I am reading a book by Billingsley (convergence of probability measures) and he makes a footnote on page 27 which I am struggling to understand. I'll explain the setup below.


Suppose $(X\_n,Y\_n)$ are random elements of $S\times S$, where $S$ is a metric space. Then since the projections $(x,y)\mapsto x$ and $(x,y)\mapsto y$ are continuous, we have that $X$ and $Y$ are random elements of $S$. 


The footnote says that the reverse implication ($X\_n$ and $Y\_n$ random elements of $S\implies$ $(X\_n,Y\_n)$ random element of $S\times S$) holds if $S$ is separable, but not in complete generality. This is what I am struggling to understand. I can't see where separability would be used here.


If we take a measurable set $A\times B\subset S\times S$ and look at the inverse image wrt $(X\_n,Y\_n)$, we just get $(X\_n^{-1}(A),Y\_n^{-1}(B))$ which is measurable by measurability of $X\_n$ and $Y\_n$. Since such measurable sets form a generating $\pi$-system of the product $\sigma$-algebra, this should conclude the result. However I haven't used any notion of separability here, so I am assuming I am mistaken somewhere.


Any help will be appreciated!



Measurable functions in product space
The problem is that the Borel $\sigma$-algebra of the product space $S\times S$ need not be the same as the product $\sigma$-algebra $B(S)\otimes B(S)$ unless $S$ is separable (or more generally, a second-countable topological space). To make your argument work, you want to deal with $B(S)\otimes B(S)$ but measurability is an issue related to $B(S\times S)$.


