Is the space of random continuous functions separable? Let $\Omega$ and $X$ be compact metric spaces with $P$ a probability measure on $\Omega$. 
Then we get the following definition of random continuous functions from "Random Probability Measures on Polish Spaces, H. Crauel page 21": let $f : X \times \Omega \to \mathbb{R}$ be a function such that:


*

*for all $x \in X$ the map $\omega \mapsto f(x, \omega)$ is measurable,

*for all $\omega \in \Omega$ the map $x \mapsto f(x, \omega)$ is continuous and bounded,

*$\omega \mapsto \sup\{|f (x, ω)| : x \in X\}$ is integrable with respect to P (it is
measurable by separability of X).


If $f$ and $g$ are both functions satisfying (1)-(3) then identify f and g if $P(\{\omega : f(\cdot, \omega) \not= g(\cdot, \omega) \}$.
Finally, a random continuous function is (the equivalence class of)
a function $f : X \times \Omega \to \mathbb{R}$ satisfying (1)–(3) above. The set of all random continuous functions is a linear space, denoted by $C_\Omega(X)$ and a norm can be defined by
$$
|f|_\infty = \int \sup_{x \in X}|f(x, \omega)|dP(\omega).
$$
My question: is $C_\Omega(X)$ separable?
 A: Yes, this space is separable! Let us first show that $f$ is measurable according to the product $\sigma$-algebra. 
Proof: $\Omega$ is a separable space. Thus, let $(x_n)_{n \in \mathbb{N}}$ be some dense set. Let be $B_{1/n}(x_i)$, $i=1,\ldots,i_n$ be a finite cover of $X$. For minimal $i=1,\ldots,i_n$ with $x \in B_{1/n}(x_i)$ we define $f_n(x,w) = f(x_i,w)$. In other words
$$f_n(x,w) = \sum_{i=1}^{i_n} 1_{A_{i,n}}(x) f(x_i,w),$$
where $A_{i,n} = B_{1/n}(x_i) \setminus (B_{1/n}(x_1) \cup \ldots B_{1/n}(x_{i-1}))$. This is measurable. And if $x \in A_{i(x),n}$, then $|x-x_{j(x,n)}| < 1/n$. And thus $f_n(x,w) \rightarrow f(x,w)$, because of the continuity. So, $f$ is measurable.

This proof gives already rise for a proof of the separability!

Note that, since $\sup_{x \in X} |f_n(x,w)|  \leq \sup_{x \in X} |f(x,w)|$, we get by the dominated convergence theorem also $|f_n-f|_\infty \rightarrow 0$. So it remains to approximate $f_n$ by a fixed set of countable functions, which satify (1) and (2). One problem is that $f_n$ does not satisfy (2).
So, we need to modifty our argument: Since $(B_{1/n}(x_i))$ is an open cover and $X$ compact, we find a finite number of partitions of unity, i.e. $h_{i_1,n}, \ldots h_{i_n,n}$ non-negative continuous functions such that


*

*$\sum_{j=1}^{i_n} h_{i_j,n} = 1$

*$\{ x \in X : h_{i_j,n} >0\}$ is contained in $B_{1/n}(x_{i_j})$.


Define now
$$f_n(x,w) = \sum_{j=1}^{i_n} h_{i_j,n}(x) f(x_{i_j},w).$$
This functions are according to the product $\sigma$-algebra. Moreover, for fixed $(x,w) \in X \times \Omega$, we have
$$|f_n(x,w)-f(x,w)| \leq \sum_{j=1}^{i_n} h_{i_j,n}(x) |f(x_i,w) - f(x,w)| \leq \sup_{i,|x-x_i| \leq 1/n} |f(x_i,w)-f(x,w)|.$$
By continuity, the last term converges to zero. The same argument as before shows also that $|f_n - f|_\infty \rightarrow 0$.

We have showed that functions of the form $\sum_{i=1}^n f_i(x) g_i(w)$ with $f_i \in C(X)$ and $g_i \in L^1(\Omega)$ are dense.

Let $(u_i)_{i \in \mathbb{N}}$ be a dense set in $C(X)$ and $(v_i)_{i \in \mathbb{N}}$ a dense set in $L^1(\Omega)$, then finite sums of products $u_i v_j$ are dense in $C_\Omega(X)$.


*

*Fix $n \in \mathbb{N}$ with $\|f -f_n\| \leq \varepsilon$.

*Take $u_{a_1}, \ldots, u_{a_n}$ with $\|h_{i_j,n} - u_{a_j}\| < \varepsilon/ i_n$.

*Also take $v_{b_1}, \ldots, v_{b_n}$ with $\|f(x_{i_j},\cdot) - v_{b_j}\| < \varepsilon/ i_n$. (Note that $|v_{b_j}|_\infty \leq \varepsilon + |f|_\infty$.)


Define 
$$g = \sum_{j=1}^{i_n} u_{a_j} v_{b_j}.$$
Then
\begin{align}
|f_n-g|_\infty &\leq \sum_{j=1}^{i_n} \int \sup_{x \in X} |u_{a_j}(x) - h_{i_j,n}(x)| |v_{b_j}(w)| \mathop{d \mathbb{P}(w)} + \sum_{j=1}^{i_n} \int |v_{b_j}(w) -f(x_{i_j},w)| \mathop{d \mathbb{P}(w)} \\
& \leq \varepsilon (1+ |f|_\infty) + \varepsilon.
\end{align}
