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I'm pretty sure this is false and that I should use Cantor's function (its extension) as a counterexample, but I don't know how.

peenktuna
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    See here: https://math.stackexchange.com/questions/671173/a-basic-doubt-on-continuous-image-of-a-measurable-set-measurable – Oscar Aug 05 '24 at 09:22
  • For the benefit of OP, let me mention that David Mitra (in a comment) gave a counter-example in above link. – Kavi Rama Murthy Aug 05 '24 at 09:25
  • @Oscar thank you! – peenktuna Aug 05 '24 at 09:31
  • @geetha290krm thank you too! – peenktuna Aug 05 '24 at 09:32
  • actually I'm sorry @Oscar, but I dont understand how the counterxample given by David Mitra works in this case, because it proves that $\exists D$ nonmeasurable (hence non borel) such that $\varphi^{-1}(D)$ is measurable, but I don't know if $\varphi^{-1}(D)$ is a borel set with certainty, right? – peenktuna Aug 05 '24 at 09:57
  • @Oscar: But your answer only demonstrates that the set $\phi^{-1}(D)$ is Lebesgue measurable, by invoking the completeness of the Lebesgue measure. That does not show that the set is Borel measurable. – Lee Mosher Aug 05 '24 at 13:10
  • @LeeMosher That's a good point. I completely missed the fact that we require the set to be Borel. I will edit the answer. Thanks! – Oscar Aug 05 '24 at 13:19
  • It is good question. In general the statement is false. However, if $f$ is injecitve then images of Borel sets are Borel again. If $f$ is just continuous, then images of Borel sets are analytic and thus still Lebesgue measurable, but not necessarily Borel. – dialegou Aug 05 '24 at 13:21

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Here is a brief counterexample (although the construction is more complicated than just using Cantor's function).

As is somewhat more well known, under the projection function $$p : [0,1] \times [0,1] \to [0,1] \qquad p(x,y)=x $$ there exists a Borel subset $B \subset [0,1] \times [0,1]$ such that $p(B)$ is not Borel. See this math.stackexchange question for an example.

Suppose now that $g : [0,1] \to [0,1] \times [0,1]$ is a continuous surjective function, for example the Peano curve. Let $D = g^{-1}(B)$ and let $f = p \circ g : [0,1] \to [0,1]$. Since $f$ is continuous, it follows that $D \subset [0,1]$ is a Borel set, but $f(D)=p(B)$ is not a Borel set.

For a bit of historical background, see this answer in the same post as above. In brief, Lebesgue himself made the mistake of asserting that the continuous image of a Borel set is a Borel set, and Souslin discovered a counterexample. This gave rise to a new branch of mathematics, descriptive set theory and the theory of analytic sets and the projective hierarchy, which is all closely tied to logic. The set $p(B)$ is an example of an analytic set which is not Borel.

Lee Mosher
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  • Thank you a lot! The historical background is really interesting too, I'll be looking into it, thanks again. – peenktuna Aug 05 '24 at 14:22