Let $(S,d)$ be a metric space and let $\mathcal{B}$ the Borel-sigma-algebra of $S$, i.e. the smallest sigma-algebra generated by the metric-topology on $S$.
Consider a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and consider two random elements $X,Y$ in $S$, i.e. $$X,Y: \Omega \to S$$ are $\mathcal{F},\mathcal{B}$-measurable maps.
My book then claims:
If $(S,d)$ is separable, then $d(X,Y)$ is $\mathcal{F}$-measurable.
Is this necessary?
I mean, the map $d: S \times S \to \mathbb{R}$ is continuous where $S \times S$ has the product metric. Thus $d$ is $\mathcal{B}(S \times S) = \mathcal{B} \otimes \mathcal{B}$-measurable. The map $\omega \mapsto (X(\omega), Y(\omega))$ is $\mathcal{F}, \mathcal{B}\otimes \mathcal{B}$-measurable as well, so $d(X,Y)$ is $\mathcal{F}$-measurable as composition of measurable maps.
I don't seem to need that $S$ is separable anywhere?
Best Answer
$\mathcal B \otimes \mathcal B$ is not the Borel sigma algebra of $S \times S$ in general . It is so if $S$ is separable.