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Let $\mathcal{L}$ denote the $\sigma$-algebra of Lebesgue measurable sets on $\mathbb{R}$. Then, if memory serves, there is an example (and of course, if there is one, there are many) of a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that is not measurable in the sense that $f:(\mathbb{R},\mathcal{L})\rightarrow (\mathbb{R},\mathcal{L})$ is measurable, but unfortunately, I was not able to recall the example. Could somebody please enlighten me?

Note that this is not in contradiction with the usual "Every continuous function is measurable.", because in this statement it is implicit that the co-domain is equipped with the Borel sets, not the Lebesgue measurable sets.

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    $\begingroup$ I don't understand your question, a continuous function is always measurable since preimage of open sets are open (open sets are borel sets then open sets are measurable). $\endgroup$ Commented Aug 29, 2013 at 22:05
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    $\begingroup$ @GastónBurrull The $\sigma$-algebra on the codomain is the $\sigma$-algebra of Lebesgue-measurable sets. Is it impossible that a Lebesgue null-set has non-measurable preimage under a continuous map? $\endgroup$ Commented Aug 29, 2013 at 22:27
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    $\begingroup$ @GastónBurrull That the preimages of open sets are open shows that a continuous function $f\colon\mathbb{R}\to\mathbb{R}$ is a measurable function $f\colon (\mathbb{R},\mathcal{B})\to (\mathbb{R},\mathcal{B})$, where $\mathcal{B}$ is the Borel $\sigma$-algebra. It is then also a fortiori measurable $(\mathbb{R},\mathcal{L})\to (\mathbb{R},\mathcal{B})$, but if that implies it's also $(\mathbb{R},\mathcal{L})\to (\mathbb{R},\mathcal{L})$ measurable, that's a non-trivial result. $\endgroup$ Commented Aug 29, 2013 at 22:45
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    $\begingroup$ @DanielFischer I never saw before this definition of measurable function! Thanks $\endgroup$ Commented Aug 30, 2013 at 0:02
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    $\begingroup$ @GastónBurrull Ah. A function $f \colon (X,\mathcal{X}) \to (Y,\mathcal{Y})$ between two measure spaces is called measurable if $(\forall S \in \mathcal{Y})(f^{-1}(S) \in \mathcal{X})$. $\endgroup$ Commented Aug 30, 2013 at 0:07

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The standard example is given by the function $g(x)=f(x)+x$, where $f$ is the devil's staircase function of Cantor. It turns out that the function $g$ is a homeomorphism from $[0,1]$ onto $[0,2]$ and has the property that $\mu(g(C))=1$ (where $C$ is the Cantor set). Pick a non measurable $A\subset g(C)$. First note that $B=g^{-1}(A)$ is measurable since $B\subset C$. It follows that $g^{-1}$ is continuous, $B$ is Lebesgue measurable but $(g^{-1})^{-1}(B)$ is non measurable.

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  • $\begingroup$ why couldn't you use the Cantor function directly? It is also a homeomorphism, and $f^{-1}(B)$ is non-measurable. $\endgroup$
    – Rodrigo
    Commented Jan 1, 2016 at 16:37
  • $\begingroup$ How it proves that g is nonmeasurable? $\endgroup$
    – Arun
    Commented Apr 18, 2017 at 4:35
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    $\begingroup$ @Rodrigo: the example ends up showing that $g^{-1} $ is the counterexample. The Cantor function is not a homeomorphism, since it is not injective, so the argument does not apply. $\endgroup$ Commented Jun 3, 2018 at 12:25
  • $\begingroup$ @Arun I think $g$ is measurable in this case. It's only $g^{-1}$ that is nonmeasurable. This is because $g$ does not send all measure zero sets to measure zero images. That is, $g: ([0, 1], \mathscr{L}) \rightarrow ([0, 2], \mathscr{L})$ is measurable, but $\exists A \subset [0, 1] : \ell(A) = 0 \land \ell(g(A)) \neq 0$ so $g^{-1}$ is not necessarily measurable. (Indeed, this answer proved it isn't.) $\endgroup$
    – kdbanman
    Commented Feb 7, 2020 at 18:22
  • $\begingroup$ To connect my comment with the answer better, take $C$ to be the Cantor set, then $\ell(C) = 0$ and $\ell(g(C)) > 0$. So my existential statement is true for our $g$, and hence $g^{-1}$ is not necessarily measurable. $\endgroup$
    – kdbanman
    Commented Feb 7, 2020 at 18:37

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