We're assuming that $A \subset \mathbb{R^n}$ is a closed rectangle. I just want to know if my proof is correct.
Proof: Let $B_n = \{x: f(x) > \frac{1}{n}\}$ Since $\{x : f(x) \neq 0\} = \bigcup\limits_{n\in\mathbb{N}}B_n$
it suffices to prove that each $B_n$ has measure $0$ (or in this case, I'll prove that $B_n$ has content $0$). Choose a partition $P$ of $A$ such that $U(f, P) < \epsilon/n$. If we let $K$ be the set of subrectangles of $P$ that intersect $B_n$, then for each $S \in K$,
$M_S(f) := \sup\{f(x) : x \in S\} > \frac{1}{n}$ So
$\sum\limits_{S \in K}\frac{1}{n}v(S) \le U(f, P) < \epsilon/n$
And the result follows since $K$ is a cover of $B_n$ by closed rectangles.
In case my terminology isn't clear:
-A subset $A \subset \mathbb{R^n}$ has measure $0$ if for every $\epsilon > 0$, there is a cover $\{U_1, U_2,...\}$ of $A$ by closed rectangles such that $\sum\limits_{i = 1}^{\infty} v(U_i) < \epsilon$.
-For a rectangle $A = [a_1, b_1] \times ...\times [a_m, b_m] \subset \mathbb{R^n}$, $v(A) = (b_1 - a_1)...(b_m - a_m) $
-$A$ has content $0$ if for every $\epsilon > 0$, there is a finite cover $\{U_1,U_2,...,U_n\}$ of $A$ by closed rectangles such that $\sum\limits_{i = 1}^{n} v(U_i) < \epsilon$.