When is the continuous image of a measurable subset of a Polish space measurable? In page two of three of this note: http://math.iisc.ac.in/~manju/MartBM/RaoSrivastava_borelisomorphism.pdf
It is said in the proof of Proposition 2, and in the section (ii) that '$f$ is clearly bi-measurable.'
It is clear to me that $f$ is continuous and injective, and that $Z$ is closed. But why is it clear that for any measurable subset of $Z$, say $M$, that $f(M)$ is a measurable subset of $X$?
I can see that it is enough to show that for any open subset of $\prod_{n \geq 1} Z_n$, $O$, that $f(O \cap Z)$ is measurable, in an attempt to try to do so I noted that since $Z$ is closed, it is $G_{\sigma}$ (and Polish), so $O \cap Z$ is $G_{\sigma}$, and $f(O \cap Z) = (f_0 \circ \pi_0)(O \cap Z)$, and so it is enough to show (using that $f_0$ is bi-measurable) that $\pi_0(O \cap Z)$ is itself measurable, or in particular that the image of a $G_{\sigma}$ subset of $\prod_{n \geq 1} Z_n$ by $\pi_0$ is measurable. But I have been unable to show this.
 A: I don't see that the claim is immediately obvious. I will sketch an argument inducing that it can't be utterly trivial
The quoted part of the proof starts with the assumption that  $B_{0}, B_{1}, \ldots$ belong to Borel $\sigma$-algebra $\mathcal{B}$ and  for each $n$, there  are Polish spaces $Z_{n}$ and continuous bijections $f_{n}: Z_{n} \rightarrow B_{n}$ which are bimeasurable.
Then $Z$ is defined as
$$
\Bigl\{\left(z_{0}, z_{1}, \ldots\right) \in \textstyle\prod_{n} Z_{n}: f_{0}\left(z_{0}\right)=f_{1}\left(z_{1}\right)=\cdots\Bigr\}
$$
and $f: Z \rightarrow X$ is given by
$$
f\left(z_{0}, z_{1}, \ldots\right)=f_{0}\left(z_{0}\right),\qquad \left(z_{0}, z_{1}, \ldots\right) \in Z.
$$
Hence we have that $f$ is injective and continuous, $Z$ is closed and $f(Z)$ equals $\bigcap_n B_n$. Also, $f = f_0\circ \pi_0$, and therefore $\pi_0$ is injective on $Z$. For measurable  $M\subseteq  Z$, showing that $f(M)$ is measurable is equivalent to show that $\pi_0(M)$ is, because $f_0$ is bimeasurable.
Now, I claim that in general, a description of $\pi_0(M)$ is not bounded in complexity starting with a description of $M$. More precisely, even for a closed $M\subseteq  X_0 \times X_1$  where $X_i$ are Polish, and such that $\pi_0\restriction M$ is injective, $\pi_0(M)$ can be arbitrarily complex.
This holds since every nonempty Borel subset $B$ of a Polish space $X$ is of the form $g(F)$ for some  continuous $g:\mathcal{N} \to X$, where $\mathcal{N}$ is the Baire space, and $F\subseteq \mathcal{N}$ is closed, and $g\restriction F$ is injective (see, e.g, (13.7) in Kechris' book). By considering the graph of $g$ we obtain my observation.
In any case, I suggest you contact the authors. They are very kind and perhaps they have a spare moment to answer this particular question.
