Wednesday, October 31, 2012

Separable random functions


$\newcommand{\si}{\sigma}$ $\newcommand{\es}{\mathscr{S}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bsT}{\boldsymbol{T}}$ First, what is a random function? To  define it we need a parameter space $\bsT$, $\newcommand{\eS}{\mathscr{S}}$ a  probability space $(\Omega, \eS, P)$, and a target space $X$. Roughly speaking  a  random function $\bsT\to X$ is defined to be a choice of probability measure (and underlying $\si$-algebra of events)  on $X^{\bsT}=$ the space of functions $\bsT\to X$.


In applications $X$ is a metric space and $\bsT$ is a  locally closed subset of some Euclidean space $\bR^N$. (Example to keep in mind:  $\bsT$ an open subset of $\bR^N$ or $\bsT$ a properly embedded submanifold of $\bR^N$. Often $X$ is a vector space.)

A random function on $\bsT$ is then a function

$$ f:\bsT\times\Omega\to X,\;\; \bsT\times\Omega\ni (t,\omega)\mapsto f(t,\omega) \in X, $$

such that, for any $t\in\bsT$,   the   correspondence

$$ \Omega\ni \omega\mapsto f_t(\omega) :=f(t,\omega)\in X $$

is measurable  with respected to the $\si$-algebra of Borel subsets  of $X$. In other words,  a random function on $\bsT$ is a family of random variables (on the same probability space) parameterized by $\bsT$.


Observe that we have a natural  map $\Phi: \Omega\to X^{\bsT}$,

$$\Omega\ni \omega\mapsto f_\omega\in X^\bsT,\;\;f_\omega(t)=f(t,\omega). $$

The pushforward via $\Phi$ of $(\eS,P)$ induces  structure of probability space on $X^{\bsT}$.  The functions $f_\omega$, $\omega\in \Omega$ are called the   sample functions  of the given random function.

Let us observe that  there are certain  properties of functions which a priori may not  measurable subsets of $\Omega$.  For example the set of $\omega$'s such that $f_\omega$ is continuous on $\bsT$   may not be measurable if $\bsT$   is uncountable.  To deal  with such issues  we will restrict our attention to certain  classes of  random functions, namely the  separable ones.

Definition 1.   Suppose that  $\bsT$ is a locally closed subset  of $\bR^N$ and $X$ is a Polish space, i.e., a  complete, separable  metric space. Fix a countable, dense subset $S\subset \bsT$. A random function $f:\bsT\times\Omega\to  X$ is called $S$-separable    if  there exists a negligible subset $N\subset  \Omega$, with the following property: for any closed subset $F\subset X$, any open subset $U\subset \bsT$ the  symmetric  difference of the sets

$$ \Omega(U,F):=\Bigl\{  \omega\in \Omega;\;\;  f_\omega(t)\in F,\forall t\in U\;\Bigr\},\;\;\Omega_S(U, F):=\Bigl\{  \omega\in \Omega;\;\; f_\omega(t)\in F,\;\;\forall t\in U\cap S\;\Bigr\} \tag{1}\label{1}$$

 is a subset of $N$, i.e.,

$$\Omega(U,S)\setminus \Omega_S(U,S),\;\;\Omega_S(U,F)\setminus \Omega(U,F)\subset N. $$


Definition 2.  Let $\bsT$ and $X$ be as in  Definition 1.  A random function  $g: \bsT\times \Omega\to X$ is called a  version of the  random function $f:\bsT\times \Omega\to X$  if

$$ P(g_t=f_t)=1,\;\;\forall t\in\bsT. $$


Let me   give an application of separability.      We say that a random function $g:\bsT\times \Omega\to X$ is  a.s.  continuous if

$$ P\bigl(\;\lbrace \omega;\;\; f_\omega: \bsT\to X\;\;\mbox{is continuous} \rbrace\;\bigr)=1. $$


Proposition 3.   Suppose that $f$ is an $S$-separable  random function $\bsT\times \Omega\to \bR$ and $g$ is a version of $f$.  If $g$ is a.s. continuous, then
\[
P\bigl(\lbrace \omega; \;\;g_\omega=f_\omega\rbrace\bigr)=1.
\]
In particular,  $f$ is a.s. continuous.

Proof.    Consider the set $N\subset \Omega$ in the definition of $S$-separability of $f$.   Define


$$\Omega_*:=\bigl\lbrace \;\omega\in \Omega\setminus N;\;\;g_\omega\;\mbox{is continuous},\;\;g(s,\omega)=f(s,\omega),\;\;\forall s\in S\;\bigr\rbrace. $$

Observe that $P(\Omega_*)=1$.   We will prove that
\[
g_\omega(t)=f_\omega(t),\;\;\forall \omega\in \Omega_*,\;\; t\in\bsT.\tag{$\ast$}\label{ast}
\]

 Fix an open set $U\subset \bsT$. For any $\omega\in\Omega_*$ set

\[
M_\omega(U, S):=  \sup_{t\in S\cap U} f_\omega(t).
\]
Invoking the definition of separability with $F=(-\infty, M_\omega(U, S)]$ we deduce  that
\[
f_\omega(t)\leq M_\omega(U,S),\;\;\forall t\in U,
\]
so that
\[
\sup_{t\in U} f_\omega(t)\leq \sup_{t\in S\cap U} f_\omega(t)\leq \sup_{t\in U} f_\omega(t).
\]
In other words, for any open set $U\subset\bsT$ we have
\[
\sup_{t\in U}f_\omega(t)=\sup_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{2}\label{2}.
\]
A variation of the above argument shows
\[
\inf_{t\in U}f_\omega(t)=\inf_{t\in S\cap U} f_\omega(t),\;\;\forall \omega\in\Omega_*\tag{3}\label{3}.
\]
Now let $\omega\in\Omega_*$,  $t_0\in T$. Given   $\newcommand{\ve}{\varepsilon}$ $\ve>0$, choose  an neighborhood $U=U(\ve \omega)$ of $t_0$ such that
\[
 g_\omega(t_0)-\ve\leq g_\omega(t)\leq g_\omega(t_0)+\ve,\;\;\forall t\in U(\ve,\omega).
\]
Since $g_\omega(t)=f_\omega(t)$ for $t\in S\cap U$ we deduce from (\ref{2}) and (\ref{3}) that
\[
g_\omega(t_0)-\ve \leq \inf_{t\in U(\ve,\omega)} f(t) \leq \sup_{t\in U(\ve,\omega)} f_\omega(t)\leq g_\omega(t_0)+\ve.
\]
In particular, we deduce
\[
g_\omega(t_0)-\ve \leq f_\omega(t_0)\leq g_\omega(t_0)+\ve,\;\;\forall \ve>0.
\]
This proves (\ref{ast}).     Q.E.D.


We have the following result.

Theorem 4.  Suppose that $f:\bsT\times \Omega\to X$ is a random function, where $\bsT$ are Polish spaces. If $X$ is compact, then  $f$ admits a separable version.


Proof.     We follow the approach in Gikhman-Skhorohod. $\DeclareMathOperator{\cl}{\mathbf{cl}}$  Fix a countable dense subset $S\subset \bsT$. $\newcommand{\eV}{\mathscr{V}}$ Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$ and with rational radii. For any  $\omega\in\Omega$  and any  open set $U\subset \bsT$ we set

$$ R(U,\omega):=\cl\bigl\lbrace f_\omega(t);\;\;t\in S\cap U\,\bigr\rbrace, $$

$$ R(t,\omega)=\bigcap_{t\in V\in\eV} R(V,\omega), $$

where $\cl$ stands for the closure of a set.   Observe that $R(t,\omega)\neq\emptyset$ because it is the intersection of a family of compact sets such that any finitely many sets in the family have nonempty intersection.


Lemma 5.  The following statements are equivalent.

(a) The random function  $f$ is $S$-separable.

(b) There exists $N\subset \Omega$ such that $P(N)=0$ and for any $\omega\in\Omega\setminus N$ and any $t\in\bsT$ we have $f_\omega(t)\in R(t,\omega)$.

Proof of the lemma.  (a) $\Rightarrow$ (b)   We know that $f$ is $S$-separable. Choose $N$ as in the definition of $S$-separability.   Fix $\omega_0\in \Omega\setminus N$ and $t_0\in \bsT$.   For any ball $V\in \eV$ that contains $t_0$  we have

\[
\bigl\lbrace \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in S\cap B\;\bigr\}=\bigl\lbrace\; \omega\in \Omega\setminus N;\;\;f_\omega(t)\in R(V,\omega_0)\;\;\forall t\in B\;\bigr\rbrace.
\]
Observe that $\omega_0$ belongs to the set in the left-hand-side of the above equality and so it must belong to the set in the right-hand-side.  Hence

\[
f_{\omega_0}(t)\in R(B,\omega_0),\;\;\forall t\in B
\]
and therefore $f_{\omega_0}(t_0)\in R(B,\omega_0)$ for any $B\in\eV$ that contains $t_0$. Thus $f_{\omega_0}(t_0)\in R(t_0,\omega_0)$ which finishes the proof of the implication (a) $\Rightarrow$ (b).


(b) $\Rightarrow$ (a)   Set $\Omega_*=\Omega\setminus N$.  Suppose that $F\subset X$ is closed.  For any $B\in\eV$  and $\omega\in \Omega_*$ we have

$$ f_\omega(t)\in F\;\;\forall t\in S\cap B \Leftrightarrow F\supset R(B,\omega). $$

Since $R(t,\omega)\subset R(B,\omega)$ for any $t\in B$ we deduce  that

$$ \Omega(B,F):=\bigl\{ \omega\in \Omega_*;\;\; f_\omega(t)\in F\;\;\forall t\in B\;\bigr\} =\bigl\{ \omega\in \Omega_*;\;\; f+\omega(t)\in F\;\;\forall t\in S\cap B\;\bigr\}=\Omega_S(B,F). $$


If $U$  an open set then we can write $U$ as a countable union of balls in $\eV$

$$U=\bigcup_n B_n. $$

Then

$$ \Omega(U,F)=\bigcap_n \Omega(B_n, F)=\bigcap_n\Omega_S(B_n, F)= \Omega_S(U,F). $$

This finishes the proof of the lemma.  q.e.d.



Lemma 6.  For any  Borel set $B\subset X$ there exists a countable subset $C_B\subset \bsT$ such that  for any $t\in\bsT$ the set

$$N(t, B):=\bigl\{ \omega\in \Omega; \;\;f_\omega(\tau)\in B,\;\;\forall \tau\in C_B,\;\;f_\omega(t)\in \bsT\setminus B\;\bigr\} $$

has probability $0$.


Proof of the lemma.   We construct $C_B$ recursively.  Choose $\tau_1\in\bsT$ arbitrarily and set $C_B^1:=\{\tau_1\}$.   Suppose that we have  constructed  $C_B^k=\{\tau_1,\dotsc,\tau_k\}$.    Set



$$ N_k(t):=\bigl\{\omega;\;\; g_\omega(\tau)\in B\;\;\forall \tau\in C_B^k,\;\;f_\omega(t)\in\Omega\setminus B\;\bigl\},\;\;p_k=\sup_{t\in\bsT} P\bigl(\;N_k(t)\;\bigr). $$

Observe that  $p_1\geq p_2\geq \cdots \geq p_k$. If $p_k=0$   we stop and we set $C_B:=C_B^k$.

If this is not the case, there exists $\tau_{k+1}\in \bsT$ such that

$$P\bigl(N_k(\tau_{k+1}\bigr)\geq \frac{1}{2}p_k. $$

Set $ C_B^{k+1}:=C_B^k\cup\{\tau_{k+1}\}$.  Observe that the events $N_1(\tau_2),\dotsc , N_k(\tau_{k+1})$ are mutually  exclusive  and thus

$$ 1\geq \sum_{j=1}^k P(N_j(\tau_{j+1})) \geq \frac{1}{2}\sum_{j=1}^k p_{j+1}. $$

Hence $\lim_{n\to\infty} p_n=0$. Now  set

$$N(t, B):=\bigcap_{k\geq 1} N_k(t). $$
                                                                                                                                                          q.e.d.


Lemma 7.   $\newcommand{\eB}{\mathscr{B}}$ Suppose that $\eB_0$ is a countable family  of Borel subsets of $X$ and $\eB$  is the family obtained by taking the intersections of  all the subfamilies of $\eB_0$.  Then there exists a countable subset $C\subset \bsT$, and for each $t$ a subset $N(t)$ of probability zero  such that for any $B\in \eB$ we have

$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t). $$

Proof.   For any $t\in \bsT$ we define

$$ C:=\bigcup_{B\in\eB_0}C_B, \;\; N(t) :=\bigcup_{B\in\eB_0} N(t,B), $$

where $C_B$  and $N(t,B)$ are  constructed as in Lemma 6. Clearly $C$ is countable.

If $B'\in\eB$ and $B\in\eB_0$ are such that $B\supset B'$, then

$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset \bigl\{ \omega;\;\; f_\omega(\tau)\in B\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B\;\bigr\}\subset N(t,B)\subset N(t). $$

If

$$ B'=\bigcap_{k\geq 1} B_k,\;\;B_k\in\eB_0\;\;\forall k, $$

then


$$ \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B'\;\bigr\}\subset \bigcup_{k\geq 1} \bigl\{ \omega;\;\; f_\omega(\tau)\in B'\;\;\forall \tau\in C\;\; f_\omega(t)\not\in B_k\;\bigr\}\subset \bigcup_{k\geq 1} N(t, B_k)\subset N(t). $$
                                                                                                                                                     q.e.d.


The proof of Theorem 4 is now within reach.  Suppose that $S$ is a countable and dense  set of points in $\bsT$ and $D$ is a countable dense subset of $X$.  Denote by $\eV$ the collection of open balls in $\bsT$ centered at points in $S$. Denote by $\eB_0=\eB_0(D)$ the collection of open balls with  in $X$ of rational radii centered at points in $D$.  As in Lemma 7, denote by $\eB$ the collection of sets obtained by taking intersections of arbitrary families in   $\eB_0$. Clearly $\eB$ contains all the closed subsets of $X$.



Fix a ball $V\in \eV$.  Lemma 7  applied to the restriction of $f$ to $V$ implies the existence of a countable set

$$C(V)\subset V$$
  
and  of a family of  negligible sets

$$N_V(t)\subset \Omega,\;\;t\in V $$

such that  for any $B\in eB$

$$\{ \omega;\;\;  f_\omega(\tau)\in B,\;\;\forall \tau\in C,\;\;f_\omega(t)\in V\setminus B\;\bigr\}\subset N_V(t). $$

Set

$$C=\bigcup_{V\in\eV}C(V),$$

while for $t\in  \bsT$ we set

$$ N(t):=\bigcup_{\eV\niV\ni t}N_V(t).

Clearly $C$ is both countable and dense in $\bsT$. We can now construct a $C$-separable version  $\tilde{f}$ of $f$. Define


  • $\tilde{f}_\omega(t)= f_\omega(t)$  if $t\in C$ or $\omega\not\in N(t)$
  • If $\omega\in N(t)$ and $t\in \bsT\setminus C$ we assign  $\tilde{f}^V_\omega(t)$  an arbitrary value in $R(t,\omega)$.


By construction $\tilde{f}$ is a version of $f$ because for any $t\in \bsT$

$$\{\omega;\;\;\tilde{f}_\omega(t)\neq f_\omega(t)\;\}\subset N(t). $$

Since $f_\omega(\tau)=\tilde{f}_\omega(\tau)$ for any $\tau\in C$, $\omega\in \Omega$ sets $R(t,\omega)$, defined as  as in Lemma 5, are the same for both   functions $\tilde{f}$ and $f$. By construction $\tilde{f}_\omega(t)\in R(t,\omega)$,   $\forall t,\omega$.    Q.E.D.



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