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This question seeks to expand the proof given in the answer to this question.

I am weak in topology, and am wondering if someone can provide a proof of why a homeomorphic image of a non-borel set is non-borel?

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möbius
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  • What is a "topological map"? – Wojowu Aug 20 '15 at 12:25
  • Given that the linked thread describes the use of a homeomorphism between $[0,1]$ and $[0,2]$, constructed using the Cantor function, I assume that topological map = homeomorphism here. – Jyrki Lahtonen Aug 20 '15 at 12:33
  • @Wojowu I have edit the post to say "topological mapping" by which I mean some mapping $f$ from one topological space $X$ to another topological space $Y$; i.e. $f : X \to Y$. – möbius Aug 20 '15 at 12:33
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    Möbius, the claim is obviously false in that generality. And I'm positive the answerer there did not mean that. – Jyrki Lahtonen Aug 20 '15 at 12:34
  • @JyrkiLahtonen Okay I will edit the post to be more specific, as per your comment. – möbius Aug 20 '15 at 12:36
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    Hint: A homeomorphism is continuous, bijective, and has a continuous inverse. Now, continuous functions are Borel measurable. Assume $f(B)$ is measurable for a homeomorphism $f$. Could $B$ be non-measurable? – user251257 Aug 20 '15 at 12:51

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Since a homeomorphism $f: X \to Y$ is continuous, it is also Borel-measurable. The same is true of $f^{-1}: Y \to X$.

Let $\mathcal{B}_{x}$ and $\mathcal{B}_{y}$ be the Borel $\sigma$-algebras on $X$ and $Y$ respectively.

Since $f$ is measurable we know that for every $B_{1} \in \mathcal{B}_{y}$ we have that $f^{-1}(B_{1}) \in \mathcal{B}_{x}$, and that for every $B_{2} \in \mathcal{B}_{x}$ we have that $f(B_{2}) \in \mathcal{B}_{y}$.

Let $B_{0} \subset X$ be a non-Borel set. Then $B_{0} \notin \mathcal{B}_{x}$. Since for every $B_{1} \in \mathcal{B}_{y}$ we have that $f^{-1}(B_{1}) \in \mathcal{B}_{x}$, it must be that there is no set in $\mathcal{B}_{y}$ that is mapped to $B_{0}$ (i.e. no Borel set in $Y$ can be mapped by $f$ to a non-Borel set in $X$).

The same holds if $B_{0} \subset Y$ is not Borel; no Borel set in $X$ can be mapped by $f$ to a non-Borel set in $Y$.

By the fact that $f$ is bijective, this implies that no non-Borel set $B_{0} \subset X$ can be mapped to a Borel set $f(B_{0})$ as otherwise $f^{-1} (f (B_{0}))= B_{0}$ must be Borel (since borel sets are mapped to borel sets), which is a contradiction. The same holds for non-borel sets $B_{0} \subset Y$.

Therefore, a homeomorphic image of a non-borel set must be non-borel.

möbius
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