Suppose that $X,Y,Z$ are metric spaces, equipped with the Borel $\sigma$-algebra.

Assume that $f:X\rightarrow Y$ is a homeomorphism, and that $g:Y\rightarrow Z$ is a measureable map (with respect to the Borel $\sigma$-algebra).

Is it necessarily true that $g\circ f$ is measureable?

Note: this is not as easy as it seems, the obvious proof is not working, let $U$ be open in $Z$, then $(g\circ f)^{-1}(U) = f^{-1}\circ g^{-1}(U)$, $g^{-1}(U)$ is a measureable set, is it necessarily true that a pre-image of a Borel set under a homeomorphism is measureable?


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    $\begingroup$ Measure theory is a great hole in my knowledge, but aren't the Borel sets the $\sigma$-algebra-closure of the closed/open sets? A homeomorphism should therefore preserve the property of being a Borel set, as it preserves the property of being open/closed. $\endgroup$ – Theo Bendit Oct 2 '17 at 13:07
  • $\begingroup$ @TheoBendit, that is correct and you should post it as an answer. More generally, OP, $f: X \to Y$ is a homeomorphism, which means that $X$ and $Y$ are topologically identical, with $f$ playing the role of "identitfy function"; which means that all properties deriving from that topology should also be identical as they are related by $f$, including in particular the Borel sigma algebra. $\endgroup$ – Mees de Vries Oct 2 '17 at 13:11
  • $\begingroup$ I understand the Idea, thanks Theo and Mees, I will try to formalize it. $\endgroup$ – Yanko Oct 2 '17 at 13:25
  • $\begingroup$ @yanko you are right, my bad $\endgroup$ – Alvin Lepik Oct 2 '17 at 15:17

Here's the result mentioned in the comments:

Let $X$ and $Y$ be topological spaces. If $f:X\to Y$ is continuous and $B$ is a Borel subset of $Y$, then $f^{-1}(B)$ is a Borel subset of $X$.

It's proof relies on the fact that the preimage of a function respects complements, unions, and intersections, as well as the fact that the preimage of an open set under a continuous function is open. From this we obtain that every continuous function is Borel measurable. Then the question follows immediately from the fact that the composition of Borel measurable functions is Borel measurable. In particular, we see that either composition $f\circ g$ and $g\circ f$ will work, provided that the domains are setup so that the composition is defined.

However, one must be careful when considering the Lebesgue measure on $\mathbb{R}$. This is because when we say a function $f:\mathbb{R}\to\mathbb{R}$ is Lebesgue measurable, we mean that $f$ is Lebesgue-to-Borel measurable; i.e., $f^{-1}(B)$ is Lebesgue measurable for every Borel subset $B$ of $\mathbb{R}$. Now if $f$ is Lebesgue measurable and $g$ is Borel measurable (in particular, continuous), we know that $g\circ f$ is Lebesgue measurable (by the same argument and the fact that Borel sets are Lebesgue measurable). But $f\circ g$ need not be, even if $g$ is assumed to be continuous.

Edit: It's not immediately clear why the composition of Borel measurable functions is Borel measurable (or why the result above is needed to conclude that continuous functions are Borel measurable) if one defines a Borel measurable function as one such that the preimage of open sets are Borel, as opposed to one such that the preimage of Borel sets are Borel. However, these definitions are equivalent:

If $f:X\to Y$ is such that $f^{-1}(U)$ is Borel for every open $U\subseteq Y$, then $f^{-1}(B)$ is Borel for every Borel $B\subseteq Y$. (The converse is trivial because open sets are Borel sets).

Proof: Let $\mathcal{M}:=\{E\subseteq Y \mid f^{-1}(E)\ \text{is Borel}\}$. By hypothesis, $\mathcal{M}$ contains the open sets. It is also a $\sigma$-algebra because the Borel subsets of $X$ form a $\sigma$-algebra and the preimage respects unions, intersections, and complements. Since the Borel $\sigma$-algebra $\mathcal{B}_Y$ is the smallest $\sigma$-algebra containing the open sets, we conclude $\mathcal{B}_Y\subseteq\mathcal{M}$. Therefore $f^{-1}(B)$ is Borel for every Borel subset $B$ of $Y$.

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    $\begingroup$ A composition of Borel measureable functions is Borel? wow I didn't know that... Thanks John! $\endgroup$ – Yanko Oct 2 '17 at 15:30
  • $\begingroup$ @yanko Yep, because $(f\circ g)^{-1}(B)=g^{-1}(f^{-1}(B))$, so if $B$ is Borel and $f,g$ are Borel measurable, then $f^{-1}(B)$ and thus $(f\circ g)^{-1}(B)$ are Borel. $\endgroup$ – John Griffin Oct 2 '17 at 15:44
  • $\begingroup$ wait, but, the definition of Borel measureable is that the pre-image of an open set is Borel measureable, $f^{-1}(B)$ is not necessarily open, it is Borel.. $\endgroup$ – Yanko Oct 2 '17 at 15:44
  • $\begingroup$ @yanko Yeah I just looked it up and you are correct. I am mistaken. Sorry! Let me modify my answer. $\endgroup$ – John Griffin Oct 2 '17 at 15:46
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    $\begingroup$ Amazing thank you so much John!!! $\endgroup$ – Yanko Oct 2 '17 at 15:57

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