# function continuous ae but not borel measurable

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?

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.
• 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 $\lim_{\delta\to 0}\sup_{y \in (x\pm\delta)}f(y) = 0$? Jan 3, 2019 at 15:42
• @Bermudes $f(x)=0$ on the complement of $C$. Jan 3, 2019 at 15:43
• 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 $C$. Jan 3, 2019 at 15:43
• Oh sure, $C$ is closed, didn't think of that. Thanks a lot! Jan 3, 2019 at 15:44