That's a question from MSE ([here](https://math.stackexchange.com/questions/1004179/space-of-borel-measurable-maps)) 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?](https://math.stackexchange.com/questions/369859/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).



Space of Borel measurable maps
With the $\sigma$-algebra specified to be the trace of the product $\sigma$-algebra, as in yadaddy's comment: No.


For let $X=[0,1]$ and $Y=\{0,1\}$, let $\mathcal A$ be the product $\sigma$-algebra on $Y^X$, and let $\mathcal B$ denote the trace of $\mathcal A$ on $M(X,Y)$. Then $\mathcal B$ is not countably generated (and hence not standard Borel).


Proof: Let $\mathcal F$ be a countable subset of $\mathcal B$, so 
${\mathcal F} = \{ E \cap M(X,Y) : E \in {\mathcal E}\} $ for some countable set ${\mathcal E} \subseteq \mathcal{A}$. The definition of the product $\sigma$-algebra implies that there is a countable set $T \subseteq X$ such that 
we have $E = \pi\_T^{-1}[\pi\_T^{}[E]]$ for every $E\in \mathcal E$, where $\pi\_T^{}$ denotes the coordinate projection from $Y^X$ to $Y^T$. Now take some $x\_0\in X\setminus T$. Then $A := \{f \in M(X,Y) : f(x\_0)=0\}\in {\mathcal B}$.
On the other hand, for every $F\in\mathcal F$, we have $f\in F$ iff $g \in \mathcal F$, where $g$ denotes the function obtained from $f$ by changing its value at $x\_0$; this remains true if we replace $\mathcal F$ by $\sigma(\mathcal F)$, and thus we get $ A \notin \sigma(\mathcal F)$.


