Let $\displaystyle A\subseteq[0,1], |A|=\aleph_0$,
f(x)={0 if $\displaystyle x\notin A$, 1 if $\displaystyle x\in A$}
Prove that f is not integrable in [0,1], or show a negative proof and prove that it is integrable in [0,1].
Just because your perturbing the zero function at a countable number of places does not mean the set of all discontinuities created is countable, and thus necessarily of measure zero. The indicator function for the rationals differs from the zero function at countably many places but the set of discontinuities introduced has measure one.