That's a question from MSE (here) that did not receive any answer for some days. I migrate it to MO.

Let $X$ and $Y$ be two standard Borel spaces and consider the set $M(X,Y)$ of measurable maps $f: X \to Y$. Is $M(X,Y)$ also standard Borel?

First of all, the cardinality of $M(X,Y)$ is $\mathfrak{c} = 2^{\aleph_0}$ for uncountable $X$ and $Y$ (see Cardinality of the borel measurable functions?) - so this doesn't contradict the Borel ismorphism theorem.

In Srivastava, "A course on Borel sets", he considers the space of $B(X,Y) \subseteq M(X,Y)$ of Baire functions, i.e. continuous functions and closed under pointwise limit. Then he states the Lebesgue – Hausdorff theorem that $B(X,Y) = M(X,Y)$ for metrizable $X$. But I haven't found a theorem or note in the book that says that $B(X,Y)$ is standard Borel.

Moreover, he also states that any Borel measurable function can be made continuous by taking a finer topology on $X$ that doesn't change the Borel $σ$-algebra of $X$, i.e. $X$ is still standard Borel. But I don't see, how to apply this theorem.

Of course, if we have a measure $\mu$ on the domain then we can for example consider the quotient space $\mathcal{L}^0$ that identifies $\mu$-a.e. equal Borel measurable maps. The corresponding Ky-Fan metric that makes $\mathcal{L}^0$ Polish can of course be seen as a pseudo-metric on $M$.

I somehow doubt that $M$ can always be standard Borel, since this question is so natural, but does not seem to appear in Srivastavas book (or I just oversaw some simple implication).

A fortiori, there is no standard Borel structure on $M(X,Y)$ with measurable evaluation function. $\endgroup$1more comment