r)<|\Delta X_{s}|\leq a}\right.$
$\displaystyle\left.-\sum_{s\leq.}\left(\left(f_{n}(X^{n}_{s-}+\Delta
X^{n}_{s})-f_{n}(X^{n}_{s-}))\right)-\Delta
X^{n}_{s}f_{n}^{\prime}(X^{n}_{s-})\right)1_{L(r)<|\Delta X^{n}_{s}|\leq
a}\right]_{t}$ $\displaystyle=\left[\sum_{s\leq.}\left(f_{n}(X^{n}_{s-}+\Delta
X_{s})-f_{n}(X^{n}_{s-}+\Delta
X^{n}_{s})-\Delta(X^{n}-X)_{s}f_{n}^{\prime}(X^{n}_{s-})\right)1_{L(r)<|\Delta
X_{s}|\leq a}\right]_{t}$ $\displaystyle\leq\sum_{s\leq
t}\left(f_{n}(X^{n}_{s-}+\Delta X_{s})-f_{n}(X^{n}_{s-}+\Delta
X^{n}_{s})-\Delta(X^{n}-X)_{s}f_{n}^{\prime}(X^{n}_{s-})\right)^{2}1_{L(r)<|\Delta
X_{s}|\leq a}$ $\displaystyle\leq
K\left(\frac{\sqrt{c}}{4\sqrt{K}}+M\frac{\sqrt{c}}{4\sqrt{K}M}\right)^{2}<c.$

Assuming a) we have for some $\theta_{1}(s,\omega),\theta_{1}(s,\omega)$,

$\displaystyle\left[\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}-\tilde{\mu}_{n})(ds,dx)\right]_{t}^{\frac{1}{2}}\leq$
$\displaystyle\left(\sum_{s\leq
t}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{1}(s)\Delta
X_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}(\Delta X_{s})^{2}1_{L(r)<|\Delta
X_{s}|\leq a}+\right.$ $\displaystyle\left.\sum_{s\leq
t}\left(f_{n}^{\prime}(X^{n}_{s-}+\theta_{2}(s)\Delta
X^{n}_{s})-f_{n}^{\prime}(X^{n}_{s-})\right)^{2}(\Delta
X^{n}_{s})^{2}1_{L(r)<|\Delta X^{n}_{s}|\leq a}\right)^{\frac{1}{2}}\leq$
$\displaystyle\left(\sum_{s\leq t}4U^{2}(\Delta X_{s})^{2}1_{L(r)<|\Delta
X_{s}|\leq a}+4U^{2}\sum_{s\leq t}(\Delta X^{n}_{s})^{2}1_{L(r)<|\Delta
X^{n}_{s}|\leq a}\right)^{\frac{1}{2}}$ $\displaystyle\leq
2U\left([X]_{t}+[X^{n}]_{t}\right)^{\frac{1}{2}}\leq
2U\left(\sqrt{[X]_{t}}+\sqrt{[X^{n}]_{t}}\right),$

the right-most side is u.i. since $\sqrt{[X^{n}]_{t}}$ is. Assuming b)

$\displaystyle{\mathbb{E}}\left[\left[\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\tilde{\mu}-\tilde{\mu}_{n})(ds,dx)\right]_{t}^{\frac{1}{2}}1_{E}\right]$
$\displaystyle\leq$
$\displaystyle{\mathbb{E}}\left[\tilde{C}(M\vee((X^{n})^{*}_{t}+X^{*}_{t}))\left(\sum_{s\leq
t}(\Delta X_{s})^{2}1_{L(r)<|\Delta X_{s}|\leq a}+\sum_{s\leq t}(\Delta
X^{n}_{s})^{2}1_{L(r)<|\Delta X^{n}_{s}|\leq
a}\right)^{\frac{1}{2}}1_{E}\right]$ $\displaystyle\leq$
$\displaystyle\sqrt{{\mathbb{E}}\left[\tilde{C}^{2}(M\vee((X^{n})^{*}_{t}+X^{*}_{t}))^{2}1_{E}\right]}\sqrt{{\mathbb{E}}\left[\left([X]_{t}+[X^{n}]_{t}\right)1_{E}\right]},$

so this term is u.i. as well.
Since the process $\int_{0}^{.}\int_{L(r)<|x|\leq
a}\left(\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))\right)-xf_{n}^{\prime}(X^{n}_{s-})\right)(\nu-\nu_{n})_{c}(ds,dx)$
is continuous and of finite variation it has zero quadratic variation so the
final term gives no contribution. We have thus shown that when assuming a) the
right-hand side of (A.1.3) has an $L^{1}$ limit less or equal
$D^{\prime}\tilde{\epsilon}$ for some $D^{\prime}\in{\mathbb{R}}^{+}$.
Similarly the right-hand side of (A.1.3) has an $L^{2}$ limit less or equal
$D^{\prime\prime}\tilde{\epsilon}$ for some
$D^{\prime\prime}\in{\mathbb{R}}^{+}$ when assuming b). This takes care of
term four of (A.1.2).

For term five of (A.1.2) we note that

$\displaystyle\left(\sum_{s\leq t}\left(\int_{a<|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)\mu(\\{s\\},dx)\right)^{2}\right)^{\frac{1}{2}}$
$\displaystyle\leq$ $\displaystyle\sum_{s\leq t}\left|\int_{a<|x|\leq
a}\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right)\mu(\\{s\\},dx)\right|$
$\displaystyle\leq$ $\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}\left|f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right|\mu(\\{s\\},dx).$
(A.1.27)

Take $W\in{\mathbb{R}}^{+}$ so large that
${\mathbb{P}}\left(\mu([0,t]\times[-a,a]\setminus[-r,r])\geq
W\right)<\epsilon$. Also let $n^{\prime}$ be so large and $\delta^{\prime}>0$
be so small that $|x-y|<\delta^{\prime}$ and $n\geq n^{\prime}$ implies that
$\max\left(|f_{n}(x)-f(y)|,|f_{n}^{\prime}(x)-f^{\prime}(y)|\right)<\frac{c}{6W}$.
Define

$\displaystyle
E_{5}(n)=A_{R}(n)\cap\left\\{(X^{n}-X)^{*}_{t}\leq\delta\right\\}\cap\left\\{\sum_{s\leq
t}\int_{|x|\leq
r}|x|\mu(\\{s\\},dx)<\frac{c}{8M}\right\\}\cap\left\\{\mu([0,t]\times[-a,a]\setminus[-r,r])<W\right\\},$

then for large $n$ we have ${\mathbb{P}}\left(E_{5}(n)^{c}\right)<4\epsilon$.
Let us now restrict our attention to this subset. Similarly to term three of
(A.1.2) we may split the right-hand side of (A.1.3)

$\displaystyle\sum_{s\leq t}\int_{|x|\leq
r}\left|f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)-f_{n}(X^{n}_{s-})+f(X_{s-})-x(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-}))\right|\mu(\\{s\\},dx)$
$\displaystyle\leq$ $\displaystyle\sum_{s\leq t}\int_{|x|\leq
r}|x|\left(\left|f^{\prime}(X_{s-}+\theta_{1}x)\right|+\left|f_{n}^{\prime}(X^{n}_{s-}+\theta_{2}x)\right|+\left|f^{\prime}(X_{s-})\right|+\left|f_{n}^{\prime}(X^{n}_{s-})\right|\right)\mu(\\{s\\},dx)$
$\displaystyle\leq$ $\displaystyle\sum_{s\leq t}\int_{|x|\leq
r}4M|x|\mu(\\{s\\},dx)<4M\frac{c}{8M}=\frac{c}{2},$

while

$\displaystyle\sum_{s\leq t}\int_{r<|x|\leq
a}\left|\left(f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x))-f_{n}(X^{n}_{s-})+f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right|\mu(\\{s\\},dx)$
$\displaystyle\leq$ $\displaystyle\sum_{s\leq t}\int_{r<|x|\leq
a}\left(\left|f_{n}(X^{n}_{s-}+x)-f(X_{s-}+x)\right|+\left|f(X_{s-})-f_{n}(X^{n}_{s-})\right|+\left|f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right|\right)\mu(\\{s\\},dx)$
$\displaystyle\leq$ $\displaystyle
3\frac{c}{10W}\mu([0,t]\times[-a,a]\setminus[-r,r])<3\frac{c}{6W}\cdot
W=\frac{c}{2},$

and therefore

$\displaystyle\sum_{s\leq t}\int_{|x|\leq
a}\left|\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-}+x)-f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right|\mu(\\{s\\},dx)<c,$

on $E_{5}(n)$ so term five converges to zero in probability. Assuming a) we
can bound the right-most side of (A.1.3) by $3U\sum_{s\leq t}\int_{|x|\leq
a}|x|\mu(\\{s\\},dx)\in L^{1},$ and use dominated convergence. Assuming b)
then

$\displaystyle{\mathbb{E}}\left[\sum_{s\leq t}\int_{|x|\leq
a}\left|\left(f_{n}(X^{n}_{s-}+x)-f_{n}(X^{n}_{s-}))+f(X_{s-}+x)-f(X_{s-})\right)-x\left(f_{n}^{\prime}(X^{n}_{s-})-f^{\prime}(X_{s-})\right)\right|\mu(\\{s\\},dx)1_{E}\right]$
$\displaystyle\leq$ $\displaystyle{\mathbb{E}}\left[\sum_{s\leq
t}\int_{|x|\leq
a}|x|\left(\left|f^{\prime}(X_{s-}+\theta_{1}x)\right|+\left|f_{n}^{\prime}(X^{n}_{s-}+\theta_{2}x)\right|+\left|f^{\prime}(X_{s-})\right|+\left|f_{n}^{\prime}(X^{n}_{s-})\right|\right)\mu(\\{s\\},dx)1_{E}\right]$
$\displaystyle\leq$
$\displaystyle{\mathbb{E}}\left[2C\left(2a+2(X^{n})^{*}_{t}+2X^{*}_{t}\right)\sum_{s\leq
t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)1_{E}\right]$ $\displaystyle\leq$
$\displaystyle
2C\sqrt{{\mathbb{E}}\left[\left(2a+2(X^{n})^{*}_{t}+2X^{*}_{t}\right)^{2}1_{E}\right]}\sqrt{{\mathbb{E}}\left[\left(\sum_{s\leq
t}\int_{|x|\leq a}|x|\mu(\\{s\\},dx)\right)^{2}1_{E}\right]},$

which shows u.i.
For term six of (A.1.2), take $r$ so small that

${\mathbb{P}}\left(\left\\{\sup_{n}\sum_{s\leq t}\int_{|x|\leq