Problem: Suppose $h:S\to\mathbb{R}$ is integrable on the set $S\subset\mathbb{R}^n$. Also $h=0$ everywhere except on a set of measure 0. How to show that $\int_S h=0$?
Attempt: I am trying to prove that $\underset{P}\sup\left\{L(h,P)\right\}=\underset{P}\inf\left\{U(h,P)\right\}=0$. I want to prove this by showing that for all positive $\epsilon$, I can produce a partition $P$ such that either $L(h,P)>-\epsilon$ or $U(h,P)<\epsilon$. However, I have trouble justifying that such a partition exist. Any suggestions?
Note that $h$ vanishes everywhere in $S$ except on a set $E \subset S$ with measure $0$.
Take any partition $P$ of $S$. If $R$ is any subrectangle, then $R$ is not a subset of $E$ since it has non-zero measure. Hence, $h$ vanishes at some point in $R$ and $L(h,P) \leqslant 0 \leqslant U(h,P)$.
This is true for every partition $P$. Therefore,
$$\sup_{P}L(h,P) \leqslant 0 \leqslant \inf_{P} U(h,P).$$
Since the integral exists
$$\int_S h = \sup_{P}L(h,P) = \inf_{P} U(h,P)= 0.$$