# If a function defined on a (Lebesgue) measurable set is continuous almost everywhere, then is that function necessarily (Lebesgue) measurable?

Here is my attempt at a proof: Suppose $$E \subseteq \Bbb R$$ is a Lebesgue measurable set, and $$f:E \rightarrow \Bbb R$$ is continuous almost everywhere on E. Let $$A$$ denote the set of all points in $$E$$ at which $$f$$ is discontinuous, and let $$s \in \Bbb R$$. Since $$f^{-1}((s,\infty))=(f^{-1}((s,\infty))\cap A) \cup (f^{-1}((s,\infty)) \cap (E \setminus A))$$, if we can show each element of the union is measurable, we are done. $$A$$ has measure zero, so in particular it has zero outer measure. This implies $$f^{-1}((s,\infty)) \cap A$$ has zero outer measure (by the monotonicity of outer measure) and is therefore measurable. For the second part, the restriction of $$f$$ to $$E \setminus A$$ is continuous and hence measurable, so $$f^{-1}((s,\infty)) \cap (E \setminus A)$$ is a measurable set. So $$f^{-1}((s,\infty))$$ is a finite union of measurable sets, and is therefore measurable. Is this correct? I welcome any comments or critcism.