# Generalized stochastic processes revisited

Michael Oberguggenberger111Unit for Engineering Mathematics, University of
Innsbruck, Technikerstr. 13, A-6020 Innsbruck, Austria,
<EMAIL_ADDRESS>
Abstract. The paper addresses the question whether a random functional, a map
from a set $E$ into the space of real-valued measurable functions on a
probability space, has a measurable version with values in $\mathbb{R}^{E}$.
Similarly, one may ask whether linear random functionals have versions in the
algebraic dual. Most importantly, it can be asked which locally convex
topological vector spaces $E$ have the “regularity property” that any linear
random functional on $E$ has a version with values in the dual $E^{\prime}$,
an important issue in the theory of generalized stochastic processes. It has
been shown by Itô and Nawata that this is the case when $E$ is nuclear.
However, the question of uniqueness has only been partially answered. We build
up a framework where these and related questions can be clarified in terms of
spaces and mappings. We study classes of spaces $E$ (beyond nuclear spaces)
with the said regularity property, prove a seemingly new uniqueness result and
exhibit various examples and counterexamples.
Keywords: Generalized stochastic processes, random functionals, regular
versions, nuclear spaces.
AMS Subject Classification: 60G20, 46A99

##  1 Introduction

This paper takes up the question of relating different notions of generalized
stochastic processes to each other, mainly motivated by processes with values
in the space of Schwartz distributions
${\mathcal{D}}^{\prime}(\mathbb{R}^{n})$. Gel’fand and Vilenkin [6]
(1961/1964) considered such “generalized random functions” as families of
random variables indexed by test functions
$\varphi\in{\mathcal{D}}(\mathbb{R}^{n})$, which are linear in $\varphi$ and
continuous in distribution on ${\mathcal{D}}(\mathbb{R}^{n})$. On the other
hand, Ullrich [27] (1956) introduced “random Schwartz distributions” as weakly
measurable maps on a probability space with values in
${\mathcal{D}}^{\prime}(\mathbb{R}^{n})$; this notion has been widely adopted
(e.g. [11]). Further, “random distributions” – continuous linear functionals
on ${\mathcal{D}}(\mathbb{R}^{n})$ with values in a Hilbert space of second
order random variables – were introduced by Itô [13] (1953).

It has been established by Itô and Nawata in [14, 15] that any generalized
random function in the sense of Gel’fand-Vilenkin has a version which is a
random Schwartz distribution in the sense of Ullrich. This nontrivial result
rests on the nuclearity of ${\mathcal{D}}(\mathbb{R}^{n})$. Also, the
linearity of generalized random functionals implies that continuity in
distribution is equivalent with continuity in probability. This shows that Itô
linear functionals also belong to the Gel’fand-Vilenkin class.

The aim of the paper is to systematically study the relations between these
classes of stochastic processes in terms of spaces and mappings, with and
without linearity and/or continuity.

Let $(\Omega,\Sigma,P)$ be a probability space and $E$ a set. A _random
functional_ [14, Section 2.3] is a map $Z:E\times\Omega\to\mathbb{R}$ such
that the map $\omega\to Z(e,\omega)$ is measurable for every $e\in E$. The
random functional $Z$ can be viewed from two sides: either as a map
$X:\Omega\to\mathbb{R}^{E}$ or as a map $Y:E\to\mathbb{R}^{\Omega}$. This
dichotomy underlies the difference in interpretation of Ullrich vs. Gel’fand-
Vilenkin. We start by observing that the map

$\kappa:L^{0}(\Omega\\!:\\!\mathbb{R}^{E})\to(L^{0}(\Omega))^{E},\qquad\kappa
X(e)(\omega)=X(\omega)(e)$

is well-defined, where $L^{0}(\Omega\\!:\\!F)$ denotes the space of
(equivalence classes of) measurable maps from $(\Omega,\Sigma)$ into a
measurable space $(F,{\mathcal{F}})$. It turns out that the map $\kappa$ is
always surjective, but not necessarily injective. On the next level, we let
$E$ be a linear space with algebraic dual $E^{\ast}$ and consider the
restriction of $\kappa$ as

$\kappa_{0}:L^{0}(\Omega\\!:\\!E^{\ast})\to{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$

and study its surjectivity/injectivity, where
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$ denotes the space of linear maps from
$E$ to $L^{0}(\Omega)$. The central issue is adding continuity, yielding the
map

$\overline{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$

where $E^{\prime}$ denotes the sequentially continuous dual of $E$ and
${\mathcal{L}}_{\rm c}(E\\!:\\!L^{0}(\Omega))$ the space of sequentially
continuous linear maps from $E$ to $L^{0}(\Omega)$. A more complete diagram
relating the mentioned spaces can be found in Section 5.

Let us temporarily say that a locally convex topological vector space has the
_surjectivity property_ , if the map $\overline{\kappa}$ is surjective for
whatever probability space $(\Omega,\Sigma,P)$. The mentioned result of Itô
and Nawata says that every nuclear space $E$ has the surjectivity property. We
exhibit several examples where the surjectivity property fails when $E$ is not
nuclear. We also investigate whether the surjectivity property is hereditary.
For example, we show that if $E=\sum_{\gamma\in\Gamma}E_{\gamma}$ is the
locally convex direct sum of spaces $E_{\gamma}$ with the surjectivity
property, then $E$ has the surjectivity property (and conversely). If $\Gamma$
is uncountable and each $E_{\gamma}$ is nuclear, $E$ provides an example of a
space with the surjectivity property which is not nuclear.

The result of Itô and Nawata also gives a certain uniqueness of the regular
version of a process ${\mathcal{L}}_{\rm c}(E\\!:\\!L^{0}(\Omega))$ in
$L^{0}(\Omega\\!:\\!E^{\prime})$, but falls short of proving injectivity of
the map $\overline{\kappa}$. We show that $\overline{\kappa}$ is injective
when $E$ is sequentially separable. We also exhibit various example of the
failure of injectivity. A crucial observation is that equivalence of
measurable functions in $L^{0}(\Omega\\!:\\!E^{\prime})$ is stronger than
equivalence of function values in ${\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$.

The results of the paper provide some progress on the “enticing open question”
posed by Pilipović and Seleši in [21, p. 252].

The plan of the paper is as follows. In Section 2 we describe the initial
question of the paper in more detail and introduce some notation. Random
functionals are addressed in Section 3, linear random functionals in Section
4. The central parts are Sections 5 and 6. In Section 5, continuous linear
random functionals are considered. Various notions of continuity are related
to each other, and the relations of the resulting spaces are worked out. The
role of sequential separability is also pinned down.

Section 6 starts out by recalling the central result of Itô and Nawata that
any continuous linear random functional on a nuclear space $E$ has a version
in $E^{\prime}$. Combining it with our injectivity result, we obtain that
$\overline{\kappa}$ is a bijection when $E$ is a sequentially separable
nuclear space. It is also shown, in particular, that the surjectivity property
is inherited by locally convex direct sums, complemented subspaces and by
countable regular inductive limits. Section 7 provides various
counterexamples, in particular, simple examples showing that the theorem of
Itô and Nawata does not hold without the nuclearity assumption, in general.
The mapping properties between the various spaces considered in Section 5 are
further analyzed. Finally, the Appendix serves to record the theorem of Itô
and Nawata, which originally slightly goes beyond nuclearity of the space $E$
itself. We also record the version of Walsh [28] who proves the surjectivity
property for complete, countably Hilbertian nuclear spaces.

We end the introduction with a few references to the literature. The issue
whether a linear random functional (of Gel’fand-Vilenkin-type) on some
topological vector space has a version (of Ullrich-type) in the continuous
dual has been pointed out in a number of papers, e.g. [2], [19], who refer to
the simplified special version of Itô’s proof in Walsh [28, Corollary 4.2]. A
detailed comparison of Itô-type random distributions with Ullrich-type
functionals can be found in [21], together with many examples and
counterexamples, mainly on ${\mathcal{K}}\\{M_{p}\\}$-spaces (see also [25]
for this setting).

Generalized stochastic processes can also be defined as measures on the dual
space. This topic, together with characteristic functionals, remains outside
the scope of this paper. The relation is extensively discussed in Itô’s
monograph [14] and at many other places, e.g. [10]. Here the question whether
such a measure on the algebraic dual gives probability one to the continuous
dual (and then allows one to construct Ullrich-type-versions of the
corresponding process) has been extensively discussed in [3]. Similar
questions arise with the earlier concept of “weak distributions” in the sense
of Segal [24], introduced as random variables on the continuous dual; see also
[7], [8].

##  2 Motivation and notations

In order to motivate the exposition, we first formulate the concepts of
Gel’fand-Vilenkin, Ullrich and Itô in more detail; for the notation of spaces
of test functions and Schwartz distributions, see e.g. [12].

Gel’fand and Vilenkin define a “generalized random function” in [6, Section
III.1.2.] as a stochastic process $X$ indexed by
${\mathcal{D}}={\mathcal{D}}(\mathbb{R}^{n})$ such that

(i) For all $\varphi\in{\mathcal{D}}$, $X(\varphi)$ is an $\mathbb{R}$-valued
random variable;

(ii) $X$ is linear in the sense that for all $\varphi,\psi\in{\mathcal{D}}$,
$\alpha,\beta\in\mathbb{R}$, $X(\alpha\varphi+\beta\psi)$ has the same
distribution as $\alpha X(\varphi)+\beta X(\psi)$;

(iii) if $\varphi_{n}\to\varphi$ in ${\mathcal{D}}$, then $X(\varphi_{n})\to
X(\varphi)$ in distribution.

Ullrich [27] starts by introducing the $\sigma$-algebra on
${\mathcal{D}}^{\prime}$ generated by the sets
$\\{T\in{\mathcal{D}}^{\prime}:T(\varphi)<c\\}$, $\varphi\in{\mathcal{D}}$,
$c\in\mathbb{R}$. Given a measurable space $(\Omega,\Sigma)$, a “random
Schwartz distribution” is a measurable transformation
$X:\Omega\to{\mathcal{D}}^{\prime}$. If $(\Omega,\Sigma,P)$ is a probability
space, this is the same as saying that $X$ is a family of random variables
$\\{X(\varphi,\omega):\varphi\in{\mathcal{D}}\\}$ such that
$X(\cdot,\omega)\in{\mathcal{D}}^{\prime}$ for almost all $\omega\in\Omega$,
see Hida [11], last paragraph in Section 1.3.(ii).

Finally, Itô [13] introduces “random distributions” as continuous linear
functionals on ${\mathcal{D}}$ with values in the Hilbert space
${\mathcal{H}}$ of random variables with mean zero and finite variance.

We set out to analyze the relations between these definitions, replacing
${\mathcal{D}}$ by $E$, where we will successively take for $E$ a set, a
linear space, and a locally convex topological vector space.

In the sequel, $(\Omega,\Sigma,P)$ will denote a probability space. Given a
measurable space $(F,{\mathcal{F}})$, the space $L^{0}(\Omega\\!:\\!F)$
denotes the set of (equivalence classes) of measurable maps
$(\Omega,\Sigma)\to(F,{\mathcal{F}})$. In case $F=\mathbb{R}$, we will take
${\mathcal{F}}={\mathcal{B}}(\mathbb{R})$, the Borel $\sigma$-agebra on
$\mathbb{R}$ and simply write $L^{0}(\Omega)$ for
$L^{0}(\Omega\\!:\\!\mathbb{R})$. We will always assume that all
representatives of measurable maps are defined on all of $\Omega$ and are
finite-valued.

Let $E$ be a set. The space $\mathbb{R}^{E}$ of maps $E\to\mathbb{R}$ will be
equipped with the Kolmogorov $\sigma$-algebra
${\mathcal{B}}_{K}(\mathbb{R}^{E})$, which is generated by the sets

$\pi^{-1}_{e}(B),B\in{\mathcal{B}}(\mathbb{R}),e\in E,$

where $\pi_{e}$ denotes the projection on the e-th slot. (This
$\sigma$-algebra is often referred to as the product $\sigma$-algebra, but it
is different from the $\sigma$-algebra generated by the product topology,
unless $E$ is countable.)

When $G,H$ are sets, we shall occasionally use the notation
${\mathcal{M}}(G\\!:\\!H)$ for the family of all maps $G\to H$, that is,
${\mathcal{M}}(G\\!:\\!H)=H^{G}$.

##  3 Random functionals

A _random functional_ on a set $E$ is a map $Z:E\times\Omega\to\mathbb{R}$
such that the map $\omega\to Z(e,\omega)$ is measurable for every $e\in E$. As
noted in the Introduction, the random functional $Z$ can be viewed from two
sides: either as a map $X:\Omega\to\mathbb{R}^{E}$ or as a map
$Y:E\to\mathbb{R}^{\Omega}$. We will build up a framework yielding a canonical
linear map that allows one to connect the two viewpoints via

$\kappa:L^{0}(\Omega\\!:\\!\mathbb{R}^{E})\to(L^{0}(\Omega))^{E}.$

We begin by defining the map $\kappa$ on single representatives as

$\kappa:{\mathcal{M}}(\Omega\\!:\\!\mathbb{R}^{E})\to({\mathcal{M}}(\Omega\\!:\\!\mathbb{R}))^{E},\quad\kappa
X(e)(\omega)=X(\omega)(e).$

######  Lemma 3.1.

$X:\Omega\to\mathbb{R}^{E}$ is
$(\Omega,\Sigma)-(\mathbb{R}^{E},{\mathcal{B}}_{K}(\mathbb{R}^{E}))$-measurable
if and only if $\kappa X(e)$ is
$(\Omega,\Sigma)-(\mathbb{R},{\mathcal{B}}(\mathbb{R}))$-measurable for all
$e\in E$.

_Proof._ Since the sets $\pi^{-1}_{e}(B)$, $B\in{\mathcal{B}}(\mathbb{R})$,
$e\in E$ generate the Kolmogorov $\sigma$-alebra, $X$ is measurable if and
only if all sets $\\{\omega\in\Omega:X(\omega)\in\pi^{-1}_{e}(B)\\}$
($B\in{\mathcal{B}}(\mathbb{R})$, $e\in E$) belong to $\Sigma$. On the other
hand, $\kappa X(e)$ is measurable if and only if all sets
$\\{\omega\in\Omega:\kappa X(e)(\omega)\in B\\}$
($B\in{\mathcal{B}}(\mathbb{R})$) belong to $\Sigma$. But

$\displaystyle\\{\omega\in\Omega:\kappa X(e)(\omega)\in
B\\}=\\{\omega\in\Omega:X(\omega)(e)\in B\\}$
$\displaystyle=\\{\omega\in\Omega:\pi_{e}X(\omega)\in
B\\}=\\{\omega\in\Omega:X(\omega)\in\pi^{-1}_{e}(B)\\},$

so the assertion follows. $\Box$

In the next step, we go over to equivalence classes. Let $X_{1},X_{2}$ be
measurable maps $\Omega\to\mathbb{R}^{E}$. Then $X_{1}$ and $X_{2}$ are
strictly equivalent [15, p. 258] (or strict versions of each other), if and
only if

$\exists\Omega_{1}\subset\Omega,P(\Omega_{1})=1\quad{\rm such\
that}\quad\forall\omega\in\Omega_{1},X_{1}(\omega)=X_{2}(\omega).$

The latter assertion means that for all $\omega\in\Omega_{1}$ and all $e\in
E$, $X_{1}(\omega)(e)=X_{2}(\omega)(e)$. On the other hand, if $Y_{1},Y_{2}$
are maps from $E$ into ${\mathcal{M}}(\Omega\\!:\\!\mathbb{R})$ such that
$Y_{1}(e)$ and $Y_{2}(e)$ are measurable for each $e\in E$, equivalence takes
place in $L^{0}(\Omega)$, meaning that

$\forall e\in E\ \exists\Omega_{1,e}\subset\Omega,P(\Omega_{1,e})=1\quad{\rm
such\
that}\quad\forall\omega\in\Omega_{1,e},Y_{1}(e)(\omega)=Y_{2}(e)(\omega).$

In the language of stochastic processes indexed by $E$, this means that
$Y_{1}$ is a version of $Y_{2}$.

######  Lemma 3.2.

If $X_{1}$ is a strict version of $X_{2}$, then $\kappa X_{1}$ is a version of
$\kappa X_{2}$.

_Proof._ By assumption, there is a subset $\Omega_{1}$ of full measure such
that for all $\omega\in\Omega_{1}$ and all $e\in E$,
$X_{1}(\omega)(e)=X_{2}(\omega)(e)$. Since $\kappa
X_{i}(e)(\omega)=X_{i}(\omega)(e)$, this in turn implies the equivalence of
$\kappa X_{1}(e)$ and $\kappa X_{2}(e)$ as measurable maps
$\Omega\to\mathbb{R}$, with $\Omega_{1,e}=\Omega_{1}$ for all $e$. $\Box$

######  Corollary 3.3.

The map $\kappa$ does not depend on the choice of representatives, thus can be
extended to a map

$\kappa:L^{0}(\Omega\\!:\\!\mathbb{R}^{E})\to(L^{0}(\Omega))^{E},$

and $\kappa$ is surjective.

_Proof._ Denoting equivalence classes of measurable maps by square brackets,
the assignment

$[X]\to\kappa[X]=[\kappa X(e)]_{e\in E}$

is well-defined by Lemma 3.2. To prove surjectivity, let
$Y\in(L^{0}(\Omega))^{E}$. For each $e\in E$, $Y(e)$ is an equivalence class
of measurable functions. Invoking the axiom of choice, we may choose a
representative $\overline{Y(e)}\in Y(e)$ for each $e\in E$. Define
$X\in{\mathcal{M}}(\Omega\\!:\\!\mathbb{R}^{E})$ by

$X(\omega)(e)=\overline{Y(e)}(\omega).$

Then $\kappa X(e)(\omega)=X(\omega)(e)=\overline{Y(e)}(\omega)$ is
$(\Omega,\Sigma)-(\mathbb{R},{\mathcal{B}}(\mathbb{R}))$-measurable. By Lemma
3.1, $X$ is
$(\Omega,\Sigma)-(\mathbb{R}^{E},{\mathcal{B}}_{K}(\mathbb{R}^{E}))$-measurable,
and $\kappa X(e)=\overline{Y(e)}\in Y(e)$. Thus $[X]\in
L^{0}(\Omega\\!:\\!\mathbb{R}^{E})$, and $\kappa[X]=Y$. $\Box$

The map $\kappa$ is not necessarily injective (Example 7.1).

From now on, unless clarity requires it, we will suppress the square bracket
notation, and – as is common practice – employ the same notation for a
representative and its class.

##  4 Linear random functionals

In this section, $E$ will be a vector space over $\mathbb{R}$, and $E^{\ast}$
its algebraic dual. The algebraic dual is equipped with the restriction of the
Kolmogorov $\sigma$-algebra $E^{\ast}\cap{\mathcal{B}}_{K}(\mathbb{R}^{E})$.
Thus a map $\Omega\to E^{\ast}$ is measurable, iff $\pi_{e}X$ belongs to
$L^{0}(\Omega)$ for all $e\in E$. The space of linear mappings $E\to
L^{0}(\Omega)$ is denoted by ${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$.

Now $X\in L^{0}(\Omega\\!:\\!E^{\ast})$ means that there is a set
$\Omega_{1}\subset\Omega$ of full measure, such that $X(\omega)$ is linear in
$E$. That is,

$\displaystyle\exists\Omega_{1},P(\Omega_{1}=1):\forall\omega\in\Omega_{1},\alpha,\beta\in\mathbb{R},e,f\in
E,$ $\displaystyle X(\omega)(\alpha f+\beta g)=\alpha X(\omega)(e)+\beta
X(\omega)(f).$

On the other hand, $Y\in{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$ means that Y is
linear in $E$ with values in $L^{0}(\Omega)$. That is,

$\displaystyle\forall\alpha,\beta\in\mathbb{R},e,f\in
E,\exists\Omega_{1},P(\Omega_{1}=1):\forall\omega\in\Omega_{1},$
$\displaystyle X(\omega)(\alpha f+\beta g)=\alpha X(\omega)(e)+\beta
X(\omega)(f).$

To be sure, in the second case $\Omega_{1}$ may depend on $\alpha,\beta,e,f$,
while in the first case it does not.

If $X\in L^{0}(\Omega\\!:\\!E^{\ast})$, then clearly $\kappa
X\in{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$. Thus the restriction of $\kappa$ to
$L^{0}(\Omega\\!:\\!E^{\ast})$ results in a map

$\kappa_{0}:L^{0}(\Omega\\!:\\!E^{\ast})\to{\mathcal{L}}(E\\!:\\!L^{0}(\Omega)),$

which is also well-defined.

######  Proposition 4.1.

The map
$\kappa_{0}:L^{0}(\Omega\\!:\\!E^{\ast})\to{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$
is surjective.

_Proof._ Let $Y\in{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$. Choose an an
algebraic basis $(e_{\lambda})_{\lambda\in\Lambda}$ of $E$. Let

$\mathbb{R}^{(\Lambda)}=\\{f\in\mathbb{R}^{\Lambda}:f(\lambda)\neq 0\ {\rm
for\ at\ most\ finitely\ many\ }\lambda\in\Lambda\\}.$

Then $E$ is isomorphic with $\mathbb{R}^{(\lambda)}$; the isomorphism $\theta$
sends $e_{\lambda}$ to the $\lambda$th unit vector in
$\mathbb{R}^{(\Lambda)}$. Its transpose $\theta^{\prime}$ is an isomorphism
between $\mathbb{R}^{\Lambda}$ and $E^{\ast}$. Choose representatives
$\overline{Y(e_{\lambda})}\in Y(e_{\lambda})$ and set

$X(\omega)=\theta^{\prime}\Big{(}\big{(}\overline{Y(e_{\lambda})}(\omega)\big{)}_{\lambda\in\Lambda}\Big{)}.$

Clearly $X(\omega)\in E^{\ast}$ for all $\omega\in\Omega$. If $e=\sum_{\rm
finite}\alpha_{\lambda}e_{\lambda}\in E$, then $X(\omega)(e)=\sum_{\rm
finite}\alpha_{\lambda}\overline{Y(e_{\lambda})}(\omega)$, so $X:\Omega\to
E^{\ast}$ is measurable. Finally,

$\big{[}\sum_{\rm
finite}\alpha_{\lambda}\overline{Y(e_{\lambda})}\big{]}=\sum_{\rm
finite}\alpha_{\lambda}\big{[}\overline{Y(e_{\lambda})}]=\sum_{\rm
finite}\alpha_{\lambda}Y(e_{\lambda})=Y(e),$

since $Y$ is linear. This shows that $[\kappa_{0}X(e)]=Y(e)$ for all $e\in E$
and so $\kappa_{0}[X]=Y$. $\Box$

The surjectivity of $\kappa_{0}$ can also be obtained as a corollary to
Proposition 6.3 (Remark 6.4). The map $\kappa_{0}$ need not be injective
(Example 7.2). However, we have the following positive result.

######  Proposition 4.2.

Suppose the dimension of $E$ is at most countable. Then
$\kappa_{0}:L^{0}(\Omega\\!:\\!E^{\ast})\to{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$
is injective.

_Proof._ An element $X\in L^{0}(\Omega\\!:\\!E^{\ast})$ is determined by its
action on an algebraic basis $\\{e_{\iota}:\iota\in I\\}$. Indeed, if $e\in E$
and $e=\sum_{\rm finite}\lambda_{\iota}e_{\iota}$, then

$X(\omega)(e)=\sum_{\rm
finite}\lambda_{\iota}X(\omega)(e_{\iota}),\qquad\kappa_{0}X(e)(\omega)=\sum_{\rm
finite}\lambda_{\iota}\kappa_{0}X(e_{\iota})(\omega).$

Now $\kappa_{0}X(e)=0$ in $L^{0}(\Omega)$ for all $e\in E$ if and only if, for
all $\iota\in I$, $\kappa_{0}X(e_{\iota})(\omega)=0$ on sets $\Omega_{\iota}$
of full measure. But then $X(\omega)(e_{\iota})=0$ for all $\iota\in I$ on
$\bigcap_{\iota\in I}\Omega_{\iota}$, which also has full measure. Thus $X=0$
in $L^{0}(\Omega\\!:\\!E^{\ast})$. $\Box$

##  5 Continuous linear random functionals

In what follows, the term _locally convex space_ will always mean a separated
(i.e., Hausdorff) locally convex topological vector space. Let $E$ be a
locally convex space. The space of _sequentially continuous linear maps_
$E\to\mathbb{R}$ will be denoted by $E^{\prime}$. It coincides with the
continuous dual of $E$, when $E$ is a _Mazur space_ [29, Definition 8-6-3]
(this happens in all application cases we have in mind). If $F$ is a
topological vector space, we denote by ${\mathcal{L}}_{c}(E\\!:\\!F)$ the
sequentially continuous linear maps from $E$ to $F$.

Let $X\in{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$ and let $e_{n}\to e$ in $E$. We
can consider various notions of continuity of $X$:

  * (dist)

$X(e_{n})\to X(e)$ in distribution. This is the formulation of Gel’fand-
Vilenkin [6].

  * (prob)

$X(e_{n})\to X(e)$ in probability. This is assumed e.g. in Itô [14] and Walsh
[28].

  * (path)

$X(e_{n})\to X(e)$ almost surely (= pathwise almost everywhere), i.e.,
$X(e_{n})(\omega)\to X(e)(\omega)$ for all $\omega$ in a set of full measure
(which may depend on $e_{n}$ and $e$).

The subspaces of ${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$, equipped with the
respective convergence, will be denoted by
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm dist}$,
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm prob}$, and
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm path}$.

######  Remark 5.1.

(a) If $X\in{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$ then $X(e_{n})\to X(e)$ iff
$X(e_{n}-e)\to 0$, by linearity. But convergence in distribution to a fixed
constant is equivalent with convergence in probability [1, Section 3, p. 27].
Thus ${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
dist}={\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm prob}$.

(b) $L^{0}(\Omega)$ is a complete metrizable topological vector space with
distance $d(f,g)=\mathrm{E}(|f-g|\wedge 1)$. Convergence in metric is
equivalent with convergence in probability [17, Section 9.3, Remark]. Thus
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm prob}={\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$, the continuous linear maps from $E$ to
$L^{0}(\Omega)$.

(c) For $p>0$, the imbedding $L^{p}(\Omega)\to L^{0}(\Omega)$ is continuous.
This is obvious for $p\geq 1$; for $0<p<1$, see [17, Section 9.3, Application
(2)].

(d) Almost sure convergence implies convergence in probability.

By (a) and (d) of Remark 5.1, we have the following chain of inclusions:

${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
path}\subset{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
prob}={\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
dist}\subset{\mathcal{L}}(E\\!:\\!L^{0}(\Omega)).$

Further, by (b),

${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm prob}={\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$

and, by (c),

${\mathcal{L}}_{\rm c}(E\\!:\\!L^{p}(\Omega))\subset{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))\quad{\rm for\ }0\leq p\leq\infty.$

Similarly, for measurable maps with values in $E^{\ast}$, we have

$L^{0}(\Omega\\!:\\!E^{\ast})_{\rm path}\subset
L^{0}(\Omega\\!:\\!E^{\ast})_{\rm prob}=L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
dist}\subset L^{0}(\Omega\\!:\\!E^{\ast}).$

Stipulating that the exceptional set arising in the definition of pathwise
convergence in $L^{0}(\Omega\\!:\\!E^{\ast})_{\rm path}$ does not depend on
$e_{n}$ and $e$, we have

$L^{0}(\Omega\\!:\\!E^{\ast})_{\rm path}=L^{0}(\Omega\\!:\\!E^{\prime}).$

The relations can be summarized in the following diagram:

$\begin{array}[]{lllllll}&&L^{0}(\Omega\\!:\\!\mathbb{R}^{E})&\stackrel{{\scriptstyle\kappa}}{{\longrightarrow}}&{\mathcal{M}}(E\\!:\\!L^{0}(\Omega))&&\\\\[4.0pt]
&&\qquad\rotatebox[origin={c}]{90.0}{$\hookrightarrow$}&&\qquad\rotatebox[origin={c}]{90.0}{$\hookrightarrow$}&&\\\\[4.0pt]
&&L^{0}(\Omega\\!:\\!E^{\ast})&\stackrel{{\scriptstyle\kappa_{0}}}{{\longrightarrow}}&{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))&&\\\\[4.0pt]
&&\quad\ \ \rotatebox[origin={c}]{90.0}{$\hookrightarrow$}\ \iota&&\quad\ \
\rotatebox[origin={c}]{90.0}{$\hookrightarrow$}\ \iota_{0}&&\\\\[4.0pt]
&&L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
dist}&\stackrel{{\scriptstyle\kappa_{1}}}{{\longrightarrow}}&{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
dist}&&\\\\[4.0pt]
&&\qquad\rotatebox[origin={c}]{90.0}{$=$}&&\qquad\rotatebox[origin={c}]{90.0}{$=$}&&\\\\[4.0pt]
&&L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
prob}&\stackrel{{\scriptstyle\kappa_{1}}}{{\longrightarrow}}&{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
prob}&=&{\mathcal{L}}_{\rm c}(E\\!:\\!L^{0}(\Omega))\\\\[4.0pt] &&\quad\ \
\rotatebox[origin={c}]{90.0}{$\hookrightarrow$}\
\iota_{1}&\\!{}^{\bar{\kappa}}\\!{\nearrow}&\quad\ \
\rotatebox[origin={c}]{90.0}{$\hookrightarrow$}\ \iota_{2}&&\\\\[4.0pt]
L^{0}(\Omega\\!:\\!E^{\prime})&=&L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
path}&\stackrel{{\scriptstyle\kappa_{2}}}{{\longrightarrow}}&{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
path}&&\end{array}$

Here $\kappa_{i}$ denotes the restriction of $\kappa$ to the respective level,
while $\bar{\kappa}=\kappa_{1}\circ\iota_{1}=\iota_{2}\circ\kappa_{2}$. All
vertical arrows are injections. The entry
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm dist}$ corresponds to the Gel’fand-
Vilenkin definition, $L^{0}(\Omega\\!:\\!E^{\prime})$ corresponds to Ullrich
and Hida. The spaces ${\mathcal{L}}_{\rm c}(E\\!:\\!L^{p}(\Omega))$
(continuous linear maps on $E$ with values in $L^{p}(\Omega)$ enter on the
right-hand side of the next to last line as subspaces of ${\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$.

The central issue of this paper is the question of “regularity”: Under what
conditions does a linear random function belonging to ${\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ have a version with values in $E^{\prime}$ – or
even an almost surely unique version?

######  Definition 5.2.

A locally convex space $E$ is said to have property (${\mathcal{R}}$), if for
whatever probability space $(\Omega,\Sigma,P)$, the map
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ is bijective.

Surjectivity means that for any $Y\in{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ there is $X\in L^{0}(\Omega\\!:\\!E^{\prime})$ such
that $\bar{\kappa}X=Y$. Thus there is $\Omega_{1}\subset\Omega$ of full
measure such that $X(\omega)\in E^{\prime}$ for all $\omega\in\Omega_{1}$ and,
for all $e\in E$, there is $\Omega_{1,e}\subset\Omega$ of full measure such
that $X(\omega)(e)=Y(e)(\omega)$ for all $\omega\in\Omega_{1,e}$. In words,
$X$ is a version of $Y$ with values in $E^{\prime}$.

Note that one can assume without loss of generality that $\Omega_{1}=\Omega$,
setting e.g. $X(\omega)=0$ for $\omega\not\in\Omega_{1}$.

Injectivity means that if $X,X^{\prime}\in L^{0}(\Omega\\!:\\!E^{\prime})$ and
$\bar{\kappa}X=\bar{\kappa}X^{\prime}$, then $X$ and $X^{\prime}$ are strict
versions of each other. In words: If for all $e\in E$ there is
$\Omega_{1,e}\subset\Omega$ of full measure such that
$X(\omega)(e)=X^{\prime}(\omega)(e)$ for all $\omega\in\Omega_{1,e}$, then
there is $\Omega_{1}\subset\Omega$ of full measure such that
$X(\omega)=X^{\prime}(\omega)$ in $E^{\prime}$.

Positive examples of spaces with property (${\mathcal{R}}$) and permanence
properties will be given in Section 6, counterexamples are in Section 7.

######  Remark 5.3.

(a) If the space $E$ has property (${\mathcal{R}})$ then one has the relations

${\mathcal{L}}_{\rm c}(E\\!:\\!L^{p}(\Omega))\subset{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))=L^{0}(\Omega\\!:\\!E^{\prime}).$

(b) If the space $E$ has property (${\mathcal{R}})$ then the map $\kappa_{1}$
is surjective, and the maps $\kappa_{2}$ and $\iota_{2}$ are both bijective.
Indeed, the surjectivity of
$\bar{\kappa}=\kappa_{1}\circ\iota_{1}=\iota_{2}\circ\kappa_{2}$ implies the
surjectivity of $\kappa_{1}$ and of $\iota_{2}$; the injectivity of
$\bar{\kappa}$ implies the injectivity of $\kappa_{2}$. But $\iota_{2}$ is
injective by definition, thus it becomes a bijection, and then the same holds
for $\kappa_{2}$.

It is not clear whether $\kappa_{1}$ is injective in this case (which in this
situation is equivalent with the surjectivity of $\iota_{1}$).

As will be observed in Section 7, the maps $\kappa_{0}$,
$\kappa_{1},\kappa_{2}$ and $\bar{\kappa}$ need not be injective, in general.
However, there is a convenient positive result.

######  Proposition 5.4.

If $E$ is a sequentially separable locally convex space, then $\bar{\kappa}$
and $\kappa_{2}$ are injective.

_Proof._ Let $X,X^{\prime}\in L^{0}(\Omega\\!:\\!E^{\prime})$ such that
$\bar{\kappa}X=\bar{\kappa}X^{\prime}$. For every $e\in E$, there is
$\Omega_{1,e}\subset\Omega$ of full measure, such that
$X(\omega)(e)=X^{\prime}(\omega)(e)$ for all $\omega\in\Omega_{1,e}$. Let
$(e_{n})_{n\in\mathbb{N}}$ be a countable dense subset of $E$. Set

$\Omega_{1}=\bigcap_{n=1}^{\infty}\Omega_{1,e_{n}}.$

Then $X(\omega)(e_{n})=X^{\prime}(\omega)(e_{n})$ for all $n\in\mathbb{N}$ and
all $\omega\in\Omega_{1}$. We may also assume that
$X(\omega),X^{\prime}(\omega)$ belong to $E^{\prime}$ for all
$\omega\in\Omega_{1}$. Let $e\in E$ and $e_{n_{j}}\to e$ in $E$. Then
$X(\omega)(e)=\lim_{j\to\infty}X(\omega)(e_{n_{j}})=\lim_{j\to\infty}X^{\prime}(\omega)(e_{n_{j}})=X^{\prime}(\omega)(e)$.
Thus $X(\omega)=X^{\prime}(\omega)$ for $\omega\in\Omega_{1}$.

The injectivity of $\bar{\kappa}=\iota_{2}\circ\kappa_{2}$ clearly implies the
injectivity of $\kappa_{2}$. $\Box$

An example of a non-separable space for which $\bar{\kappa}$ is not injective
(nor are $\kappa_{0},\kappa_{1},\kappa_{2}$) is given in Example 7.3.

##  6 Regularity theorems

We begin by stating the central regularization result.

######  Theorem 6.1.

(a) If $E$ is a nuclear locally convex space, then
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ is surjective.

(b) If in addition $E$ is sequentially separable, then $\bar{\kappa}$ is a
bijection, i.e., the space $E$ has property (${\mathcal{R}}$).

_Proof._ The surjectivity in (a) follows from Corollary A.3 in the Appendix,
the injectivity in (b) from Proposition 5.4 above. $\Box$

The fundamental part (a) is due to Itô and Nawata [14, 15]. Part (b) presents
a slight improvement over the uniqueness result of Itô and Nawata, where
almost sure uniqueness is formulated only for $\sigma$-concentrated
functionals (cf. Appendix). The observation that sequential separability is an
alternative hypothesis appears to be new.

######  Remark 6.2.

Every metrizable nuclear space is separable and, in particular, sequentially
separable [20, Theorem 4.4.10]. In general, a nuclear space need not be
separable [20, Paragraph 4.4.9]. Also, a separable nuclear space need not be
sequentially separable; an example is $\mathbb{R}^{I}$, $I=[0,1]$, see [16,
14.4.H] (its nuclearity follows from [20, Proposition 5.2.1]).

We now turn to permanence properties, showing that surjectivity of
$\bar{\kappa}$ carries over to arbitrary direct sums and to complemented
subspaces, while property (${\mathcal{R}}$) is inherited by regular inductive
limits.

_Direct sums._ Let $E_{\gamma}$, $\gamma\in\Gamma$ be an arbitrary family of
locally convex spaces and $E=\sum_{\gamma\in\Gamma}E_{\gamma}$ be its locally
convex direct sum (i.e., topologized by the locally convex inductive limit
topology with respect to the injections $\iota_{\gamma}:E_{\gamma}\to E$). Its
dual is the product space
$E^{\prime}=\prod_{\gamma\in\Gamma}E_{\gamma}^{\prime}$; it is endowed with
the product $\sigma$-algebra with respect to the individual $\sigma$-algebras
on $E_{\gamma}^{\prime}$. We can consider the maps
$\bar{\kappa}_{\gamma}:L^{0}(\Omega\\!:\\!E_{\gamma}^{\prime})\to{\mathcal{L}}_{c}(E_{\gamma}\\!:\\!L^{0}(\Omega))$
as well as
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$.
Let $Y\in{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$. We can consider
$Y|E_{\gamma}=Y\circ\iota_{\gamma}$, which clearly belongs to
${\mathcal{L}}_{c}(E_{\gamma}\\!:\\!L^{0}(\Omega))$ (since $\iota_{\gamma}$ is
continuous).

######  Proposition 6.3.

Let $E=\sum_{\gamma\in\Gamma}E_{\gamma}$. Then $\bar{\kappa}$ is surjective if
and only if all $\bar{\kappa}_{\gamma}$, $\gamma\in\Gamma$ are surjective,.

_Proof._ The if-part: By assumption, there are $X_{\gamma}\in
L^{0}(\Omega\\!:\\!E_{\gamma}^{\prime})$, $\gamma\in\Gamma$, such that
$\bar{\kappa}_{\gamma}X_{\gamma}=Y|E_{\gamma}$. As noted earlier, we may
assume that $X_{\gamma}(\omega)$ is defined for all $\omega\in\Omega$. We set
$X(\omega)=(X_{\gamma}(\omega))_{\gamma\in\Gamma}\in\prod_{\gamma\in\Gamma}E_{\gamma}^{\prime}=E^{\prime}$.
Clearly, $X$ is measurable, hence belongs to $L^{0}(\Omega\\!:\\!E^{\prime})$.

Recall that each $e\in E$ is of the form $e=\sum_{\rm finite}e_{\gamma}$ with
$e_{\gamma}\in E_{\gamma}$. Then

$\bar{\kappa}X(e)(\omega)=X(\omega)(e)=\sum_{\rm
finite}X_{\gamma}(\omega)(e_{\gamma})=\sum_{\rm
finite}\bar{\kappa}_{\gamma}X_{\gamma}(e_{\gamma})(\omega)=\sum_{\rm
finite}Y(e_{\gamma})(\omega).$

Since the sum is finite, this holds on a set of full measure. Thus
$\bar{\kappa}X(e)=Y(e)$ in $L^{0}(\Omega)$.

The only-if-part: Fix some $\gamma\in\Gamma$ and let
$Y_{\gamma}\in{\mathcal{L}}_{c}(E_{\gamma}\\!:\\!L^{0}(\Omega))$. Extend
$Y_{\gamma}$ to $Y\in{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$ by setting
$Y\circ\iota_{\delta}=0$, $\delta\neq\gamma$, and
$Y\circ\iota_{\gamma}=Y_{\gamma}$. By assumption, there is $X\in
L^{0}(\Omega\\!:\\!E^{\prime})$ such that $\bar{\kappa}X=Y$. Set
$X_{\gamma}=X\circ\iota_{\gamma}\in L^{0}(\Omega\\!:\\!E_{\gamma}^{\prime})$.
If $f\in E_{\gamma}$, then
$\bar{\kappa}_{\gamma}X_{\gamma}(f)(\omega)=\bar{\kappa}_{\gamma}X(\iota_{\gamma}(f))(\omega)=X(\omega)(\iota_{\gamma}(f))=Y(\iota_{\gamma}(f))(\omega)=Y_{\gamma}(f)(\omega)$.
Thus $\bar{\kappa}_{\gamma}X_{\gamma}=Y_{\gamma}$. $\Box$

Proposition 6.3 is of interest, because uncountable direct sums of nuclear
spaces are not nuclear [18, Remark above Corollary 28.8], yet $\bar{\kappa}$
is surjective.

######  Remark 6.4.

Let $\Lambda$ be any index set and $E=\mathbb{R}^{(\Lambda)}$. The finest
locally convex topology on $E$ coincides with the direct sum topology on
$\mathbb{R}^{(\Lambda)}$. Further, any converging sequence in $E$ is contained
in a finite dimensional subspace $F$ [22, Spplements III(2) and V(1)]. Using
coordinates in $F$, it is quite obvious that any linear map $Y:F\to
L^{0}(\Omega)$ is pathwise continuous. Thus

${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))\subset{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
path}\subset{\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm
prob}\subset{\mathcal{L}}(E\\!:\\!L^{0}(\Omega)),$

so all spaces are equal when $E$ is equipped with the finest locally convex
topology. In particular,
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))={\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$. Similarly,

$L^{0}(\Omega\\!:\\!E^{\ast})=L^{0}(\Omega\\!:\\!E^{\prime})=L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
path}$

in this case. Since $\mathbb{R}$ is nuclear, Proposition 6.3 gives that
$\kappa_{0}=\kappa_{1}=\bar{\kappa}$ is surjective, resulting in an
alternative proof of Proposition 6.3.

######  Corollary 6.5.

Surjectivity of $\bar{\kappa}$ is inherited by complemented subspaces.

_Proof._ A complemented subspace $E_{1}$ of a locally convex space possesses,
by definition, a topological complement $E_{2}$, so that $E$ equals the direct
sum $E_{1}\oplus E_{2}$. Apply Proposition 6.3. $\Box$

_Countable inductive limits._ Let $E_{1}\subset E_{2}\subset\ldots\subset
E_{n}\subset\ldots$ be a sequence of locally convex spaces with continuous
inclusion maps. Let the linear space $E$ be their union. If $E$ carries the
finest locally convex topology making all inclusions $E_{n}\subset E$
continuous, then it is called the _inductive limit_ of the spaces $E_{n}$. The
inductive limit is called _regular_ if each bounded subset of $E$ is contained
in some $E_{n}$ and bounded there [16, Section 4.5]. As above, we can consider
the mappings
$\bar{\kappa}_{n}:L^{0}(\Omega\\!:\\!E_{n}^{\prime})\to{\mathcal{L}}_{c}(E_{n}\\!:\\!L^{0}(\Omega))$
as well as
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$.

######  Proposition 6.6.

Let $E$ be the regular inductive limit of the spaces $E_{n}$. If all
$\bar{\kappa}_{n}$, $n\in\mathbb{N}$ are bijective, then $\bar{\kappa}$ is
bijective.

_Proof._ Let $Y\in{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$. This means that
if $e_{k}\to e$ in $E$, then $Y(e_{k})\to Y(e)$ in $L^{0}(\Omega)$. Fix
$n\in\mathbb{N}$. If $e_{k}\to e$ in $E_{n}$ then $e_{k}\to e$ in $E$. Thus
$Y|E_{n}\in{\mathcal{L}}_{c}(E_{n}\\!:\\!L^{0}(\Omega))$. By assumption, there
exists $X_{n}\in L^{0}(\Omega\\!:\\!E_{n}^{\prime})$ such that
$\bar{\kappa}_{n}X_{n}=Y|E_{n}$. Further, if $X_{n}^{\prime}\in
L^{0}(\Omega\\!:\\!E_{n}^{\prime})$ also satisfies
$\bar{\kappa}_{n}X_{n}^{\prime}=Y|E_{n}$, then $X_{n}=X_{n}^{\prime}$ almost
surely.

Next, observe that for all $e\in E_{n}$, $\bar{\kappa}_{n+1}X_{n+1}(e)=Y(e)$
in $L^{0}(\Omega)$. This implies that
$\bar{\kappa}_{n}(X_{n+1}|E_{n})(e)=Y(e)=\bar{\kappa}_{n}X_{n}(e)$. Thus
$\bar{\kappa}_{n}(X_{n+1}|E_{n})=\bar{\kappa}_{n}X_{n}=Y|E_{n}$. The
injectivity of $\bar{\kappa}_{n}$ implies that there is a set
$\Omega_{n}\subset\Omega$ of full measure such that
$X_{n+1}(\omega)|E_{n}=X_{n}(\omega)$ for all $\omega\in\Omega_{n}$.

Let $\Omega^{\prime}=\bigcap_{n=1}^{\infty}\Omega_{n}$. Then
$P(\Omega^{\prime})=1$ and we have obtained a sequence of functionals
$X_{n}\in L^{0}(\Omega\\!:\\!E_{n}^{\prime})$ such that
$X_{n+m}(\omega)(e)=X_{n}(\omega)(e)$ for all $m,n\in\mathbb{N}$, $e\in E_{n}$
and $\omega\in\Omega^{\prime}$. For $\omega\in\Omega^{\prime}$, the map

$X(\omega):E=\bigcup_{n=1}^{\infty}E_{n}\to\mathbb{R}$

given by $X(\omega)(e)=X_{n}(\omega)(e)$ for $e\in E_{n}$ is well-defined. It
is clearly linear. Let $e_{k}\to e$ in $E$. By regularity, there is
$m\in\mathbb{N}$ such that all $e_{k}$ and $e$ belong to $E_{m}$. The topology
of $E_{m}$ is finer than the one induced by $E$ on $E_{m}$, hence $e_{k}\to e$
in $E_{m}$, and so $X(\omega)(e_{k})=X_{m}(\omega)(e_{k})\to
X_{m}(\omega)(e)=X(\omega)(e)$. Thus $X(\omega):E\to\mathbb{R}$ is
sequentially continuous and hence belongs to $E^{\prime}$, the sequential dual
of $E$. Since $\omega\to X(\omega)(e)$ is measurable for all $e\in E$, $X$ is
measurable when $E^{\prime}$ is endowed with the Kolmogorov $\sigma$-algebra.
Summarizing, we have obtained $X\in L^{0}(\Omega\\!:\\!E^{\prime})$ such that
$\bar{\kappa}X=Y$, and $\bar{\kappa}$ is seen to be surjective.

Assume that $X^{\prime}\in L^{0}(\Omega\\!:\\!E^{\prime})$ such that
$\bar{\kappa}X^{\prime}=Y$. This means in particular for $e\in E_{n}$ that
$X^{\prime}(\omega)(e)=Y(e)(\omega)=X_{n}(\omega)(e)$ on a set of full
measure; thus $\bar{\kappa}_{n}(X^{\prime}|E_{n})=\bar{\kappa}_{n}X_{n}$. By
the assumed injectivity of $\bar{\kappa}_{n}$, there is a set
$\Omega_{n}^{\prime}\subset\Omega$ of full measure such that
$X^{\prime}(\omega)|E_{n}=X_{n}(\omega)$ for all
$\omega\in\Omega_{n}^{\prime}$. Let
$\Omega^{\prime\prime}=\bigcap_{n=1}^{\infty}\Omega_{n}^{\prime}$. Then
$P(\Omega^{\prime\prime})=1$ and $X^{\prime}(\omega)=X(\omega)$ for
$\omega\in\Omega^{\prime\prime}$, so $\kappa$ is seen to be injective. $\Box$

######  Example 6.7.

The following spaces of smooth functions have property (${\mathcal{R}}$):

${\mathcal{S}}(\mathbb{R}^{n}),\ {\mathcal{E}}(\mathbb{R}^{n}),\
{\mathcal{D}}_{K}(\mathbb{R}^{n}),\ {\mathcal{D}}(\mathbb{R}^{n})$

where $K\subset\mathbb{R}^{n}$ is compact. The first three spaces are
metrizable nuclear spaces (see e.g. [26, Section 51], [20, Section 6.2]). The
space ${\mathcal{D}}(\mathbb{R}^{n})$ is the strict inductive limit of the
spaces ${\mathcal{D}}_{K}(\mathbb{R}^{n})$, so one can either use Proposition
6.6 or invoke its nuclearity [14, Section 1.5] and sequential separability.

##  7 Counterexamples and mapping properties

Counterexamples to injectivity. The meaning of the maps
$\kappa,\kappa_{0},\kappa_{1},\kappa_{2},\bar{\kappa}$ and
$\iota,\iota_{0},\iota_{1},\iota_{2}$ can be recalled from the diagram in
Section 5.

######  Example 7.1.

The map $\kappa$ is not necessarily injective. Indeed, take
$\Omega=\mathbb{R}$ with any probability measure which has a continuous,
everywhere positive density, and take $E=\mathbb{R}$. Let

$X(\omega)(e)=\left\\{\begin{array}[]{ll}1,&e=\omega,\\\
0,&e\neq\omega.\end{array}\right.$

Then $\\{\omega\in\Omega:\forall e\in E,X(\omega)(e)=0\\}=\emptyset$, so
$[X]\neq[0]$. But $[\kappa X(e)]=0$ in $L^{0}(\Omega)$ for all $e\in E$,
because $\\{\omega\in\Omega:X(\omega)(e)\neq 0\\}=\\{e\\}$, and
$P(\\{e\\})=0$.

The process $X$ in Example 7.1 is a version of the zero operator $Y\equiv 0$
which belongs, in particular, to ${\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$. Note that $X$ is not only nonzero, but also
nonlinear, since

$\\{\omega\in\Omega:\forall
e,f,X(\omega)(e+f)=X(\omega)(e)+X(\omega)(f)\\}=\emptyset.$

Thus $X\in L^{0}(\Omega\\!:\\!\mathbb{R}^{E})$ is a version of $Y\equiv 0$
which does not belong to $L^{0}(\Omega\\!:\\!E^{\ast})$.

######  Example 7.2.

A generalization of Example 7.1 shows that again $\kappa_{0}$ need not be
injective, in general. To see this, take $\Omega=\mathbb{R}$ with a
probability measure as in Example 7.1 and set $E=\mathbb{R}^{(\mathbb{R})}$.
Equip $E^{\ast}=\mathbb{R}^{\mathbb{R}}$ with the Kolomogorov $\sigma$-algebra
${\mathcal{B}}_{K}(\mathbb{R}^{\mathbb{R}})$. Define

$X\in L^{0}(\Omega\\!:\\!E^{\ast})\ {\rm by\
}X(\omega)(e)=\pi_{\omega}(e)=e(\omega),\ e\in E.$

Then $\\{\omega\in\Omega:\forall e\in E,X(\omega)(e)=0\\}=\emptyset$, so
$[X]\neq[0]$. But $[\kappa_{0}X(e)]=0$ in $L^{0}(\Omega)$ because
$\\{\omega\in\Omega:X(\omega)(e)\neq 0\\}$ is finite for all $e\in E$, hence
has zero probability.

######  Example 7.3.

If $E$ in Example 7.2 is given the finest locally convex topology (see Remark
6.4), then $E^{\prime}=E^{\ast}$. The process $X$ in Example 7.2 belongs to
$L^{0}(\Omega\\!:\\!E^{\ast})=L^{0}(\Omega\\!:\\!E^{\prime})=L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
prob}$, and $X\neq 0$ as a member of those spaces. On the other hand,
$\kappa_{0}X(e)=\kappa_{1}X(e)=\kappa_{2}X(e)=\bar{\kappa}X(e)=0$ in
$L^{0}(\Omega)$ for all $e\in E$. This shows that $\kappa_{1}$, $\kappa_{2}$
and $\bar{\kappa}$ are not injective in this situation.

Note that $\mathbb{R}^{(\mathbb{R})}$ is not separable [16, 14.4.H], so this
example shows that the hypothesis of Proposition 5.4 cannot be dropped, in
general.

Counterexamples to surjectivity. For completeness, we state possible lack of
surjectivity of $\iota$ and $\iota_{0}$. It trivially rests on the fact that
there are locally convex spaces $E$ such that $E^{\ast}\neq E^{\prime}$.

######  Example 7.4.

There is a probability space $(\Omega,\Sigma,P)$ and a locally convex space
$E$ such that $\iota$ and $\iota_{0}$ are not surjective.

In fact, it suffices to take any space $E$ such that $E^{\prime}\neq E^{\ast}$
and a one-point probability space $\Omega$. If $X\in E^{\ast}\setminus
E^{\prime}$, there is $e_{n}\to e$ such that $X(e_{n})\not\to X(e)$. There is
$\varepsilon>0$ such that $|X(e_{n})-X(e)|>\varepsilon$ for infinitely many
$n$. As every event on a one-point probability space, this event has
probability 1. Thus
$x\in{\mathcal{L}}^{0}(\Omega\\!:\\!E^{\ast})\setminus{\mathcal{L}}^{0}(\Omega\\!:\\!E^{\ast})_{\rm
prob}$, so $\iota$ is not surjective. The same reasoning applies to
$\iota_{0}$, considering $X$ as a map $E\to L^{0}(\Omega)=\mathbb{R}$.

The next examples show that Theorem 6.1(a) does not hold in general: without
the nuclearity assumption, $\bar{\kappa}$ need not be surjective. It may even
happen that $L^{2}$-continuous elements of $L^{0}(\Omega\\!:\\!E^{\ast})$ do
not have a version in $E^{\prime}$.

######  Example 7.5.

We construct a Gaussian probability space $(\Omega,\Sigma,P)$, a normed space
$E$ and a linear random functional $X\in L^{0}(\Omega\\!:\\!E^{\ast})_{\rm
prob}$, $\bar{\kappa}X\in{\mathcal{L}}_{\rm c}(E\\!:\\!L^{2}(\Omega))$ such
that $P(\\{\omega\in\Omega:X(\omega)\not\in E^{\prime}\\})=1$.

Indeed, let $\Omega_{0}=(\mathbb{R},{\mathcal{B}}(\mathbb{R}),P_{0})$ where
$P_{0}$ is the law of a standard normal distribution ${\mathcal{N}}(0,1)$. We
let $\Omega$ be the product space $\Omega_{0}^{\mathbb{N}}$, more precisely,
$\Omega=(\mathbb{R}^{\mathbb{N}},{\mathcal{B}}_{K}(R^{\mathbb{N}}),\otimes
P_{0})$ where $P=\otimes P_{0}$ is the product probability.

Next, let $E=\mathbb{R}^{(\mathbb{N})}$ eqipped with the $\ell^{2}$-norm
(actually any $\ell^{p}$ with $1\leq p\leq 2$ would do). Since $E$ is dense in
$\ell^{2}$, we have $E^{\prime}=\ell^{2}$, while
$E^{\ast}=\mathbb{R}^{\mathbb{N}}$. The elements of $E$ are of the form

$e=\sum_{\rm finite}\alpha_{j}e_{j}$

with $\alpha_{j}\in\mathbb{R}$ and $e_{j}$ the $j$th unit vector.

Define the linear random functional $X$ by
$X(\omega)(e)=\langle\omega,e\rangle$. Clearly, $X(\omega)\in E^{\ast}$ for
all $\omega$ and

$\langle\omega,e\rangle=\sum_{\rm finite}\omega_{j}\alpha_{j}\quad{\rm
when}\quad e=\sum_{\rm finite}\alpha_{j}e_{j}.$

As a finite sum of measurable functions, $\omega\to X(\omega)(e)$ is
measurable for all $e\in E$, so $X\in L^{0}(\Omega\\!:\\!E^{\ast})$.

_The probability distribution of $X(\cdot)(e)$._ Clearly, it is Gaussian with
mean zero, $\mathrm{E}\big{(}X(\cdot)(e)\big{)}=0$ and variance

$\mathrm{V}\big{(}X(\cdot)(e)\big{)}=\mathrm{E}\big{(}X(\cdot)(e)^{2}\big{)}=\mathrm{E}\Big{(}\sum_{\rm
finite}\omega_{j}\alpha_{j}\sum_{\rm
finite}\omega_{k}\alpha_{k}\Big{)}=\sum_{\rm
finite}\alpha_{j}^{2}=\|e\|_{\ell^{2}}^{2}$

since $\mathrm{E}(\omega_{j}\omega_{k})=\delta_{jk}$.

Let $e^{(n)},e\in E$ and $e^{(n)}\to e$ in $\ell^{2}$. Then

$\mathrm{E}\Big{(}\big{(}X(\cdot)(e^{(n)})-X(\cdot)(e)\big{)}^{2}\Big{)}=\mathrm{E}\Big{(}\big{(}X(\cdot)(e^{(n)}-e)\big{)}^{2}\Big{)}=\|e^{(n)}-e\|_{\ell^{2}}\to
0,$

thus $\kappa X\in{\mathcal{L}}_{\rm c}(E\\!:\\!L^{2}(\Omega))$.

Finally, we show that for almost all $\omega\in\Omega$, $X(\omega)\not\in
E^{\prime}$. Note that

$\\{\omega\in\Omega:X(\omega)\not\in E^{\prime}\\}=\\{\omega\in\Omega:\exists
e^{(n)}\to e,X(\omega)(e^{(n)})\not\to X(\omega)(e)\\}$

Let $e^{(n)}=\alpha_{n}e_{n}$ with $\alpha_{n}\to 0$ to be specified. Then
$e^{(n)}\to 0\equiv e$ in $\ell^{2}$. Let

$A_{n}=\\{\omega\in\Omega:X(\omega)(e^{(n)})\geq
1\\}=\\{\omega\in\Omega:\omega_{n}\alpha_{n}\geq
1\\}=\\{\omega\in\Omega:\omega_{n}\geq\tfrac{1}{\alpha_{n}}\\}.$

Let

$A=\\{\omega\in\Omega:\omega\in A_{n}\ {\rm for\ infinitely\ many\ }n\\}.$

Note that the $A_{n}$ are independent. With this choice of $e^{(n)}$ and
$e\equiv 0$,

$A\subset\\{\omega\in\Omega,X(\omega)(e^{(n)})\not\to
0\\}\subset\\{\omega\in\Omega,X(\omega)\not\in E^{\prime}\\}.$

We will show that $\sum_{n=1}^{\infty}P(A_{n})=\infty$. The Borel-Cantelli-
Lemma then implies that $P(A)=1$. But

$P(A_{n})=\frac{1}{\sqrt{2\pi}}\int_{1/\alpha_{n}}^{\infty}{\rm e}\hskip
1.0pt^{-y^{2}/2}\hskip 1.0pt{\rm
d}y\geq\frac{1}{\sqrt{2\pi}}\int_{1/\alpha_{n}}^{\infty}y{\rm e}\hskip
1.0pt^{-y^{2}}\hskip 1.0pt{\rm d}y=\frac{1}{2\sqrt{2\pi}}{\rm e}\hskip
1.0pt^{-1/\alpha_{n}^{2}}.$

Choosing $\alpha_{n}=(\log n)^{-1/2}$, we have $P(A_{n})\geq n/2\sqrt{2\pi}$.
With this choice, $P(A)=1\leq P\big{(}\\{\omega\in\Omega,X(\omega)\not\in
E^{\prime}\\}\big{)}$.

######  Example 7.6.

Example 7.5 can be further exploited to show that the maps $\iota_{1}$ and
$\iota_{2}$ need not be surjective. It follows from its proof that the process
$X$ is $L^{2}$-continuous: if $e^{(n)}\to e$ in $E$, then
$X(\cdot)(e^{(n)}-e)\to 0$ in $L^{2}(\Omega)$. This implies that
$X\in{\mathcal{L}}(\Omega\\!:\\!E^{\ast})_{\rm prob}$, but it does not belong
to
${\mathcal{L}}(\Omega\\!:\\!E^{\prime})={\mathcal{L}}(\Omega\\!:\\!E^{\ast})_{\rm
path}$. Thus $\iota_{1}$ is not surjective.

The statement $P\big{(}\\{\omega\in\Omega,X(\omega)\not\in
E^{\prime}\\}\big{)}=1$ can also be read as

$P\big{(}\\{\omega\in\Omega:\exists e^{(n)}\to
e,\kappa_{2}X(e^{(n)})(\omega)\not\to\kappa_{2}X(e)(\omega)\\}\big{)}=1,$

showing that $\kappa_{2}X$ does not belong to
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm path}$, while clearly $\kappa_{1}X$
belongs to ${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm prob}$. Thus $\iota_{2}$
is not surjective either.

######  Example 7.7.

Let $\Omega=(0,1)$ with the Borel $\sigma$-algebra
$\Sigma={\mathcal{B}}(\mathbb{R})$ and the uniform probability measure $P$.
Let $E=L^{p}(0,1)$, where $1\leq p<\infty$. Define
$Y\in{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$ by $Y(e)=e$. Then there does
not exist $X\in L^{0}(\Omega\\!:\\!E^{\prime})$ such that $\bar{\kappa}X=Y$.
Indeed, such an $X$ would have to satisfy $X(\omega)(e)=e(\omega)$ on some set
of full measure $\Omega_{e}$, possibly depending on $e$. There are two reasons
why this cannot happen. First, let $(e_{n})_{n\in\mathbb{N}}$ be any
convergent sequence in $L^{p}(0,1)$, say with limit $e$. We have
$X(\omega)(e_{n})=e_{n}(\omega)$ on the set
$\Omega^{\prime}=\bigcap_{n=1}^{\infty}\Omega_{e_{n}}\cap\Omega_{e}$, which
has full measure. If $X(\omega)$ were a member of $E^{\prime}$, then
$X(\omega)(e_{n})\to X(\omega)(e)$ for $\omega\in\Omega^{\prime}$. But this
means that $e_{n}(\omega)\to e(\omega)$ almost everywhere. The argument
implies that every $L^{p}$-converging sequence converges almost everywhere,
which is not true.

Second, one could also argue that $X(\omega)$ is a point-evaluation
functional, which cannot belong to $E^{\prime}=L^{q}(0,1)$,
$\frac{1}{p}+\frac{1}{q}=1$. This is excluded because $X(\omega)$ restricted
to $C[0,1]$ is the evaluation functional (Dirac measure) at point $\omega$. It
is wellknown that the evaluation functional on $C[0,1]$ does not have an
$L^{p}$-continuous extension to $L^{p}(0,1)$.

Example 7.7 is another instance where $\bar{\kappa}$ is not surjective. We now
produce an example in which even $\kappa_{2}$ is not surjective, thus even
pathweise continuous linear random functionals need not have a version in the
dual.

######  Example 7.8.

Take again $\Omega=(0,1)$, $E=L^{\infty}(0,1)$ and
$Y\in{\mathcal{L}}_{c}(E\\!:\\!L^{0}(\Omega))$ given by $Y(e)=e$. Clearly, $Y$
belongs to ${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))_{\rm path}$. There does not
exist $X\in L^{0}(\Omega\\!:\\!E^{\prime})$ such that $\bar{\kappa}X=Y$.
Indeed, we would again have that $X(\omega)(e)=e(\omega)$, so $X(\omega)$,
restricted to $C[0,1]$, would simply be the Dirac measure at $\omega$.
According to [4, Theorem IV.8.16], the dual of $L^{\infty}(0,1)$ is equal to
the space of bounded additive set functions on the Lebesgue extension of
$\Sigma$, which vanish on sets of Lebesgue measure zero. The latter is
obviously false for the Dirac measure, which hence cannot belong to the dual
of $L^{\infty}(0,1)$.

We can summarize the mapping properties as follow.

(a) As was shown in Sections 3 and 4, the maps $\kappa$ and $\kappa_{0}$ are
always surjective.

(b) The maps $\kappa$, $\kappa_{0}$, $\kappa_{1},\kappa_{2}$ and
$\bar{\kappa}$ need not be injective (Examples 7.1, 7.2, 7.3).

(c) The maps $\iota$ and $\iota_{0}$ need not be surjective (Example 7.4).

(d) The maps $\bar{\kappa}$, $\iota_{1}$, $\iota_{2}$ and $\kappa_{2}$ are not
necessarily surjective (Examples 7.5 and 7.7, Example 7.6, Example 7.8).

It remains open whether $\kappa_{1}$ is surjective, in general. If a locally
convex space has property (${\mathcal{R}}$), then $\bar{\kappa}$ as well as
$\kappa_{2}$ and $\iota_{2}$ are bijective, and $\kappa_{1}$ is surjective
(Remark 5.3).

##  Appendix A Appendix: Regularity and nuclear spaces

The present paper relies on the theorem that for every nuclear locally convex
space the map $\bar{\kappa}$ is surjective. In a slightly more general form,
the theorem has been proved by Itô and Nawata [15] and in Itô [14]. A somewhat
less general version, simplifying Itô’s proof, is due to Walsh [28]. The
purpose of this appendix is to record Itô’s version [14] of the theorem for
reference in the paper, and to compare it with the version of Walsh [28]. The
proofs of this result are easily readable both in [14] and [28] and will not
be reproduced here.

Following Itô’s presentation [14, Section 1.2], a topology $\tau$ on a linear
space $E$ is called _multi-Hilbertian_ , if it is generated by a directed
family ${\mathcal{P}}$ of separable Hilbertian seminorms. If ${\mathcal{P}}$
is countable, $(E,\tau)$ is referred to as _countably Hilbertian_.

Let $p,q$ be two separable Hilbertian seminorms on $E$. Then $p$ is called
_Hilbert-Schmidt bounded_ by $q$ if $\sum_{n=1}^{\infty}p(e_{n})^{2}$ is
finite for some (and hence any [14, Remark 1.1.2]) orthonormal basis
$(e_{n})_{n\in\mathbb{N}}$ in $(E,q)$. Let $\tau_{1},\tau_{2}$ be multi-
Hilbertian topologies on $E$. Then $\tau_{1}$ is termed _Hilbert-Schmidt
weaker_ than $\tau_{2}$, if every $\tau_{1}$-continuous (separable Hilbertian)
seminorm $p$ is Hilbert-Schmidt bounded by some $\tau_{2}$-continuous
(separable Hilbertian) seminorm $q$. Given a multi-Hilbertian topology $\tau$
on $E$, the strongest multi-Hilbertian topology Hilbert-Schmidt weaker than
$\tau$ is called the _Kolmogorov $I$-topology of $\tau$_ and denoted by
$I(\tau)$. It is generated by the family of separable Hilbertian seminorms
which are Hilbert-Schmidt weaker than some $\tau$-continuous (separable
Hilbertian) seminorm.

For the notion of a _nuclear locally convex space_ we follow the classical
sources [9, Chap. II §2, no 1], [20, Sect. 4.1], see also [16, 23, 26].

######  Proposition A.1.

A locally convex space $(E,\tau)$ is nuclear if and only if it is multi-
Hilbertian and $I(\tau)=\tau$.

_Proof._ Let ${\mathcal{P}}$ be a directed fundamental system of seminorms
defining the topology $\tau$. For $p\in{\mathcal{P}}$, let $E_{p}$ be the
completion of the normed space $E/p^{-1}(0)$. Recall that the topology $\tau$
on $E$ is equal to the projective topology with respect to the canonical maps
$\iota^{p}:E\to E_{p}$, $p\in{\mathcal{P}}$ [18, Remark 24.5(a)]. For $p\leq
q$, the identity map induces a continuous linear map $\iota_{q}^{p}:E_{q}\to
E_{p}$ such that $\iota_{q}^{p}\circ\iota^{q}=\iota^{p}$.

According to [18, Proposition 28.4], a locally convex space is nuclear if and
only if it has a fundamental system ${\mathcal{P}}$ of Hilbertian seminorms
and for every continuous Hilbertian seminorm $p$ there exists a continuous
Hilbertian seminorm $q$, $p\geq q$ such that $\iota_{q}^{p}:E_{q}\to E_{p}$ is
a Hilbert-Schmidt operator. What is more, using [18, Lemma 29.7] and its proof
together with [18, Remark 24.5(b)], for each continuous Hilbertian seminorm
$p$, there is a countable orthonormal basis in $E_{p}^{\prime}$. In
particular, $E_{p}^{\prime}$ and consequently $E_{p}$ are separable. Thus
every nuclear space is multi-Hilbertian.

It is easy to see that $p$ is Hilbert-Schmidt bounded by $q$ if and only if
the map $\iota_{q}^{p}:E_{q}\to E_{p}$ is a Hilbert-Schmidt operator. (The
essential argument can be found e.g. in [20, Section 2.5]. To go over to the
quotient spaces, one may use [18, Remark 12.5(a)] together with [14, Theorem
1.1.1(i)].)

It follows that for a multi-Hilbertian space, the assertion that
$I(\tau)=\tau$ is equivalent with nuclearity. $\Box$

In order to formulate the central _regularization theorem_ , we recall some
terminology from [14, Section 2.3]. As in the present paper, the elements of
${\mathcal{L}}(E\\!:\\!L^{0}(\Omega))$ are called _linear random functionals_
on $E$. Let $E$ be equipped with a multi-Hilbertian topology $\tau$. The
elements of ${\mathcal{L}}_{\rm c}(E\\!:\\!L^{0}(\Omega))$ are referred to as
_$\tau$ -continuous linear random functionals_ in [14, (2) in Section 2.3]. An
$E_{\tau}^{\prime}$-valued random variable $X$ (i.e., an element of
$L^{0}(\Omega\\!:\\!E_{\tau}^{\prime})$) is called _$\sigma$ -concentrated_ if
there exists a countably Hilbertian topology $\theta$ coarser than $\tau$ such
that $P(X\in E_{\theta}^{\prime})=1$, [14, Definition 2.3.1]. Let
$Y\in{\mathcal{L}}_{\rm c}(E\\!:\\!L^{0}(\Omega))$. Then an
$E_{\tau}^{\prime}$-valued random $X$ is called a _$\tau$ -regular version_ of
$Y$ if (i) $X$ is $\sigma$-concentrated and (ii) for every $e\in E$,
$X(e)=Y(e)$ almost surely ([14, Definition 2.3.1]). In the language of Section
5, (ii) means that $\bar{\kappa}X=Y$. Collecting Theorems 2.3.1, 2.3.2 and
2.3.3 from [14] gives the central result:

######  Theorem A.2.

(The regularization theorem [14, 15]) A linear random functional $Y$ on $E$
has a $\tau$-regular version if and only if it is $I(\tau)$-continuous. It is
unique in the sense that if $X$ and $X^{\prime}$ are $\tau$-regular versions
of $Y$, then $X=X^{\prime}$ almost surely.

######  Corollary A.3.

If $E$ is a nuclear space, then
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ is surjective.

_Proof._ It follows by combining Theorem A.2 with Proposition A.1. $\Box$

Theorem A.2 states almost sure uniqueness only for $\tau$-regular, hence
$\sigma$-concentrated versions. It is not evident whether this gives
injectivity of $\bar{\kappa}$ in general.

There is an important special case of countably Hilbertian nuclear spaces.
Suppose that the topology of a countably Hilbertian nuclear space $E$ is
generated by a directed sequence of separable Hilbertian norms $p_{1}\leq
p_{2}\leq\ldots\leq p_{n}\leq\ldots$. In addition, suppose that the norms are
compatible [5, Chapter I.2.2].

The Hilbert spaces $E_{m}=E_{p_{m}}$ are defined as above (as the completion
of $E$ with respect to $p_{m}$). Due to compatibility, the maps
$\iota_{n}^{m}:E_{n}\to E_{m}$ ($n\geq m$) are injective. One obtains the
nested sequence of Hilbert spaces $E_{1}\supset E_{2}\supset\ldots
E_{n}\supset\ldots$ with continuous inclusion maps. The space $E$ is equipped
with the corresponding projective topology. The space $E$ is complete if and
only if it coincides with the intersection $E=\bigcap_{n=1}^{\infty}E_{n}$,
see [5, Theorem in Chapter I.3.2] (and, in particular, a Fréchet space). By
the argument given in the proof of Propositon A.1, each $E_{n}$ is separable;
for $E$, separability holds because every metrizable nuclear space is
separable [20, Theorem 4.4.10]. (For the separability results, one may also
refer to [5, Chapter I.6.5] and [6, Chapter I.3.4].) In particular, a nuclear
space $E=\bigcap_{n=1}^{\infty}E_{n}$ as above is countably Hilbertian and a
separable Fréchet space. This class of spaces has been considered in Gel’fand-
Vilenkin [6, Chapter I.3.2].

The proof of Walsh [28, Theorem 4.1 and Corollary 4.2] shows that
$\bar{\kappa}:L^{0}(\Omega\\!:\\!E^{\prime})\to{\mathcal{L}}_{\rm
c}(E\\!:\\!L^{0}(\Omega))$ is surjective for this narrower class.

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