Problem: Let $f$ be a real valued function defined on a measurable domain $E$. Suppose that $f$ is continuous except at a finite number of points. Is $f$ measurable? Proof: Let $D \subseteq E$ be set of all point at which $f$ is discontinuous. Since $D$ is finite, $m(D) = 0$, so $f$ is measurable on $D$. Since $f$ is continuous on $E \setminus D$, it must be measurable on $f$. Hence, $f$ is measurable on the union $(E \setminus D) \cup D =E$.

@user193319 Measurable with respect to which measure? Is the measure you're working with complete (i.e., subset of a nullset is measurable)?

@MartinSleziak I am working with the Lebesgue measure.

Oh, yes. Sorry.

So we know that both $D$ and $E\setminus D$ is measurable.

Yes, because $D$ is of zero measure, and differences of measurable sets are measurable.

Probably you should include some argument why $f|_A$ and $f|_{X\setminus A}$ measurable implies $f|_X$ measurable. (For a measurable $A$.)

This is what you're using in your proof. (But maybe you had such result before.)

Yes, that is a theorem in the book.

I think that the proof you mentioned works whenever $m(D)=0$. (In particular, this includes countable sets.)

Oh, very nice!

I did not notice a problem there. Maybe somebody else will read your message and respond if they have something to add.

Okay. Thank you for your time.