TITLE: How can one show that the topology of convergence in measure is separable?
QUESTION [7 upvotes]: Let $X$ be a polish space equiped with the borel sigma-algebra and a probability measure $\mu$. How can one show that the set of all borel measurable functions $f:X\rightarrow R $  ($R$ being the real numbers), where two a.e. equal functions are identified, equiped with the topology of convegence in measure is separable?

REPLY [2 votes]: An alternative is to use the Functional Monotone Class Theorem.
Let $\cal A$ be a countable collection of sets that generates ${\cal B}(X)$,
and put $${\cal K}=\{1_A: A\mbox{ is a finite intersection of }{\cal A}\mbox{ sets }\}.$$
Let ${\cal K}^\prime$ be the (countable!) $\mathbb{Q}$-vector space generated by $\cal K$, and 
set $${\cal M}=\{h: k^\prime_n\to h \mbox{ in probability for some }k^\prime_n\in{\cal K}^\prime\}.$$
Then $\cal K$ and $\cal M$ satisfy the conditions of the FMCT, and hence
 $\cal M$ includes all bounded ${\cal B}(X)$-measurable functions. A truncation argument now shows that any ${\cal B}(X)$-measurable function can be 
approximated in probability by a sequence in ${\cal K}^\prime$.