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

Question

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.

Answer 1

This proof looks correct to me. The only part that I had to think about to be sure on was that a continuous function on any measurable set is measurable, but that is true because the inverse image of an unbounded interval is open. So yes, it looks good to me.