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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).

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  • $\begingroup$ In your question, $M$ is just a set, not a measurable space. (Or did I miss anything?) So what should it mean that $M$ be standard Borel? $\endgroup$ – Lutz Mattner Nov 8 '14 at 17:58
  • $\begingroup$ $M$ is equipped with the trace $\sigma$-algebra of $Y^X$ with the product $\sigma$-algebra. $\endgroup$ – yadaddy Nov 8 '14 at 20:20
  • $\begingroup$ It seems to me that this $\sigma$-algebra is not countably generated, if $X$ has the power of the continuum and $Y$ has power at least two, since every set in the $\sigma$-algebra imposes conditions on the functions at only countably many points. Hence the answer to your question should be "No". $\endgroup$ – Lutz Mattner Nov 9 '14 at 22:01
  • $\begingroup$ But the same also holds for the measurable space of continuous maps, e.g. for $X = [0,1]$ the set $C([0,1],Y)$ is not an element of the product $\sigma$-algebra on $Y^{[0,1]}$ since it is not countably determined. However, the trace from the product space on $C$ is precisely the $\sigma$-algebra generated by the topology of uniform convergence on $C$. $\endgroup$ – yadaddy Nov 14 '14 at 18:30
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    $\begingroup$ I expanded my comment above to an answer. The step "$f \in \mathcal F$ iff $g\in \mathcal F$" does not work with $C(X,Y)$ in place of $M(X,Y)$, since then $g\notin C(X,Y)$. $\endgroup$ – Lutz Mattner Nov 14 '14 at 22:10
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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)$.

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  • $\begingroup$ I see, that's the point! Thanks a lot. $\endgroup$ – yadaddy Nov 15 '14 at 8:31

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