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It is easy to prove that if $f$ is a function continuous almost everywhere, then $f$ is Lebesgue-measurable by using the property that $\mathcal L$ (the Lebesgue-measure) is complete.

Though I've been wondering if the statement "every function continuous ae is Borel-measurable" is true. I feel like it's not but I have a hard time finding a counterexample. So does such a function (continuous ae and not Borel-measurable) exist?

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Your assertion is false. Let $E$ be a non-Borel subset of the Cantor set $C$. [There are only $\mathfrak c$ Borel subsets of $C$, but $2^{\mathfrak c}$ subsets of $C$, so at least one subset is not Borel.] Then the characteristic function $f$ of $E$ is continuous (at least) on the complement of the Cantor set, since $C$ is a closed set. Thus $f$ is continuous a.e. But $\{x: f(x)>0\} = E$ is not Borel. So $f$ is not Borel measurable.

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    $\begingroup$ Thanks, I tried something like that, but how do you prove that $f$ is continuous on the complement of $C$? I mean if $x \in C^\complement$, then for any $\delta > 0$: $f(x) = \inf_{y \in (x\pm\delta)}f(y) = 0$, but why is it that $\lim_{\delta\to 0}\sup_{y \in (x\pm\delta)}f(y) = 0$? $\endgroup$ – Bermudes Jan 3 at 15:42
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    $\begingroup$ @Bermudes $f(x)=0$ on the complement of $C$. $\endgroup$ – Yanko Jan 3 at 15:43
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    $\begingroup$ For every point $a$ not in $C$, there is a neighborhood of $a$ where $f$ is identically zero. This works for any closed set in place of $C$. $\endgroup$ – GEdgar Jan 3 at 15:43
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    $\begingroup$ Oh sure, $C$ is closed, didn't think of that. Thanks a lot! $\endgroup$ – Bermudes Jan 3 at 15:44
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    $\begingroup$ I will add that to the answer. $\endgroup$ – GEdgar Jan 3 at 15:45

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