Let $A$ be a rectangle in $\mathbb{R}^2$. Suppose $f: A \to \mathbb{R}$ is bounded. Suppose $f(x) = 0 $ except on $F$, where $F$ is closed and has measure zero.
Does it follow that $f$ is integrable on $A$ and $\int_A f = 0 $ ?
Update with my try:
Suppose $|f(x)| \leq M$. Since $F$ is a closed(hence) compact and is of measure zero, we can cover $F$ with finite boxes, say $R_1,...,R_n$ such that $\sum_{i=1}^n Vol(R_i) < \frac{ \epsilon}{2M} $. We will construct a partition $P$ of $A$ that satisfy $U(f,P) - L(f,P) < \epsilon $ so that we will have $f$ is integrable on $A$. We can denote $R_i = [a_i,b_i] \times [c_i,d_i] $ for all $i=1,...,n$. We use all the $a_i,b_i,c_i,d_i$ to define a partition $P$ of $A$. Let $Q$ be an arbitrary subrectangle determined by the partition $P$. Since $f(x) = 0$ on every $Q \subset A \setminus F $, we only need to worry on the $Q's$ that satisfy $N \subset Q$. Hence
$$ U(f,P) - L(f,P) = \sum_{N \subset Q} ( \sup_{x \in Q} f - \inf_{x \in Q} f)Vol(Q) = \sum_{i=1}^n ( \sup_{x \in R_i} f - \inf_{x \in R_i} f) Vol(R_i) < 2M \frac{ \epsilon}{2M} = \epsilon $$
