# Non-topological large deviations via maxitive monetary risk measures

José M. Zapata  Centro Universitario de la Defensa. Universidad Politécnica de
Cartagena. c/ Coronel López Peña S/N, Santiago de La Ribera, 30720, Murcia,
Spain<EMAIL_ADDRESS>
(Date: August 27, 2024)

###### Abstract.

In large deviations theory one usually has a family of probability measures in
some space $X$ with a topology. In this paper we study what kind of large
deviation results can be established when no topology is assumed on the
underlying measure space $X$. The main tool is duality theory of risk
measures. We prove that a monetary risk measures $\phi$ satisfies the Laplace
principle $\phi(f)={\rm ess.sup}_{X}(f-I)$ for some measurable function
$I\colon X\to[0,\infty]$ if and only $\phi$ is maxitive and continuous from
below. In the special case of the asymptotic entropy
$\psi(f)=\lim_{n\to\infty}\tfrac{1}{n}\log\int e^{nf}d\nu_{n}$, this result
can been understood as a non-topological analogue of the Varadhan’s integral
lemma. In the spirit of Bryc’s theorem, we study non-topological analogues of
exponential tightness which ensure the Laplace principle characterised in the
main result. The theory is applied to obtain non-topological large deviation
results for sequences of sublinear expectations.

_Key words:_ maxitive monetary risk measures, large deviations, asymptotic
entropy .

_AMS 2020 Subject Classification: 60F10, 47H07, 91B05._

##  1\. Introduction

Suppose that $(X,\mathcal{X})$ is a measurable space and fix some reference
probability measure $\mu$ on $\mathcal{X}$. Given a probability measure $\nu$
on $\mathcal{X}$ such that $\nu\ll\mu$, the entropic risk measure with risk
aversion $\gamma>0$ is defined as

$e_{\gamma,\nu}(f)=\tfrac{1}{\gamma}\log\int e^{\gamma f}d\mu,$

for all function $f\in L^{\infty}(X,\mathcal{X},\mu)$. Given a sequence of
risk aversions $\gamma_{n}\uparrow\infty$ and a sequence of probability
measures $(\nu_{n})_{n\in\mathbb{N}}$ on $\mathcal{X}$ with $\nu_{n}\ll\mu$,
the (upper/lower) _asymptotic entropy_ is defined by the (upper/lower) limit

$\psi(f)=\underset{n\to\infty}{\lim}e_{\gamma_{n},\nu_{n}}(f)=\underset{n\to\infty}{\lim}\tfrac{1}{\gamma_{n}}\log\int
e^{\gamma_{n}f}d\nu_{n}.$

For simplicity in the exposition, we assume for now that the limit above
exists for all $f\in L^{\infty}(X,\mathcal{X},\mu)$111As usual
$L^{\infty}(X,\mathcal{X},\mu)$ denotes the space of all (classes of
equivalence of) $\mu$-essentially bounded real-valued measurable functions on
$X$.. The asymptotic entropy plays an important role in the field of large
deviations. In fact, one of the most fundamental results in large deviations
theory is the Varadhan’s integral lemma, which give conditions under which the
asymptotic entropy can be computed by means of a rate function. More
specifically, in the special situation where $X$ is a regular topological
space and $\mathcal{X}=\mathcal{B}(X)$ is the Borel $\sigma$-field on $X$, the
Varadhan’s integral lemma asserts that the asymptotic entropy verifies the
Laplace principle222$C_{b}(X)$ denotes all real-valued continuous bounded
functions on $X$.

$\psi(f)=\underset{x\in X}{\sup}(f(x)-I(x))\quad\mbox{ for all }f\in C_{b}(X)$

whenever the sequence $(\nu_{n})_{n\in\mathbb{N}}$ satisfies the _large
deviation principle_ 333I.e., $-\inf_{x\in
A^{\circ}}I(x)\leq\liminf_{n}\tfrac{1}{\gamma_{n}}\log\nu_{n}(A)\leq\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log\nu_{n}(A)\leq-\inf_{x\in\overline{A}}I(x)$
for all $A\in\mathcal{B}(X)$. Here, $\overline{A}$ and $A^{\circ}$ denote
respectively the closure and interior of $A$.  (LDP) with rate function
$I\colon X\to[0,\infty]$. On the other hand, the Bryc’s theorem states that
the Laplace principle also holds for the minimal rate function
$I_{\min}(x)=\sup_{f\in C_{b}(X)}\\{f(x)-\phi(f)\\}$ if the sequence
$(\nu_{n})_{n\in\mathbb{N}}$ is exponentially tight444I.e., for all $M>0$
there exists $K\subset X$ compact such that
$\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log\nu_{n}(K^{c})\leq-M$. . We
refer to [3] for terminology and details. Using the language of risk analysis,
the asymptotic entropy $\psi\colon C_{b}(X)\to\mathbb{R}$ is a monetary risk
measure, that is, a monotone and translation invariant functional. Moreover,
it has the remarkable property of being maxitive, that is, $\psi(f\vee
g)\leq\psi(f)\vee\psi(g)$. Recently Kupper and Zapata [9] have shown that,
under an appropriate formulation, both Varadhan’s integral lemma and Bryc’s
theorem hold for any maxitive monetary risk measure on $C_{b}(X)$.

In these results, and usually in large deviation theory, the underlying space
$X$ is endowed with a (sufficiently regular) topology, which plays an
important role in the theory and applications; see [12]. In this article we
aim to study what type of large deviation results can be established when no
topology is assumed on the underlying measurable space $(X,\mathcal{X})$. To
reach this goal, in line with [9], we introduce a new Laplace principle for
general monetary risk measures and characterise it by means of duality theory
of risk measure. While our results apply to general maxitive monetary risk
measures on $L^{\infty}(X,\mathcal{X},\mu)$, we explain it now for easier
readability for the case of the asymptotic entropy
$\psi(f)=\lim_{n\to\infty}\tfrac{1}{\gamma_{n}}\log\int
e^{\gamma_{n}f}d\nu_{n}$, which is a particular instance of maxitive monetary
risk measure. Namely, we prove that the asymptotic entropy verifies the
following form of the Laplace principle

$\psi(f)=\underset{x\in X}{{\rm ess.sup}}(f(x)-I_{\min}(x))\quad\mbox{ for all
}f\in L^{\infty}(X,\mathcal{X},\mu),$ (1.1)

for the minimal penalty $I_{\min}={\rm ess.sup}\\{f-\psi(f)\colon f\in
L^{\infty}\\}$, if and only if the functional $\psi$ is continuous from below,
i.e. $\psi(f_{n})\uparrow\psi(f)$ whenever $f_{n}\uparrow f$ a.s.. Varadhan’s
integral lemma is a generalization of the classical Laplace’s method to
infinite dimensional spaces; for further details see remark after Theorem
4.3.1 in [3]. Thereby, our result characterising (1.1) is an alternative non-
topological extension of the Laplace’s method to infinite dimensional spaces.
In addition, we prove that the Laplace principle (1.1) is satisfied if and
only if the following upper large deviation estimate holds

$\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\nu_{n}(A)=-{\rm
ess.inf}_{A}I_{\min}\quad\mbox{ for all }A\in\mathcal{X}.$

We analyse different sufficient conditions for the Laplace principle. In the
spirit of Bryc’s theorem we study local properties on $\psi$ that guarantee
the representation (1.1). More specifically, we prove that (1.1) holds if for
every $M>0$ we can find $A\in\mathcal{X}$ such that

$\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\nu_{n}(A^{c})<-M\mbox{
and, }$ (1.2)
$\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\nu_{n}(B)\leq\frac{1}{M}-{\rm
ess.inf}_{B}I_{\min}\quad\mbox{ for all }B\subset A,\>B\in\mathcal{X}.$ (1.3)

The above condition can be understood as a non-topological analogue of
exponential tightness in standard large deviations theory, where the
requirement for $A$ to be compact is replaced by the condition (1.3); cf. [3,
Theorem 4.4.2].

Although we have outlined above our results for the special case of the
asymptotic entropy, they apply to general maxitive monetary risk measures. For
instance, we cover the generalized asymptotic entropy

$\psi(f)=\underset{n\to\infty}{\lim}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{nf})$

of a sequence of sublinear expectations $(\mathcal{E}_{n})_{n\in\mathbb{N}}$,
for which we provide new limit results.

###  1.1. Summary

We now briefly summarize the content of the individual sections of this paper.

  * •

In Section 2 we provide the setting of the paper, and introduce and
characterise the Laplace principle for general monetary risk measures. We also
analyse some simple examples.

  * •

Section 3 is devoted to the proof of the characterisation result stated in the
previous section.

  * •

In Section 4 we study local tightness conditions that ensure the Laplace
principle.

  * •

Section 5 is devoted to a moment condition that provides the Laplace principle
on unbounded functions.

  * •

In Section 6 we apply the results to obtain limit results for sequences of
sublinear expectations.

  * •

The paper concludes with an appendix, where we include some auxiliary results.

##  2\. Laplace principle for monetary risk measures

###  2.1. Notation and setting

Throughout this paper, let $(X,\mathcal{X})$ be a measurable space, and fix
some reference probability measure $\mu$ on $\mathcal{X}$. All equalities and
inequalities between measurable functions are understood in the $\mu$-almost
sure sense. We denote by $L^{p}:=L^{p}(X,\mathcal{X},\mu)$, $p\in[1,\infty]$,
the standard Lebesgue spaces of (equivalence classes modulo $\mu$-a.s.
equality of) real-valued measurable functions on $X$. In this paper we deal
with two types of essential supremums. The _essential supremum_ of a
measurable function $f\colon X\to[-\infty,\infty]$, denoted by ${\rm
ess.sup}_{X}f$, is defined as ${\rm
ess.sup}_{X}f=\inf\\{r\in\mathbb{R}\colon\mu(r\geq f)=1\\}$. The _essential
supremum_ of a family $\mathcal{F}$ of measurable functions from $X$ to
$[-\infty,\infty]$ is denoted by

${\rm ess.sup}\mathcal{F}=\underset{f\in\mathcal{F}}{{\rm ess.sup}}\>f.$

We refer to the Appendix for the definition and some properties of the
essential supremum of a family of measurable functions. We denote by
$\mathscr{M}_{1}$ the set of all probability measures $\nu$ on
$(X,\mathcal{X})$ such that $\nu\ll\mu$. We always make the convention
$(\pm\infty)\cdot 0:=0$, so that given two functions $f,g\colon
X\to[-\infty,\infty]$, the function $f1_{A}+g1_{A^{c}}$ takes the same values
as $f$ on $A\subset X$ and the same values as $g$ on $A^{c}$.

######  Definition 2.1.

A _monetary risk measure_ 555The definition provided here is in line with
actuarial science literature, which differs slightly from the one used in the
mathematical finance literature (see e.g. [4]), where $\rho$ is a monetary
risk measure if $\phi(f)=\rho(-f)$ satisfies (N),(M), and (T) above. is a
functional $\phi\colon L^{\infty}\to\mathbb{R}$ satisfying the axioms:

  * (N)

_Normalization_ : $\phi(0)=0$,

  * (M)

_Monotonicity_ : $f\leq g$ implies $\phi(f)\leq\phi(g)$,

  * (T)

_Translation invariance_ : $\phi(f+c)=\phi(f)+c$ for all $c\in\mathbb{R}$.

Next, we recall some standard terminology and concepts in risk analysis that
will be used in this paper. Let $\phi\colon L^{\infty}\to\mathbb{R}$ be a
monetary risk measure.

  * •

The _acceptance set_ of $\phi$ is defined by $\mathcal{A}:=\\{f\in
L^{\infty}\colon\phi(f)\leq 0\\}$.

  * •

We say that $\phi$ is _continuous from below_ if

$f_{n}\uparrow f\mbox{ a.s. in }L^{\infty}\quad\mbox{ implies
}\quad\phi(f_{n})\uparrow\phi(f).$

  * •

We say that $\phi$ is _convex_ if $\phi(\lambda
f+(1-\lambda)g)\leq\lambda\phi(f)+(1-\lambda)\phi(g)$ for all $f,g\in
L^{\infty}$ and $\lambda\in(0,1)$.

  * •

$\phi$ is _law-invariant_ if $\phi(f)=\phi(g)$ whenever $f$ and $g$ have the
same distribution under $\mu$.

The _convex conjugate_ $\phi^{\ast}\colon\mathscr{M}_{1}\to[0,\infty]$ of
$\phi$ is defined by

$\phi^{\ast}(\nu):=\underset{f\in L^{\infty}}{\sup}\left\\{\int
fd\nu-\phi(f)\right\\}=\sup_{f\in\mathcal{A}}\int fd\nu.$

The continuity from below of a convex risk measure $\phi$ is equivalent to its
lower semicontinuity in the weak-$\ast$ topology $\sigma(L^{\infty},L^{1})$
and implies the following representation

$\phi(f)=\underset{\nu\in\mathscr{M}_{1}}{\sup}\left\\{\int
fd\nu-\phi^{\ast}(\nu)\right\\}\quad\mbox{ for all }f\in L^{\infty};$

see [4, Theorem 4.33].666We use the terminology of [4] up to a sign change,
then continuity from above in [4] corresponds to continuity from below in the
present paper.

###  2.2. Characterisation result

The concept of of maxitive monetary risk measure was studied in [9]. This
notion is adapted to the present setting as follows.

######  Definition 2.2.

We say that a monetary risk measure $\phi$ is _maxitive_ if $\phi(f\vee
g)\leq\phi(f)\vee\phi(g)$ for all $f,g\in L^{\infty}$.777Here, $(f\vee
g)(x)=\max\\{f(x),g(x)\\}$ for all $x\in X$.

Notice that, by monotonicity, a maxitive monetary risk measure $\phi$
satisfies $\phi(f\vee g)=\phi(f)\vee\phi(g)$ for all $f,g\in L^{\infty}$. In
addition, every maxitive monetary risk measure $\phi$ is convex; see [9,
Proposition 2.1]. We call _penalty_ to any measurable function $I\colon
X\to[0,\infty]$. We denote by
$\mathcal{I}:=L^{0}(X,\mathcal{X},\mu;[0,\infty])$ the set of all penalties.
Given $I\in\mathcal{I}$ such that ${\rm ess.inf}_{X}I=0$, the _penalized
worst-case_ risk measure is defined as

$\phi^{\max}_{I}(f):={\rm ess.sup}_{X}(f-I).$

It is straightforward to verify that $\phi^{\max}_{I}$ is a maxitive monetary
risk measure. In the following we characterise all monetary risk measures
$\phi$ such that there exists some penalty $I\in\mathcal{I}$ so that

$\phi(f)=\phi^{\max}_{I}(f)\quad\mbox{ for all }f\in L^{\infty}.$

In that case, we say that $\phi$ satisfies the _Laplace principle_ with
penalty $I$.888In large deviations theory one usually deals with a _rate
function_ , which is lower semicontinuous by definition. We use instead the
term ‘penalty’ because $I$ is not in general lower semicontinuous.  Next, we
state the main result of this section. We postpone the proof to the next
section.

######  Theorem 2.3.

Let $\phi\colon L^{\infty}\to\mathbb{R}$ be a monetary risk measure, and
define $J\colon\mathcal{X}\to[-\infty,0]$ by

$J_{A}=\underset{r<0}{\inf}\phi(r1_{A^{c}}).$ (2.1)

The following conditions are equivalent.

  * (i)

$\phi$ is maxitive and continuous from below.

  * (ii)

$f\vee g,h\in\mathcal{A}$ whenever $f,g\in\mathcal{A}$,
$(h_{n})_{n\in\mathbb{N}}\subset\mathcal{A}$, and $h_{n}\uparrow h$ a.s..

  * (iii)

There exists $I\in\mathcal{I}$ such that

$\mathcal{A}=\\{f\in L^{\infty}\colon f\leq I\\}.$

  * (iv)

$\phi$ is convex, continuous from below, and there exists $I\in\mathcal{I}$
such that

$\phi^{\ast}(\nu)=\int Id\nu\quad\mbox{ for all }\nu\in\mathscr{M}_{1}.$

  * (v)

$\phi$ is maxitive and there exists $I\in\mathcal{I}$ such that

$J_{A}=-{{\rm ess.inf}}_{A}I\quad\mbox{ for all }A\in\mathcal{X}.$ (2.2)

  * (vi)

$\phi$ satisfies the Laplace principle for some penalty $I\in\mathcal{I}$,
i.e.

$\phi(f)={{\rm ess.sup}}_{X}(f-I)\quad\mbox{ for all }f\in L^{\infty}.$ (2.3)

Moreover, if one of the equivalent properties (iii)–(vi) is satisfied for
$I\in\mathcal{I}$, then one necessarily has

$I=\underset{f\in L^{\infty}}{\rm ess.sup}\\{f-\phi(f)\\}=-{\rm
ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}.$ (2.4)

Given a sequence $(\nu_{n})_{n\in\mathbb{N}}$ of probability measures in
$\mathscr{M}_{1}$, we define the (upper) _asymptotic entropy_ by

$\psi(f)=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\int e^{nf}d\nu_{n}.$

We have that $\psi$ is maxitive (see Proposition 6.5 below), and that

$J_{A}:=\underset{r<0}{\inf}\psi(r1_{A^{c}})=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\nu_{n}(A),$

(see Lemma 6.3 below; see also [9] in the case that $X$ is a topological space
and $\mathcal{X}=\mathcal{B}(X)$). Below we collect some simple examples where
the asymptotic entropy admits the Laplace principle (2.3).

######  Examples 2.4.

  1. (1)

Suppose that $\nu_{n}$ is the exponential distribution of parameter $n$, i.e.
$\nu_{n}(A)=\int_{A}ne^{-nx}dx$ for all $A\in\mathcal{B}([0,\infty))$. As an
underlying reference probability measure $\mu$ we choose any probability
measure with the same null sets as the Lebesgue measure on $[0,\infty)$. For
each $f\in L^{\infty}$,

$\displaystyle\psi(f)$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\int
e^{nf}d\nu_{n}$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\left(n\int
e^{n(f(x)-x)}dx\right)$
$\displaystyle=\underset{n\to\infty}{\lim}\tfrac{1}{n}\log\left(\int
e^{n(f(x)-x)}dx\right)$ $\displaystyle={\rm
ess.sup}_{x\in[0,\infty)}(f(x)-x),$

where we have applied in the last equality [11, Lemma 2.11]. Therefore, we
obtain (2.3) for the penalty $I$ defined by $I(x)=x$ for a.e.
$x\in[0,\infty)$.

  2. (2)

Suppose that $\xi_{1},\xi_{2},\xi_{3},\ldots$ are independent standard normal
random variables. For each $n$, consider the sample mean
$\bar{\xi}_{n}=\tfrac{1}{n}(\xi_{1}+\xi_{2}+\ldots+\xi_{n})$, which has normal
distribution $N(0,\tfrac{1}{n})$, i.e.

$\nu_{n}(A):=\mathbb{P}(\bar{\xi}_{n}\in
A)=\sqrt{\tfrac{n}{2\pi}}\int_{A}e^{-\tfrac{x^{2}n}{2}}dx.$

Consider as an underlying probability measure $\mu$ any probability measure
with the same null sets as the Lebesgue measure on $\mathbb{R}$. Then, for
each $f\in L^{\infty}$,

$\displaystyle\psi(f)$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\int
e^{nf}d\nu_{n}$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\left(\sqrt{\tfrac{n}{2\pi}}\int_{-\infty}^{\infty}e^{n(f-\tfrac{x^{2}}{2})}dx\right)$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\left(\int_{-\infty}^{\infty}e^{n(f-\tfrac{x^{2}}{2})}dx\right)$
$\displaystyle={\rm ess.sup}_{x\in\mathbb{R}}(f(x)-\tfrac{x^{2}}{2}),$

where we have applied in the last equality [11, Lemma 2.11]. Thus, we obtain
(2.3) for the penalty $I$ given by $I(x)=x^{2}/2$ for a.e. $x\in\mathbb{R}$.

######  Remark 2.5.

Suppose that the asymptotic entropy $\psi$ is continuous from below. Then,
since $\psi$ is also maxitive, it satisfies any of the equivalent conditions
in Theorem 2.3. In particular, due to (v) in Theorem 2.3 we obtain the
following upper large deviation estimate

$\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\nu_{n}(A)=-{\rm
ess.inf}_{A}I\quad\mbox{ for all }A\in\mathcal{X}.$

Suppose, moreover, that $X$ is a topological space and
$\mathcal{X}=\mathcal{B}(X)$ is the Borel $\sigma$-field. Since $\inf_{x\in
A}I(x)\leq{\rm ess.inf}_{A}I$ for $A\subset X$, we get the usual upper bound
in the large deviation principle, i.e.,

$\underset{n\to\infty}{\limsup}\tfrac{1}{n}\log\nu_{n}(A)\leq-\inf_{x\in
A}I(x)\quad\mbox{ for all }A\subset X\mbox{ closed.}$

On the other hand, the lower bound does not necessary hold. However, if the
limit $\lim_{n\to\infty}\tfrac{1}{n}\log\nu_{n}(A)$ exists and $\inf_{x\in
A}I(x)={\rm ess.inf}_{A}I$ for all open set $A\subset X$, then we obtain also
the lower bound

$\underset{n\to\infty}{\liminf}\tfrac{1}{n}\log\nu_{n}(A)\geq-\inf_{x\in
A}I(x)\quad\mbox{ for all }A\subset X\mbox{ open.}$

The equality $\inf_{x\in A}I(x)={\rm ess.inf}_{A}I$ holds for all open set
$A\subset X$ if $I$ has an upper semicontinuous representative and every $x\in
X$ admits a neighborhood base $\mathcal{U}_{x}$ such that $\mu(U)>0$ for all
$U\in\mathcal{U}_{x}$.

As a consequence of Theorem 2.3, we obtain the following result covering the
law-invariant case. We postpone the proof to Section 3.

######  Corollary 2.6.

Let $\phi\colon L^{\infty}\to\mathbb{R}$ be a monetary risk measure and $J$ be
as in (2.1). The following conditions are equivalent.

  * (i)

$J_{A}=0$ for all $A\in\mathcal{X}$ with $\mu(A)>0$,

  * (ii)

$\phi(f)={\rm ess.sup}_{X}f$ for all $f\in L^{\infty}$.

If $\mu$ is atomless and $L^{2}$ is separable, then the conditions (i)-(ii)
are also equivalent to

  * (iii)

$\phi$ is maxitive and law-invariant.

Observe that, alike (v) in Theorem 2.3, in (i) in Corollary 2.6 we do not
assume that $\phi$ is maxitive. Next, we see an example that illustrates
Corollary 2.6.

######  Example 2.7.

The two most popular monetary risk measures in financial practice are VaR and
AVaR. The monetary risk measure VaR at level $\alpha\in(0,1)$ is the
functional ${\rm VaR}_{\mu,\alpha}\colon L^{\infty}\to\mathbb{R}$ defined by

${\rm
VaR}_{\mu,\alpha}(f):=q^{-}_{\mu,\alpha}(f)=\inf\left\\{m\in\mathbb{R}\colon\mu(f\leq
m)\geq\alpha\right\\},$

where $q^{-}_{\mu,\alpha}$ is the left $\alpha$-quantile of the distribution
of $f$ under $\mu$. In addition, the monetary risk measure AVaR at level
$\alpha\in(0,1)$ is the functional ${\rm AVaR}_{\mu,\alpha}\colon
L^{\infty}\to\mathbb{R}$ defined by

${\rm AVaR}_{\mu,\alpha}(f):=\tfrac{1}{1-\alpha}\int_{\alpha}^{1}{\rm
VaR}_{\mu,\lambda}(f)d\lambda.$

We can use Corollary 2.6 to study the asymptotic behavior of AVaR. Suppose
that the sequences $(\nu_{n})_{n\in\mathbb{N}}\subset\mathcal{M}_{1}$, and
$(\alpha_{n})_{n\in\mathbb{N}}\subset(0,1)$ satisfy

$\limsup_{n\to\infty}\tfrac{\nu_{n}(A)}{1-\alpha_{n}}\geq 1\quad\mbox{ for all
}A\in\mathcal{X}\mbox{ with }\mu(A)>0.$ (2.5)

Then, we claim that

$\underset{n\to\infty}{\lim}\>{\rm AVaR}_{\nu_{n},\alpha_{n}}(f)={\rm
ess.sup}_{X}f\quad\mbox{ for all }f\in L^{\infty}.$ (2.6)

Notice that the condition (2.5) is satisfied if $\nu_{n}=\mu$ for all
$n\in\mathbb{N}$ and $\alpha_{n}\uparrow 1$. In that simple case the
conclusion is well-known; see e.g. [6].

To show (2.6), we consider the monetary risk measures

$\overline{\phi}(f):=\underset{n\to\infty}{\limsup}\>{\rm
AVaR}_{\nu_{n},\alpha_{n}}(f),\quad\underline{\phi}(f):=\underset{n\to\infty}{\liminf}\>{\rm
AVaR}_{\nu_{n},\alpha_{n}}(f).$

Fixed $r<0$ and $A\in\mathcal{X}$ with $\mu(A)>0$, we have

$\nu_{n}(r1_{A^{c}}\leq
m)=\left(1-\nu_{n}(A)\right)1_{[r,0)}(m)+1_{[0,\infty)}(m).$

Then, for $\lambda\in(0,1)$, it holds

${\rm
VaR}_{\nu_{n},\lambda}(r1_{A^{c}})=\inf\left\\{m\in\mathbb{R}\colon\nu_{n}(r1_{A^{c}}\leq
m)\geq\lambda\right\\}=r1_{(0,1-\nu_{n}(A)]}(\lambda).$

Thus,

${\rm
AVaR}_{\nu_{n},\alpha_{n}}(r1_{A^{c}})=\tfrac{r}{1-\alpha_{n}}\int_{\alpha_{n}}^{1}1_{(0,1-\nu_{n}(A)]}(\lambda)d\lambda=r\left|1-\tfrac{\nu_{n}(A)}{1-\alpha_{n}}\right|^{+}.$

Therefore, we get

$\underline{\phi}(r1_{A^{c}})=\underset{n\to\infty}{\liminf}{\rm
AVaR}_{\nu_{n},\alpha_{n}}(r1_{A^{c}})=r\underset{n\to\infty}{\liminf}\left|1-\tfrac{\nu_{n}(A)}{1-\alpha_{n}}\right|^{+}=0,$

where the last equality follows from (2.5). Consequently,

$\underline{J}_{A}=\inf_{r<0}\underline{\phi}(r1_{A^{c}})=0.$

Since the equality above holds for every $A\in\mathcal{X}$, by Corollary 2.6
it follows that

${\rm ess.sup}_{X}f=\underline{\phi}(f)\quad\mbox{ for all }f\in L^{\infty}.$

In addition, for every $f\in L^{\infty}$ we have

$\displaystyle\underline{\phi}(f)$
$\displaystyle\leq\overline{\phi}(f)\leq\overline{\phi}({\rm
ess.sup}_{X}f)=\overline{\phi}(0)+{\rm ess.sup}_{X}f={\rm ess.sup}_{X}f.$

Hence, we obtain

${\rm
ess.sup}_{X}f=\underline{\phi}(f)=\overline{\phi}(f)=\underset{n\to\infty}{\lim}\>{\rm
AVaR}_{\nu_{n},\alpha_{n}}(f)\quad\mbox{ for all }f\in L^{\infty},$

and then (2.6) follows.

##  3\. Proof of the characterisation result

In this section we prove Theorem 2.3. Before we proceed with the proof, we
need to collect some preliminary results. Throughout, let $\phi\colon
L^{\infty}\to\mathbb{R}$ be a monetary risk measure with acceptance set
$\mathcal{A}$. We define the _minimal penalty_ $I_{\min}$ associated with
$\phi$ by

${I_{\min}}=\underset{f\in L^{\infty}}{{\rm
ess.sup}}\\{f-\phi(f)\\}=\underset{f\in\mathcal{A}}{{\rm ess.sup}}\>f.$

######  Lemma 3.1.

Let $I\in\mathcal{I}$. Then,

$\phi(f)\geq{\rm ess.sup}_{X}(f-I)\quad\mbox{ for all }f\in L^{\infty}$ (3.1)

if and only if $I\geq I_{\min}$.

###### Proof.

If (3.1) holds, then

$I\geq f-\phi(f)\quad\mbox{ for all }f\in L^{\infty}.$

Taking the essential infimum over all $f\in L^{\infty}$, we obtain that $I\geq
I_{\min}$. Conversely, suppose that $I\geq I_{\min}$. Given $f\in L^{\infty}$,
it follows that $\phi(f)\geq f-I_{\min}\geq f-I$, and taking the essential
infimum on $X$ we obtain (3.1). ∎

We define, as in (2.1), the _concentration function_
$J\colon\mathcal{X}\to[-\infty,0]$ by

$J_{A}:=\underset{r<0}{\inf}\phi(r1_{A^{c}}).$

The proof of the following lemma is easy to find and therefore omitted.

######  Lemma 3.2.

Suppose that $A,B\in\mathcal{X}$. The following holds true:

  1. (1)

$J_{A}=-\infty$ if $\mu(A)=0$,

  2. (2)

$J_{A}=0$ if $\mu(A)=1$,

  3. (3)

$J_{A}\leq J_{B}$ whenever $A\subset B$,

  4. (4)

If $\phi$ is maxitive, then $J_{A\cup B}\leq J_{A}\vee J_{B}$.

Both $I_{\min}$ and $J$ are related by means of the following.

######  Proposition 3.3.

$I_{\min}$ has the representation

$-I_{\min}={{\rm ess.inf}}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}.$

In particular, $J_{A}\geq-{\rm ess.inf}_{A}I_{\min}$ for all
$A\in\mathcal{X}$.999For $\mu(A)=0$, by convention we have ${\rm
ess.inf}_{A}I_{\min}=\infty$.

###### Proof.

Fix $A\in\mathcal{X}$ and $M\in\mathbb{N}$. By definition of $I_{\min}$ we
have

$I_{\min}\geq-M1_{A^{c}}-\phi(-M1_{A^{c}}).$

Thus, multiplying101010Recall that $(\pm\infty)0=0$ by convention. by
$-1_{A}$,

$-I_{\min}1_{A}\leq\phi(-M1_{A^{c}})1_{A}.$

Letting $M\to\infty$, we have

$-I_{\min}1_{A}\leq J_{A}1_{A}.$

Since $-I_{\min}\leq 0$, it follows that $-I_{\min}\leq J_{A}1_{A}$. Taking
the essential infimum over all $A\in\mathcal{X}$, we obtain

$-I_{\min}\leq{\rm ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}.$

To prove the other inequality, fix $f\in L^{\infty}$. Suppose that $(s_{n})$
is a sequence of simple functions so that $s_{n}\uparrow f$. For a fixed
$n\in\mathbb{N}$, we have that

$s_{n}=\sum_{k=1}^{m}a_{k}1_{A_{k}}.$

for a finite partition $(A_{k})\subset\mathcal{X}$ of $X$. Given
$r\in\mathbb{R}$ with $r<-\|f\|_{\infty}$, for each $k\in\\{1,2,\ldots,m\\}$
we have

$\phi(f)\geq\phi(a_{k}1_{A_{k}}+r1_{A_{k}^{c}})=a_{k}+\phi((r-a_{k})1_{A_{k}^{c}}).$

Letting $r\to-\infty$, we obtain

$\phi(f)\geq a_{k}+J_{A_{k}}\quad\mbox{ for all }k.$

Multiplying by $1_{A_{k}}$ and summing over all $k$, we get

$\phi(f)\geq s_{n}+\sum_{k=1}^{m}J_{A_{k}}1_{A_{k}}\geq s_{n}+{\rm
ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}.$

Letting $n\to\infty$ results in

$-{\rm ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}\geq f-\phi(f).$

Finally, taking the essential supremum over all $f\in L^{\infty}$, we obtain

$-{\rm ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}\geq I_{\min}.$

This completes the proof. ∎

######  Lemma 3.4.

If $I\in\mathcal{I}$, then $I={\rm ess.sup}\Big{\\{}({\rm
ess.inf}_{A}I)1_{A}\colon A\in\mathcal{X}\Big{\\}}$.

###### Proof.

Suppose that $(s_{n})$ is a sequence of non-negative simple functions such
that $s_{n}\uparrow I$ a.s.. For fixed $n\in\mathbb{N}$, we have that

$s_{n}=\sum_{k}a_{k}1_{A_{k}}$

for a finite partition $(A_{k})\subset\mathcal{X}$ of $X$ and $a_{k}\geq 0$.
Then,

$s_{n}\leq\sum_{k}({{\rm ess.inf}_{A_{k}}}I)1_{A_{k}}\leq{\rm
ess.sup}\Big{\\{}({\rm ess.inf}_{A}I)1_{A}\colon A\in\mathcal{X}\Big{\\}}.$

Letting $n\to\infty$, we get

$I\leq{\rm ess.sup}\Big{\\{}({\rm ess.inf}_{A}I)1_{A}\colon
A\in\mathcal{X}\Big{\\}}\leq I,$

and the assertion follows. ∎

######  Proposition 3.5.

Suppose that $I\in\mathcal{I}$. The following holds true:

  1. (1)

If $J_{A}\leq-{\rm ess.inf}_{A}I$ for all $A\in\mathcal{X}$, then
$I_{\min}\geq I$.

  2. (2)

If $J_{A}\geq-{\rm ess.inf}_{A}I$ for all $A\in\mathcal{X}$, then
$I_{\min}\leq I$.

###### Proof.

Suppose for instance that $J_{A}\leq-{\rm ess.inf}_{A}I$ for all
$A\in\mathcal{X}$. Then, by Lemma 3.4 we have

$\displaystyle I$ $\displaystyle={\rm ess.sup}\Big{\\{}({\rm
ess.inf}_{A}I)1_{A}\colon A\in\mathcal{X}\Big{\\}}$ $\displaystyle{\leq}{\rm
ess.sup}\Big{\\{}-J_{A}1_{A}\colon A\in\mathcal{X}\Big{\\}}$
$\displaystyle=-{{\rm ess.inf}}\Big{\\{}J_{A}1_{A}\colon
A\in\mathcal{X}\Big{\\}}$ $\displaystyle=I_{\min},$

where the last equality follows from Proposition 3.3. ∎

Now, we turn to the proof of Theorem 2.3.

###### Proof.

The proofs of $(i)\Leftrightarrow(ii)$ and $(iii)\Rightarrow(ii)$ are easy to
find, and therefore omitted.

$(v)\Rightarrow(vi)$: To show (2.3), we first assume that $f\in L^{\infty}$ is
a simple function, that is, $f=\sum_{i=1}^{N}a_{i}1_{\\{f=a_{i}\\}}$ with
$a_{1}<a_{2}<\ldots<a_{N}$. In that case,

$f=\underset{1\leq i\leq N}{\bigvee}\left(a_{i}1_{\\{a_{i}\geq
f>a_{i-1}\\}}+r1_{\\{a_{i}\geq f>a_{i-1}\\}^{c}}\right)$

for $r<-\|f\|_{\infty}$, where we define $a_{0}=-\infty$. Since $\phi$ is
maxitive, we have

$\phi(f)=\underset{1\leq i\leq N}{\bigvee}\phi(a_{i}1_{\\{a_{i}\geq
f>a_{i-1}\\}}+r1_{\\{a_{i}\geq f>a_{i-1}\\}^{c}})=\underset{1\leq i\leq
N}{\bigvee}\Big{(}a_{i}+\phi\left((r-a_{i})1_{\\{a_{i}\geq
f>a_{i-1}\\}^{c}}\right)\Big{)}.$

Letting $r\to-\infty$, we get

$\displaystyle\phi(f)$ $\displaystyle=\underset{1\leq i\leq
N}{\bigvee}\left(a_{i}+J_{\\{a_{i}\geq f>a_{i-1}\\}}\right)$
$\displaystyle=\underset{1\leq i\leq N}{\bigvee}\left(a_{i}-{\rm
ess.inf}_{\\{a_{i}\geq f>a_{i-1}\\}}I\right)$ $\displaystyle=\underset{1\leq
i\leq N}{\bigvee}{\rm ess.sup}_{\\{a_{i}\geq
f>a_{i-1}\\}}\left(a_{i}-I\right)$ $\displaystyle={\rm
ess.sup}_{X}\left(f-I\right).$

Finally, for general $f\in L^{\infty}$, we take a sequence $(s_{n})$ of simple
functions such that $\|s_{n}-f\|_{\infty}\to 0$. Every monetary risk measure
on $L^{\infty}$ is Lipschitz continuous; see [4, Lemma 4.3]. Then, since
$\phi_{I}^{\max}(s_{n})=\phi(s_{n})$, we have111111Recall that
$\phi_{I}^{\max}(f)={\rm ess.sup}_{X}(f-I)$, by definition.

$|\phi(f)-\phi_{I}^{\max}(f)|\leq|\phi(f)-\phi(s_{n})|+|\phi_{I}^{\max}(s_{n})-\phi_{I}^{\max}(f)|\leq\|f-s_{n}\|_{\infty}+\|f-s_{n}\|_{\infty}.$

Letting $n\to\infty$, we get

$\phi(f)=\phi_{I}^{\max}(f)={\rm ess.sup}_{X}\left(f-I\right),$

This proves (vi).

$(vi)\Rightarrow(v)$: If (vi) holds, then $\phi$ is clearly maxitive. Fix
$A\in\mathcal{X}$. For every $r<0$ we have

$\phi(r1_{A^{c}})={\rm ess.sup}_{X}(r1_{A^{c}}-I)\geq{\rm ess.sup}_{X}(-\infty
1_{A^{c}}-I)=-{\rm ess.inf}_{A}I.$

Letting $r\to-\infty$ it follows that $J_{A}\geq-{\rm ess.inf}_{A}I$. To prove
the other inequality, fix $M>0$, and set $\alpha:=({\rm ess.inf}_{A}I)\wedge
M\in[0,\infty)$. We have that $\mu\left(A\cap\\{I<\alpha\\}\right)=0$.
Consequently, given $r<0$, it holds

$r1_{A^{c}}-I\leq r\quad\mbox{ on }\\{I<\alpha\\}.$

Moreover, since $r1_{A^{c}}\leq 0$, we have

$r1_{A^{c}}-I\leq-\alpha\quad\mbox{ on }\\{I\geq\alpha\\}.$

Choosing $r<-\alpha$, it follows that $r1_{A^{c}}-I\leq-\alpha$. Hence,

$J_{A}\leq\phi(r1_{A^{c}})={\rm ess.sup}_{X}(r1_{A^{c}}-I){\leq}-\alpha=(-{\rm
ess.inf}_{A}I)\vee-M.$

Letting $M\to\infty$, we obtain $J_{A}\leq-{\rm ess.inf}_{A}I$, and (v)
follows.

$(i)\Rightarrow(iv)$: Since $\phi$ is maxitive, then $\phi$ is convex; see [9,
Proposition 2.1]. Since $\phi$ is maxitive, we have that $f\vee
g\in\mathcal{A}$ for all $f,g\in\mathcal{A}$. Moreover, by definition of
$I_{\min}$ we have that $I_{\min}={\rm ess.sup}\mathcal{A}$. Then, it follows
from Theorem A.1 that there exists a sequence $f_{1}\leq f_{2}\leq\ldots$ in
$\mathcal{A}$ such that $f_{n}\uparrow I_{\min}$ a.s.. Fix
$\nu\in\mathscr{M}_{1}$. Since $\nu\ll\mu$, we have that $f_{n}\uparrow
I_{\min}$ $\nu$-a.s.. Hence, by monotone convergence we have

$\displaystyle\phi^{\ast}(\nu)$ $\displaystyle=\sup_{f\in\mathcal{A}}\int
fd{\nu}\geq\sup_{n}\int f_{n}d{\nu}=\int I_{\min}{d\nu}.$

On the other hand, since $f\leq I_{\min}$ ($\mu$-a.s. and $\nu$-a.s.) for all
$f\in\mathcal{A}$, we have

$\displaystyle\phi^{\ast}(\nu)$ $\displaystyle=\sup_{f\in\mathcal{A}}\int
f{d\nu}\leq\int I_{\min}{d\nu}.$

This proves (iv).

$(iv)\Rightarrow(vi)$: Since $\phi$ is convex and continuous from
below121212Recall that the definition of monetary risk measure is taken from
[4] up to a sign change. Thus, continuity from below, as defined here,
corresponds to continuity from above in [4]. , by [4, Theorem 4.33] one has

$\displaystyle\phi(f)$
$\displaystyle=\underset{\nu\in\mathscr{M}_{1}}{\sup}\left\\{\int
fd\nu-\phi^{\ast}(\nu)\right\\}$
$\displaystyle=\underset{\nu\in\mathscr{M}_{1}}{\sup}\left\\{\int fd\nu-\int
Id\nu\right\\}=\underset{\nu\in\mathscr{M}_{1}}{\sup}\left\\{\int(f-I)d\nu\right\\}$
$\displaystyle={\rm ess.sup}_{X}(f-I),$

where the last equality follows from Proposition A.2. This proves (vi).

$(vi)\Rightarrow(iii)$: From the representation (2.3) we obtain

$\mathcal{A}=\left\\{f\in L^{\infty}\colon f\leq I\right\\}.$

Finally we argue that if any of the properties (iii)–(vi) is satisfied for
$I\in\mathcal{I}$, then $I$ verifies (2.4). First, if $J$ satisfies (v) for
$I\in\mathcal{I}$, then $I=I_{\min}$ by Proposition 3.5. Hence (2.4) follows
in view of Proposition 3.3. Second, if (vi) is satisfied for
$I\in\mathcal{I}$, then we have shown above that (v) is satisfied for the same
$I$, hence again $I=I_{\min}$ by Proposition 3.5. Third, if (iv) is satisfied
for $I\in\mathcal{I}$, we have proven above that (vi) (and in turn (v)) is
satisfied for the same $I$, obtaining the same conclusion. Finally, if (iii)
is satisfied for $I\in\mathcal{I}$, then necessarily

$I={\rm ess.sup}\mathcal{A}=I_{\min}.$

This completes the proof of Theorem 2.3. ∎

Finally, we prove Corollary 2.6.

###### Proof.

$(i)\Rightarrow(ii)$: By monotonicity and translation invariance, we have that

$\phi(f)\leq\phi\left({\rm ess.sup}_{X}f\right)=\phi(0)+{\rm
ess.sup}_{X}f={\rm ess.sup}_{X}f.$

On the other hand, if $J_{A}=0$ for all $A\in\mathcal{X}$ with $\mu(A)>0$, by
Proposition 3.3 it follows that

$I_{\min}=-{\rm ess.inf}\\{J_{A}1_{A}\colon A\in\mathcal{X}\\}=0,$

where above we have applied that, if $\mu(A)=0$, then $J_{A}1_{A}=-\infty\cdot
0=0$ by convention. Then,

${\rm ess.sup}_{X}f={\rm ess.sup}_{X}(f-I_{\min})\leq\phi(f),$

and then (ii) follows.

$(ii)\Rightarrow(i)$: Using the definition of $I_{\min}$, we have

$0\leq I_{\min}=\underset{f\in L^{\infty}}{{\rm
ess.sup}}\\{f-\phi(f)\\}=\underset{f\in L^{\infty}}{{\rm ess.sup}}\\{f-{\rm
ess.sup}_{X}f\\}\leq 0.$

Hence, $I_{\min}=0$. Then, given $A\in\mathcal{X}$ with $\mu(A)>0$, we obtain

$J_{A}=-{\rm ess.inf}_{A}I_{\min}=0,$

by the implication $(vi)\Rightarrow(v)$ in Theorem 2.3.

$(ii)\Rightarrow(iii)$: It is clear. In fact, we need to assume neither that
$\mu$ is atomless nor that $L^{2}$ is separable.

$(iii)\Rightarrow(ii)$: We assume that $\mu$ is atomless and $L^{2}$ is
separable. Suppose that $\phi$ is maxitive and law-invariant. Since $\phi$ is
law-invariant, $\mu$ is atomless and $L^{2}$ is separable, due to [4, Remark
4.64] (this result is due to [7]), we have that $\phi$ is continuous from
below. Then, by Theorem 2.3 we have

$\phi(f)={\rm ess.sup}_{X}(f-I_{\min})\quad\mbox{ for all }f\in L^{\infty}.$

It suffices to prove that $I_{\min}=0$. By contradiction, suppose that
$\mu(I_{\min}>0)>0$. We have

$0=-\phi(0)=-{\rm ess.sup}_{X}(-I)={\rm ess.inf}_{X}I.$

Hence, we can find rational numbers $q_{1},q_{2}\in\mathbb{Q}$,
$0<q_{1}<q_{2}$, such that

$\mu(I_{\min}>q_{2})>0\quad\mbox{ and }\quad\mu(q_{1}>I_{\min})>0.$

Take $\varepsilon>0$ so that $\varepsilon<\mu(I_{\min}>q_{2})$ and
$\varepsilon<\mu(q_{1}>I_{\min})$. Since $\mu$ is atomless, we can find
$A,B\in\mathcal{X}$ such that $A\subset\\{I_{\min}>q_{2}\\}$,
$B\subset\\{I_{\min}<q_{1}\\}$, and $\mu(A)=\mu(B)=\varepsilon$. Take
$q\in(q_{1},q_{2})$. We have that

$\phi(q1_{A})={\rm ess.sup}_{X}(q1_{A}-I_{\min})\leq{\rm
ess.sup}_{X}((q-q_{2})1_{A}-I_{\min}1_{A^{c}})\leq 0.$

On the other hand,

$\phi(q1_{B})={\rm ess.sup}_{X}(q1_{B}-I_{\min})\geq{\rm
ess.sup}_{X}((q-q_{1})1_{A}-I_{\min}1_{A^{c}})=q-q_{1}>0.$

Since $\phi$ is law-invariant, and $q1_{A}$ and $q1_{B}$ have the same law
(recall that $\mu(A)=\mu(B)$), we have

$0\geq\phi(q1_{A})=\phi(q1_{B})>0,$

a contradiction. ∎

##  4\. Tightness

In this section we consider a monetary risk measure $\phi\colon
L^{\infty}\to\mathbb{R}$. We are interested in finding local properties on
$\phi$ that guarantee the representation (2.3) characterised in Theorem 2.3.

######  Definition 4.1.

We say that $\phi$ is _tight_ if for all $M>0$ there exists $A\in\mathcal{X}$
such that

  1. (1)

$J_{A^{c}}<-M$,

  2. (2)

$J_{B}\leq\frac{1}{M}-{\rm ess.inf}_{B}I_{\min}$ for all $B\subset A$,
$B\in\mathcal{X}$.

######  Theorem 4.2.

Let $\phi\colon L^{\infty}\to\mathbb{R}$ be a maxitive monetary risk measure.
If $\phi$ is tight, then

$\phi(f)={\rm ess.sup}_{X}(f-I_{\min})\quad\mbox{ for all }f\in L^{\infty}.$

###### Proof.

Due to Theorem 2.3 it is enough to prove that $J_{B}=-{\rm ess.inf}I_{\min}$
for all $B\in\mathcal{X}$. Fixed $B\in\mathcal{X}$, we have that
$J_{B}\geq-{\rm ess.inf}I_{\min}$ due to Proposition 3.3. As for the other
inequality, fix $M>0$ and take $A\in\mathcal{X}$ as in Definition 4.1. Then,
since $\phi$ is maxitive, we have

$\displaystyle J_{B}$ $\displaystyle\leq J_{A\cap B}\vee J_{A^{c}}$
$\displaystyle\leq\left(M^{-1}-{\rm ess.inf}_{A\cap B}I_{\min}\right)\vee(-M)$
$\displaystyle\leq\left(M^{-1}-{\rm ess.inf}_{B}I_{\min}\right)\vee(-M).$

Letting $M\to\infty$ yields $J_{B}\leq-{\rm ess.inf}_{B}I_{\min}$, which
completes the proof. ∎

######  Lemma 4.3.

Suppose $A\in\mathcal{X}$ satisfies that $J_{B}=-{\rm ess.inf}_{B}I$ for all
$B\subset A$, $B\in\mathcal{X}$. Then $I1_{A}=I_{\min}1_{A}$.

###### Proof.

Define $\tilde{I}=I1_{A}+\infty 1_{A^{c}}$. For every $C\in\mathcal{X}$, it
holds

$J_{C}\geq J_{A\cap C}=-{\rm ess.inf}_{A\cap C}I=-{\rm ess.inf}_{C}\tilde{I}.$

Then, it follows by Proposition 3.5 that $\tilde{I}\geq I_{\min}$. Since
$\tilde{I}1_{A}=I1_{A}$, we have that $I1_{A}\geq I_{\min}1_{A}$. On the other
hand, due to Proposition 3.3 we have

$\displaystyle I_{\min}$ $\displaystyle={\rm ess.sup}\\{-J_{B}1_{B}\colon
B\in\mathcal{X}\\}$ $\displaystyle\geq{\rm ess.sup}\\{-J_{B}1_{B}\colon
B\in\mathcal{X},\>B\subset A\\}$ $\displaystyle={\rm ess.sup}\\{({\rm
ess.inf}_{B}I)1_{B}\colon B\in\mathcal{X},\>B\subset A\\}$ $\displaystyle={\rm
ess.sup}\\{({\rm ess.inf}_{B}\tilde{I})1_{A\cap B}\colon B\in\mathcal{X}\\}$
$\displaystyle={\rm ess.sup}\\{({\rm ess.inf}_{B}\tilde{I})1_{B}1_{A}\colon
B\in\mathcal{X}\\}$ $\displaystyle=1_{A}\cdot{\rm ess.sup}\\{({\rm
ess.inf}_{B}\tilde{I})1_{B}\colon B\in\mathcal{X}\\}$
$\displaystyle=1_{A}\tilde{I},$

where the last equality follows from Lemma 3.4. Then, it follows that
$I1_{A}\leq I_{\min}1_{A}$, and the assertion follows. ∎

We have the following refinement of Theorem 4.2, which allows for an explicit
representation of the minimal penalty $I_{\min}$.

######  Corollary 4.4.

Suppose that $\phi$ is a maxitive monetary risk measure. Let
$(A_{k})_{k\in\mathbb{N}}$ be a sequence of elements of $\mathcal{X}$ such
that

  1. (1)

$A_{k}\uparrow X$ as $k\to\infty$.

  2. (2)

$J_{A_{k}^{c}}\downarrow-\infty$ as $k\to\infty$.

  3. (3)

For each $k\in\mathbb{N}$ there exists $I_{k}\in\mathcal{I}$ such that

$J_{B}=-{\rm ess.inf}_{B}I_{k}\quad\mbox{ for all }B\subset
A,\>B\in\mathcal{X}.$

Then $I_{k+1}1_{A_{k}}=I_{k}1_{A_{k}}$ for all $k\in\mathbb{N}$, and

$\phi(f)={\rm ess.sup}_{X}(f-I)\quad\mbox{ for all }f\in L^{\infty},$

where $I=\sum_{k=1}^{\infty}I_{k}1_{B_{k}}$, $B_{1}=A_{1}$ and
$B_{k}=A_{k}\setminus A_{k-1}$ for $k>1$.

###### Proof.

Lemma 4.3 together with the condition (3) implies that $I_{k}=I_{\min}$ on
$A_{k}$. Then, the condition (3) is rewritten as $J_{B}=-{\rm
ess.inf}_{B}I_{\min}$ for all $B\subset A$. This, along with the condition (2)
implies that $\phi$ is tight. Hence, due to Theorem 4.2 we have

$\phi(f)={\rm ess.sup}_{X}(f-I_{\min})\quad\mbox{ for all }f\in L^{\infty}.$

Finally, since $I_{k}=I_{\min}$ on $B_{k}$, we get
$I_{\min}=\sum_{k=1}^{\infty}I_{\min}1_{B_{k}}=\sum_{k=1}^{\infty}I_{k}1_{B_{k}}$.
∎

##  5\. Representation on unbounded functions

In this section we extend to unbounded functions the Laplace principle (2.3)
characterised in Theorem 2.3. Throughout, we consider a monetary measure
131313i.e., for all $f,g\in L^{0}$ $\phi$ satisfies: (N) $\phi(0)=0$, (M)
$\phi(f)\leq\phi(g)$ whenever $f\leq g$, and (T) $\phi(f+c)=\phi(f)+c$.

$\phi\colon L^{0}\to[-\infty,\infty]$

defined on the space $L^{0}$ of all (classes of equivalence of) real-valued
measurable functions on $X$, and consider the minimal penalty $I_{\min}$ of
its restriction $\phi|_{L^{\infty}}$ on $L^{\infty}$, i.e.

$I_{\min}=\underset{f\in L^{\infty}}{{\rm ess.sup}}\\{f-\phi(f)\\}.$

######  Remark 5.1.

Every monetary risk measure $\phi_{0}$ on $L^{\infty}$ can be extended to a
monetary risk measure on $L^{0}$ by setting

$\bar{\phi}_{0}(f):=\underset{N\to\infty}{\lim}\underset{M\to\infty}{\lim}\phi_{0}((f\wedge
N)\vee(-M))\in[-\infty,\infty].$

Moreover, if $\phi_{0}$ is maxitive, then $\bar{\phi}_{0}$ is maxitive as
well.

We define the space

$L^{\phi}:=\\{f\in L^{0}\colon\mbox{ there exists }t>1\mbox{ such that
}\phi(tf)<\infty\\}.$

Similar spaces are introduced also in [14], in a topological context. Note
that $L^{\infty}\subset L^{\phi}$.

######  Lemma 5.2.

If $f\in L^{\phi}$, then

$\underset{M\to\infty}{\lim}\underset{h\in L^{0}_{+}}{\inf}\phi(f1_{\\{f\geq
M\\}}-h1_{\\{f<M\\}})=-\infty,$

where $L^{0}_{+}=\\{f\in L^{0}\colon f\geq 0\\}$.

###### Proof.

Suppose that $f\in L^{\infty}$ and take $t>1$ such that $\phi(tf)<\infty$· Fix
$M\in\mathbb{N}$ and define $g=\exp(f-M)$. Given $h\in L^{0}_{+}$ with
$-h\leq\log(g^{t})$, we have

$\displaystyle-M+\phi(f1_{\\{f\geq M\\}}-h1_{\\{f<M\\}})$
$\displaystyle=\phi(-M+f1_{\\{f\geq M\\}}-h1_{\\{f<M\\}})$
$\displaystyle=\phi(\log(g)1_{\\{g\geq 1\\}}-h1_{\\{g<1\\}})$
$\displaystyle\leq\phi(\log(g^{t}))$ $\displaystyle=\phi(t(f-M))$
$\displaystyle=-Mt+\phi(tf).$

Taking the infimum over all $h\in L^{0}_{+}$ with $-h\leq\log(g^{t})$, we have

$\underset{h\in L^{0}_{+}}{\inf}\phi(f1_{\\{f\geq
M\\}}-h1_{\\{f<M\\}})\leq(1-t)M+\phi(tf).$

Since $t>1$ and $\phi(tf)<\infty$, the right hand side converges to $-\infty$
as $M\to\infty$. Then, we obtain the conclusion by letting $M\to\infty$. ∎

######  Theorem 5.3.

Suppose that $\phi\colon L^{0}\to[-\infty,\infty]$ is maxitive141414i.e.,
$\phi(f\vee g)\leq\phi(f)\vee\phi(g)$ for all $f,g\in L^{0}$. and the
restriction $\phi|_{L^{\infty}}$ on $L^{\infty}$ is continuous from below,
then

$\phi(f)={\rm ess.sup}_{X}(f-I_{\min})\quad\mbox{ for all }f\in L^{\phi}.$

###### Proof.

Fix $M,N\in\mathbb{N}$. Since $\phi$ is maxitive, we have that

$-M\vee\phi(f)=\phi((-M)\vee f)\leq\phi\big{(}(-M)\vee(f\wedge
N)\big{)}\vee\left(\underset{h\in L^{0}_{+}}{\inf}\phi(f1_{\\{f\geq
N\\}}-h1_{\\{f<N\\}})\right).$

Since $\phi$ is continuous from below on ${L^{\infty}}$, due to Theorem 2.3 we
have

$\displaystyle(-M)\vee\phi(f)$ $\displaystyle\leq\Big{(}{\rm
ess.sup}_{X}\big{(}(-M)\vee(f\wedge
N)-I_{\min}\big{)}\Big{)}\vee\left(\underset{h\in
L^{0}_{+}}{\inf}\phi(f1_{\\{f\geq N\\}}-h1_{\\{f<N\\}})\right)$
$\displaystyle\leq\Big{(}{\rm ess.sup}_{X}\big{(}(-M)\vee
f-I_{\min}\big{)}\Big{)}\vee\left(\underset{h\in
L^{0}_{+}}{\inf}\phi(f1_{\\{f\geq N\\}}-h1_{\\{f<N\\}})\right).$

Letting $N\to\infty$ it follows by Lemma 5.2 that

$\displaystyle(-M)\vee\phi(f)$ $\displaystyle\leq{\rm
ess.sup}_{X}\big{(}(-M)\vee f-I_{\min}\big{)}$ $\displaystyle=\Big{(}{\rm
ess.sup}_{X}(f-I_{\min})\Big{)}\vee\Big{(}{\rm
ess.sup}_{X}(-M-I_{\min})\Big{)}.$

Now, letting $M\to\infty$, we get

$\phi(f)\leq{\rm ess.sup}_{X}(f-I_{\min}).$

On the other hand, since $\phi$ is maxitive

$\displaystyle(-M)\vee\phi(f)$ $\displaystyle=\phi((-M)\vee f)$
$\displaystyle\geq\phi((-M)\vee(f\wedge N))$ $\displaystyle\geq{\rm
ess.sup}_{X}\big{(}(-M)\vee(f\wedge N)-I_{\min}\big{)}$
$\displaystyle=\Big{(}{\rm
ess.sup}_{X}\big{(}-M-I_{\min}\big{)}\Big{)}\vee\Big{(}{\rm
ess.sup}_{X}\big{(}f\wedge N-I_{\min}\big{)}\Big{)}.$

Letting $M\to\infty$, we get

$\phi(f)\geq{\rm ess.sup}_{X}\big{(}f\wedge N-I_{\min}\big{)}.$

Letting $N\to\infty$, we finally obtain

$\phi(f)\geq{\rm ess.sup}_{X}\big{(}f-I_{\min}\big{)},$

where we have applied that the essential supremum commutes with countable
supremums. This completes the proof. ∎

##  6\. Large deviation results for sublinear expectations

A monetary risk measure $\mathcal{E}\colon L^{\infty}\to\mathbb{R}$ is called
a _sublinear expectation_ (or _coherent risk measure_ , using the language of
risk analysis) if $\mathcal{E}$ is positively homogeneous, i.e.,
$\mathcal{E}(af)=a\mathcal{E}(f)$ for all $a\geq 0$, and sublinear, i.e.,
$\mathcal{E}(f+g)\leq\mathcal{E}(f)+\mathcal{E}(g)$.

######  Example 6.1.

Suppose that $\mathcal{P}$ is a non-empty sets of probability measures in
$\mathscr{M}_{1}$. Then $\mathcal{E}(f):=\sup_{\nu\in\mathcal{P}}\int fd\nu$
is a sublinear expectation.

Throughout, we extend any sublinear expectation $\mathcal{E}$ to $L^{0}_{+}$
by setting
$\mathcal{E}(h)=\lim_{N\to\infty}\mathcal{E}(h\wedge N)$ for $h\in L^{0}_{+}$.
Such an extension is also positively homogeneous and sublinear on $L^{0}_{+}$.
Given a sublinear expectation $\mathcal{E}$ and $\gamma>0$, we define the
entropic risk measure $e_{\mathcal{E},\gamma}$ by

$e_{\mathcal{E},\gamma}(f):=\tfrac{1}{\gamma}\log\mathcal{E}(e^{\gamma f}).$

Given a sequence of sublinear expectations
$(\mathcal{E}_{n})_{n\in\mathbb{N}}$ and a sequence of real numbers
$(\gamma_{n})\subset(0,\infty)$ such that $\gamma_{n}\uparrow\infty$, we
define the upper asymptotic entropy as

$\overline{\psi}(f):=\underset{n\to\infty}{\limsup}\>e_{\mathcal{E}_{n},\gamma_{n}}(f)=\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f}).$

Similarly, we define the lower asymptotic entropy $\underline{\psi}$ by
considering the limit inferior. Both functionals
$\overline{\psi},\underline{\psi}$ are monetary risk measures. In addition, we
consider the respective associated concentrations $\overline{J},\underline{J}$
as in (2.1), and the respective minimal penalties
$\overline{I},\underline{I}$. The following lemma is well-known in large
deviations theory and often referred to as the principle of the largest term;
see e.g. [3, Lemma 1.2.15] and [10, Exercise 14.8].

######  Lemma 6.2.

If $(a_{n}),(b_{n})$ are two $[0,\infty]$-valued sequences and
$\gamma_{n}\uparrow\infty$, then

  1. (1)

$\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log(a_{n}+b_{n})=\left(\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log
a_{n}\right)\vee\left(\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log
b_{n}\right)$.

  2. (2)

$\liminf_{n\to\infty}\tfrac{1}{\gamma_{n}}\log(a_{n}+b_{n})\leq\left(\liminf_{n\to\infty}\tfrac{1}{\gamma_{n}}\log
a_{n}\right)\vee\left(\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log
b_{n}\right)$.

######  Lemma 6.3.

For every $A\in\mathcal{X}$, it holds

$\overline{J}_{A}=\underset{n\to\infty}{\limsup}\frac{1}{n}\log\mathcal{V}_{n}(A),\quad\underline{J}_{A}=\underset{n\to\infty}{\liminf}\frac{1}{n}\log\mathcal{V}_{n}(A),$
(6.1)

where $\mathcal{V}_{n}(A)=\mathcal{E}_{n}(1_{A})$ is the capacity associated
with $\mathcal{E}_{n}$. Moreover,

$\overline{I}=-\underset{A\in\mathcal{X}}{{\rm
ess.inf}}\Big{(}\limsup_{n\to\infty}\tfrac{1}{n}\log\mathcal{V}_{n}(A)\Big{)}1_{A},\quad\underline{I}=-\underset{A\in\mathcal{X}}{{\rm
ess.inf}}\Big{(}\liminf_{n\to\infty}\tfrac{1}{n}\log\mathcal{V}_{n}(A)\Big{)}1_{A}.$
(6.2)

###### Proof.

We prove the second equality in (6.1); the first one is even easier. Fix
$r<0$. If $A\in\mathcal{X}$, we have

$\displaystyle\underline{\psi}(r1_{A^{c}})$
$\displaystyle=\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}(r1_{A^{c}})})$
$\displaystyle=\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(1_{A}+e^{\gamma_{n}r}1_{A^{c}})$
$\displaystyle\geq\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(1_{A})$
$\displaystyle\geq\underset{n\to\infty}{\liminf}\frac{1}{n}\log\mathcal{V}_{n}(A).$

Letting $r\to-\infty$, we get
$\underline{J}_{A}\geq\underset{n\to\infty}{\liminf}\frac{1}{n}\log\mathcal{V}_{n}(A)$.

On the other hand, by Lemma 6.2 we have

$\displaystyle\underline{\psi}(r1_{A^{c}})$
$\displaystyle=\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(1_{A}+e^{\gamma_{n}r}1_{A^{c}})$
$\displaystyle\leq\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\left(\mathcal{V}_{n}(A)+e^{\gamma_{n}r}\mathcal{V}_{n}(A^{c})\right)$
$\displaystyle=\left(\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{V}_{n}(A)\right)\vee\left(\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\left(e^{\gamma_{n}r}\mathcal{V}_{n}(A^{c})\right)\right)$
$\displaystyle=\left(\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{V}_{n}(A)\right)\vee\left(r+\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{V}_{n}(A^{c})\right).$

Letting $r\to-\infty$, we get

$\underline{J}_{A}\leq\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f}1_{A}).$
(6.3)

Finally, (6.2) follows from Proposition 3.3. ∎

######  Lemma 6.4.

For every $f,g\in L^{0}$, it holds

$\overline{\psi}(f\vee
g)\leq\overline{\psi}(f)\vee\overline{\psi}(g),\quad\underline{\psi}(f\vee
g)\leq\underline{\psi}(f)\vee\overline{\psi}(g).$

###### Proof.

We prove the second inequality; the first one is even easier. If $f,g\in
L^{0}$, then by Lemma 6.2 we have

$\displaystyle\underline{\psi}(f\vee g)$
$\displaystyle=\liminf_{n\to\infty}\frac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}(f\vee
g)})$
$\displaystyle\leq\liminf_{n\to\infty}\frac{1}{\gamma_{n}}\log\Big{(}\mathcal{E}_{n}(e^{\gamma_{n}f}1_{\\{f\geq
g\\}})+\mathcal{E}_{n}(e^{\gamma_{n}g}1_{\\{f<g\\}})\Big{)}$
$\displaystyle=\Big{(}\liminf_{n\to\infty}\frac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f}1_{\\{f\geq
g\\}})\Big{)}\vee\Big{(}\limsup_{n\to\infty}\frac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}g}1_{\\{f<g\\}})\Big{)}$
$\displaystyle\leq\underline{\psi}(f)\vee\overline{\psi}(g).$

The proof is complete. ∎

As in Section 5, we define the space

$L^{\overline{\psi}}:=\left\\{f\in L^{0}\colon\mbox{ there exists }t>1\mbox{
s.t.
}\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}tf})<\infty\right\\}.$

######  Proposition 6.5.

Suppose that $\overline{\psi}|_{L^{\infty}}$ is continuous from below. Then,
it holds

$\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f})={\rm
ess.sup}_{X}(f-\overline{I})\quad\mbox{ for all }f\in L^{\overline{\psi}},$
(6.4)

Moreover, if $-{\rm
ess.inf}_{A}\overline{I}\leq\liminf_{n\to\infty}\tfrac{1}{n}\log\mathcal{V}_{n}(A)$
for all $A\in\mathcal{X}$, then $\overline{\psi}(f)=\underline{\psi}(f)$ for
all $f\in L^{\infty}$ and the limit superior in (6.4) is in fact a limit.

###### Proof.

Since $\overline{\psi}$ is maxitive by Lemma 6.4, and
$\overline{\psi}|_{L^{\infty}}$ is continuous from below, we obtain the
representation (6.4) as a consequence of Theorem 5.3. If $-{\rm
ess.inf}_{A}\overline{I}\leq\liminf_{n\to\infty}\tfrac{1}{n}\log\mathcal{V}_{n}(A)$
for all $A\in\mathcal{X}$, we have that the limit superior in (6.4) is a limit
as a consequence of Proposition A.3. ∎

######  Definition 6.6.

Set $L^{\infty}_{+}=\\{f\in L^{\infty}\colon f\geq 0\\}$. We call
$\mathcal{E}\colon L^{\infty}_{+}\to\mathbb{R}\cup\\{\infty\\}$ a _non-linear
integral_ if

  1. (1)

$\mathcal{E}(f)\leq\mathcal{E}(g)$ whenever $f\leq g$,

  2. (2)

$\mathcal{E}(af)=a\mathcal{E}(f)$ for all $a\geq 0$, $a\in\mathbb{R}$.

  3. (3)

$\mathcal{E}(f)=0$ implies that $f=0$.

######  Lemma 6.7.

Let $\mathcal{E}\colon L^{\infty}_{+}\to\mathbb{R}\cup\\{\infty\\}$ be a non-
linear integral and $f\in L_{+}^{\infty}$ such that
$\mathcal{E}(f^{\gamma_{n}})^{1/\gamma_{n}}<\infty$ for all $n$. Then
$\lim_{n\to\infty}\mathcal{E}(f^{\gamma_{n}})^{1/\gamma_{n}}=\|f\|_{\infty}.$

###### Proof.

If $f=0$ the result is trivial. Assume $f\neq 0$. Fix $\varepsilon>0$ and set
$A_{\varepsilon}:=\\{|f|\geq\|f\|_{\infty}-\varepsilon\\}$. Then

$\infty>\mathcal{E}(f^{\gamma_{n}})^{1/\gamma_{n}}\geq\mathcal{E}(1_{A_{\varepsilon}}(\|f\|_{\infty}-\varepsilon)^{\gamma_{n}})^{1/{\gamma_{n}}}=(\|f\|_{\infty}-\varepsilon)\mathcal{E}(1_{A_{\varepsilon}})^{1/\gamma_{n}}.$

It follows that $\mathcal{E}(1_{A_{\varepsilon}})<\infty$. In addition, since
$\mu(A_{\varepsilon})>0$ we have that $\mathcal{E}(1_{A_{\varepsilon}})>0$ due
to the condition (3). Taking above the limit superior as $n\to\infty$ results
in

$\underset{n\to\infty}{\limsup}\mathcal{E}(f^{\gamma_{n}})^{1/\gamma_{n}}\geq\|f\|_{\infty}.$

On the other hand,

$\mathcal{E}(f^{\gamma_{n}})^{1/{\gamma_{n}}}=\mathcal{E}(f^{\gamma_{n}-1}f)^{1/\gamma_{n}}\leq\|f\|_{\infty}^{\tfrac{\gamma_{n}-1}{\gamma_{n}}}\mathcal{E}(f)^{1/\gamma_{n}}.$

Since $\mathcal{E}(f)\neq 0$ (because $f\neq 0$), taking the limit inferior as
$n\to\infty$ yields
$\underset{n\to\infty}{\liminf}\mathcal{E}(f^{n})^{1/n}\leq\|f\|_{\infty}.$
The proof is complete. ∎

######  Theorem 6.8.

Let $\mathcal{E}\colon L^{\infty}_{+}\to\mathbb{R}\cup\\{\infty\\}$ be a non-
linear integral such that there exists a sequence
$(h_{n})_{n\in\mathbb{N}}\subset L_{+}^{\infty}$ so that for each
$n\in\mathbb{N}$

$\mathcal{V}_{n}(A)=\mathcal{E}(h_{n}1_{A})\quad\mbox{ for all
}A\in\mathcal{X}.$ (6.5)

In addition, let $(A_{k})_{k\in\mathbb{N}}\subset\mathcal{X}$ such that

  1. (1)

$A_{k}\uparrow X$ as $k\to\infty$.

  2. (2)

$\limsup_{n\to\infty}\tfrac{1}{\gamma_{n}}\log\mathcal{V}_{n}(A_{k}^{c})\to-\infty$
as $k\to\infty$.

  3. (3)

For all $k\in\mathbb{N}$ there exists $I_{k}\in\mathcal{I}$ such that

$\frac{\|\log(h_{n}e^{\gamma_{n}I_{k}})1_{A_{k}}\|_{\infty}}{\gamma_{n}}\to
0\quad\mbox{ as }n\to\infty.$

Then, $I_{k+1}1_{A_{k}}=I_{k}1_{A_{k}}$ for all $k\in\mathbb{N}$, and

$\underset{n\to\infty}{\lim}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f})={\rm
ess.sup}_{X}(f-I)\quad\mbox{ for all }f\in L^{\infty},$ (6.6)

where $I=\sum_{k=1}^{\infty}I_{k}1_{B_{k}}$, $B_{1}=A_{1}$, and
$B_{k}=A_{k}\setminus A_{k-1}$ for $k>1$.

###### Proof.

We show that $(A_{k})_{k\in\mathbb{N}}$ verifies the hypothesis of Corollary
4.4. Fix $k\in\mathbb{N}$ and $\delta\in(0,1)$. Due to (3) we have that for
$n$ large enough

$e^{-\delta\gamma_{n}}\leq h_{n}e^{\gamma_{n}I_{k}}\leq
e^{\delta\gamma_{n}}\quad\mbox{ on }A_{k}.$

Then, if $B\subset A_{k}$, we have

$\displaystyle\overline{J}_{B}$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{V}(B)$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(h_{n}1_{B})$
$\displaystyle=\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(h_{n}e^{\gamma_{n}I_{k}}e^{-\gamma_{n}I_{k}}1_{B})$
$\displaystyle\leq\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(e^{\delta\gamma_{n}}e^{-\gamma_{n}I_{k}}1_{B})$
$\displaystyle=\delta+\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(e^{-\gamma_{n}I_{k}}1_{B}).$

Similarly, we obtain

$\underline{J}_{B}\geq-\delta+\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(e^{-\gamma_{n}I_{k}}1_{B}).$

Letting $\delta\downarrow 0$, we get

$\underset{n\to\infty}{\liminf}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(e^{-\gamma_{n}I_{k}}1_{B})\leq\underline{J}_{B}\leq\overline{J}_{B}\leq\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}(e^{-\gamma_{n}I_{k}}1_{B}).$

Due to Lemma 6.7 the liminf and limsup above are in fact limits and

$\overline{J}_{B}=\underline{J}_{B}=\underset{n\to\infty}{\lim}\log\mathcal{E}\left((e^{-I_{k}}1_{B})^{\gamma_{n}}\right)^{\tfrac{1}{\gamma_{n}}}=\log\left({\rm
ess.sup}_{B}e^{-I_{k}}\right)=-{\rm ess.inf}_{B}I_{k}.$ (6.7)

This shows that the condition (3) in Corollary 4.4 is verified. Since (1) and
(2) above correspond respectively to the conditions (1) and (2) in Corollary
4.4, as a consequence we obtain

$\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\mathcal{E}_{n}(e^{\gamma_{n}f})={\rm
ess.sup}_{X}(f-I)\quad\mbox{ for all }f\in L^{\infty},$ (6.8)

where $I=\sum_{k=1}^{\infty}I_{k}1_{B_{k}}$, $B_{1}=A_{1}$ and
$B_{k}=A_{k}\setminus A_{k-1}$ for $k>1$.

Finally, if we prove that $-{\rm ess.inf}_{A}I\leq\underline{J}_{A}$ for all
$A\in\mathcal{X}$, then the limsup in (6.8) will be a limit by Proposition
A.3. By contradiction assume that $\underline{J}_{A}<\beta<-{\rm
ess.inf}_{A}I$ for some $A\in\mathcal{X}$ and $\beta\in\mathbb{R}$. This
implies that $\mu(A\cap\\{I<-\beta\\})>0$. Then, since $A_{k}\uparrow X$,
there exists $k\in\mathbb{N}$ such that $\mu(B)>0$ where $B=A_{k}\cap
A\cap\\{I<-\beta\\}$. Since (6.7) is valid for any $B\subset A_{k}$, we have

$\beta>\underline{J}_{A}\geq\underline{J}_{B}=\overline{J}_{B}=-{\rm
ess.inf}_{B}I\geq\beta,$

a contradiction. ∎

######  Example 6.9.

Consider $X=[0,\infty)$ endowed with the Borel $\sigma$-field
$\mathcal{B}([0,\infty))$. Let $\nu_{n}$ be the probability measure defined by

$\nu_{n}(A)=\tfrac{n^{2}}{n+1}\int_{A}(x+1)e^{-nx}dx\quad\mbox{ for
}A\in\mathcal{B}([0,\infty)).$

As a reference probability measure, we take any Borel probability measure
$\mu$ with the same null sets as the Lebesgue measure $\lambda$ on
$[0,\infty)$. We claim that

$\underset{n\to\infty}{\lim}\tfrac{1}{n}\log\int e^{nf}d\nu_{n}={\rm
ess.sup}_{X}(f-x)\quad\mbox{ for all }f\in L^{\infty}.$ (6.9)

Observe that the above representation cannot be derived using the same
strategy as in Examples 2.4. Instead we use Theorem 6.8. For each
$k\in\mathbb{N}$, we set $A_{k}:=[0,k]$. Then, for $k\in\mathbb{N}$, we have

$\displaystyle\overline{J}_{A_{k}^{c}}$
$\displaystyle=\limsup_{n\to\infty}\frac{1}{n}\log\nu_{n}(A_{k}^{c})$
$\displaystyle=\limsup_{n\to\infty}\frac{1}{n}\log\left(\tfrac{n^{2}}{n+1}\int_{k}^{\infty}(x+1)e^{-nx}dx\right)$
$\displaystyle=\limsup_{n\to\infty}\frac{1}{n}\log\left(\int_{k}^{\infty}(x+1)e^{-nx}dx\right)$
$\displaystyle=\limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{e^{-kn}}{n^{2}}+\frac{e^{-kn}\left(k+1\right)}{n}\right)$
$\displaystyle=-k+\limsup_{n\to\infty}\frac{1}{n}\log\left(\frac{1}{n^{2}}+\frac{\left(k+1\right)}{n}\right)$
$\displaystyle=-k.$

This proves that $\overline{J}_{A_{k}^{c}}\to-\infty$ as $k\to\infty$.

On the other hand, if we set $h_{n}(x):=\tfrac{n^{2}}{n+1}(x+1)e^{-nx}$ and
$I_{k}(x)=x$, then for each $k\in\mathbb{N}$ we have

$\displaystyle\underset{n\to\infty}{\lim}\tfrac{1}{n}\|\log(h_{n}e^{nI_{k}})1_{A_{k}}\|_{\infty}$
$\displaystyle=\underset{n\to\infty}{\lim}\underset{x\in[0,k]}{{\rm
ess.sup}}\tfrac{1}{n}\left|\log\left(\tfrac{n^{2}}{n+1}(x+1)\right)\right|$
$\displaystyle=\underset{n\to\infty}{\lim}\tfrac{1}{n}\log\left(\tfrac{n^{2}}{n+1}(k+1)\right)=0.$

Then (6.9) follows as a consequence of Theorem 6.8.

###  6.1. Distorted entropic risk measures

We finish the section by analysing a particular example. A distortion function
is a non-decreasing function $g\colon[0,1]\to[0,1]$ with $g(0)=0$ and
$g(1)=1$. The _distortion risk measure_ $\mathcal{E}_{\mu,g}$ associated with
the distortion function $g$ is defined by

$\mathcal{E}_{\mu,g}(f)=\int_{0}^{\infty}g\left(\mu(f>x)\right)dx-\int_{-\infty}^{0}\left[1-g\left(\mu(f>x)\right)\right]dx.$

In particular, if $f\geq 0$, we have

$\mathcal{E}_{\mu,g}(f)=\int_{0}^{\infty}g\left(\mu(f>x)\right)dx.$

If $g$ is concave, then $d_{\mu,g}$ is a coherent risk measure which is law-
invariant; see [13]. Given a sequence $(g_{n})_{n\in\mathbb{N}}$ of concave
distortion functions the corresponding asymptotic entropy

$\psi(f)=\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\int_{0}^{\infty}g_{n}\left(\mu(e^{\gamma_{n}f}>x)\right)dx,$

is maxitive by Proposition 6.5. Straightforward verification shows that $\psi$
is law-invariant. Then, as a consequence of Corollary 2.6 we obtain the
following.

######  Proposition 6.10.

Let $(g_{n})_{n\in\mathbb{N}}$ be a sequence concave distortion functions, and
$(\gamma_{n})_{n\in\mathbb{N}}\subset(0,\infty)$ satisfy
$\gamma_{n}\uparrow\infty$. Suppose that $\mu$ is atomless and $L^{2}$ is
separable. Then,

$\underset{n\to\infty}{\limsup}\tfrac{1}{\gamma_{n}}\log\int_{0}^{\infty}g_{n}\left(\mu(e^{\gamma_{n}f}>x)\right)dx={\rm
ess.sup}_{X}f$ (6.10)

for all $f\in L^{\infty}$.

### Final remarks on maxitive integration

Maxitive monetary risk measures are related to maxitive integrals. A set
function $\theta\colon\mathcal{X}\to[-\infty,0]$ is said to be a
_concentration_ if $\theta_{\emptyset}=-\infty$, $\theta_{X}=0$, and
$\theta_{A}\leq\theta_{B}$ for all $A\subset B$. We say that $\theta$ is
maxitive if $\theta(A\cup B)\leq\theta(A)\vee\theta(B)$. Consider a
concentration $\theta$ such that $\theta(A)=\theta(B)$ whenever
$\mu(A\setminus B)=0=\mu(B\setminus A)$. The _maxitive integral_ of $f\in
L^{\infty}$ is defined by

$\int^{\max}fd\theta:=\underset{c\in\mathbb{R}}{\sup}\\{c+\theta_{\\{f\geq
c\\}}\\}.$ (6.11)

The maxitive integral (6.11) was introduced in [2]. It is closely related to
the Maslov integral in idempotent analysis [15], and can be obtain as a
transformation of the Shilkret integral [16]. In addition, maxitive monetary
risk measures can be interpreted as lineal functionals in the context of
idempotent analysis. Then, a monetary risk measure $\phi\colon
L^{\infty}\to\mathbb{R}$ is maxitive if and only if its associated
concentration $J$ is maxitive and $\phi$ admits the representation

$\phi(f)=\int^{\max}fdJ\quad\mbox{ for all }f\in L^{\infty}.$

The proof follows by straightforward adaptation to the present setting of the
proof of [9, Proposition 2.2] (see also [2, Corollary 6]).

##  Appendix A Auxiliary definitions and results

###  A.1. Essential supremum of a family of measurable functions

In this subsection we denote by
$\bar{L}^{0}:=L^{0}(X,\mathcal{X},\mu;[-\infty,\infty])$ the set of all
(equivalence classes modulo $\mu$-a.s. equality of) measurable functions from
$X$ to $[-\infty,\infty]$. Given $\mathcal{F}\subset\bar{L}^{0}$, we denote by
$U(\mathcal{F})=\\{g\in\bar{L}^{0}\colon f\leq g,\>\forall
f\in\mathcal{F}\\}$.

######  Theorem A.1.

[4, Theorem A.32] Let $\mathcal{F}$ be a non-empty subset of $\bar{L}^{0}$.
Then, there exists a unique element $f^{\ast}\in U(\mathcal{F})$ such that
$f^{\ast}\leq g$ for all $g\in U(\mathcal{F})$. If, in addition, $\mathcal{F}$
satisfies that $f\vee g\in\mathcal{F}$ whenever $f,g\in\mathcal{F}$, then
there exists a sequence $f_{1}\leq f_{2}\leq\ldots$ in $\mathcal{F}$ such that
$f_{n}\uparrow f^{\ast}$ a.s..

Given non-empty $\mathcal{F}\subset\bar{L}^{0}$, the measurable function
$f^{\ast}$ characterised in Theorem A.1 is called the _essential supremum_ of
$\mathcal{F}$ and is denoted by ${\rm ess.sup}\mathcal{F}$. For the empty set
we define ${\rm ess.sup}\emptyset=-\infty$ by convention. The _essential
supremum_ of $\mathcal{F}$ is defined by ${\rm ess.inf}\mathcal{F}=-{\rm
ess.sup}\\{-f\colon f\in\mathcal{F}\\}$.

######  Proposition A.2.

Suppose that $f\colon X\to\mathbb{R}\cup\\{-\infty\\}$ is a measurable
function which is bounded from above. Then

$\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int fd\nu={\rm ess.sup}_{X}f.$

###### Proof.

Clearly, $\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int fd\nu\leq{\rm
ess.sup}_{X}(f)$. Suppose by contradiction that

$\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int fd\nu<{\rm ess.sup}_{X}f.$ (A.1)

Then there exists $A\in\mathcal{X}$ with $\mu(A)>0$ such that

$\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int fd\nu<f\quad\mbox{ on }A.$ (A.2)

Since $\mu(\cdot|A)\in\mathscr{M}_{1}(\mu)$, we have

$\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int f(x)\nu\geq\int f(x)\mu(dx|A).$

On the other hand, (A.2) implies that

$\sup_{\nu\in\mathscr{M}_{1}(\mu)}\int f(x)\nu<\int f(x)\mu(dx|A),$

a contradiction. ∎

###  A.2. Pairs of monetary risk measures

In this subsection we consider two monetary risk measures
$\overline{\phi},\underline{\phi}\colon L^{\infty}\to\mathbb{R}$ such that
$\underline{\phi}(f)\leq\overline{\phi}(f)$ for all $f\in L^{\infty}$. In
addition we define the respective minimal penalty functions
$\overline{I},\underline{I}$ as in (2.4), and the respective concentrations
$\overline{J},\underline{J}$ as in (2.1).

######  Proposition A.3.

Let $I\in\mathcal{I}$ and suppose that $\overline{\phi}$ has the
representation

$\overline{\phi}(f)={\rm ess.sup}_{X}(f-I)\quad\mbox{ for all }f\in
L^{\infty}.$ (A.3)

If $-{\rm ess.inf}_{A}I\leq\underline{J}_{A}$ for all $A\in\mathcal{X}$, then
$\overline{\phi}(f)=\underline{\phi}(f)$ for all $f\in L^{\infty}$ and
$I=\overline{I}=\underline{I}$.

###### Proof.

First observe that (A.3) implies that $I=\overline{I}$ due to Theorem 2.3.
Suppose that $-{\rm ess.inf}_{A}\overline{I}\leq\underline{J}_{A}$ for all
$A\in\mathcal{X}$. By Proposition 3.5 we have that
$\overline{I}\geq\underline{I}$. In addition, it follows from the respective
definitions of $\overline{I},\underline{I}$ that
$\overline{I}\leq\underline{I}$. Hence, $\overline{I}=\underline{I}$. Finally,
if $f\in L^{\infty}$, we have

${\rm
ess.sup}_{X}(f-\overline{I})=\overline{\phi}(f)\geq\underline{\phi}(f)\geq{\rm
ess.sup}_{X}(f-\underline{I})={\rm ess.sup}_{X}(f-\overline{I}).$

It follows that $\overline{\phi}(f)=\underline{\phi}(f)$ for all $f\in
L^{\infty}$. ∎

## References

  * [1] C. D. Aliprantis, K. C. Border, _Infinite dimensional analysis_. Springer, Berlin, third edition, 2006.
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