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).

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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

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)$.

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

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.