$\newcommand{\si}{\sigma}$ $\newcommand{\es}{\mathscr{S}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bsT}{\boldsymbol{T}}$ 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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