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?
Best Answer
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.