If $E[f(X)g(Y)]=E[f(X)g(X)]$ then $E[h(X,Y)]=E[h(X,X)]$. Let E be a metrizable locally compact topological space, let $C_0(E)$ be the set of all continuous real functions on $E$ that tend to $0$ at infinity, and let $X$ and $Y$ be E-valued random variables.
Is the assertion that if $E[f(X)g(Y)]=E[f(X)g(X)]$ for every $f,g \in C_0(E)$ then $E[h(X,Y)]=E[h(X,X)]$ for any bounded Borel measurable function $h$ on $E\times E$ true?
If true, how can we prove it using the monotone class theorem?

Note that disjoint compact subsets $K, L$  of $E$ have disjoint open neighbourhoods $U$ and $V$ respectively; if $f$ and $g$ are continuous functions with $0 \le f, g \le 1$, $f = 1$ on $K$ and $0$ outside $U$ and
$g = 1$ on $L$ and $0$ outside $V$,
$\mathbb E[f(X) g(X)] = 0$ since $f(x) g(x)$ is identically $0$ while $\mathbb E[f(X) g(Y)] \ge \mathbb P(X \in K, Y \in L)$. Thus from $\mathbb E[f(X) g(X)] = \mathbb E[f(X) g(Y)]$ we get $\mathbb P(X \in K, Y \in L) = 0$.  From this we should be able to conclude $X = Y$ almost surely.  If so, $h(X,X) = h(X,Y)$ a.s., and then their expectations are the same.
