Spivak Calculus on Manifolds - Problem 3-18 
$\textbf{3.18}$ If $f: A \longrightarrow \mathbb{R}$ is non-negative and $\int_A f = 0$, show that $\{ x \in A : f(x) \neq 0 \}$ has measure $0$. Hint: prove that $\left\{ x \in A : f(x) > \frac{1}{n} \right\}$ has content $0$.

I know that the set $\{ x \in A : f(x) \neq 0 \}$ has measure $0$. Hint: prove that $\{ x ; f(x) > \frac{1}{n} \}$ has content $0$ for each $n$, then the same set has measure $0$ and observing that $$\{ x \in A : f(x) \neq 0 \} = \bigcup_{i = 1}^{\infty} \left\{ x \in A : f(x) > \frac{1}{n} \right\}$$ i.e, $\{ x \in A : f(x) \neq 0 \}$ is a countable union of set with measure $0$, then $\{ x \in A : f(x) \neq 0 \}$ has measure $0$, therefore is sufficient prove the hint. 
I think to prove the hint I need to use the fact that a function $f: A \longrightarrow \mathbb{R}$ is discontinuous in $x \in A$ if and only if $o(f,x) > 0$ (theorem $1-10$ of Spivak's book) and use the fact that $f$ is integrable if and only if $B := \left\{ x \in A : f \  \text{is not continuous at} \ x \right\}$ has measure $0$ (theorem $3-8$ of Spivak's book). I'm trying relate that $f(x) > \frac{1}{n}$ with $o(f,x) > \frac{1}{n}$, but I don't know how to do this. I would like to receive a hint about how to do this.
Thanks in advance!

$\textbf{P.S.:}$
1) $A \subset \mathbb{R}^n$ is a rectangle;
2) A set with content $0$ is an set $A$ such that for every $\varepsilon > 0$, we can find a finite cover $\{ U_1, \cdots, U_n \}$ of $A$ by closed rectangles such that $\sum_{i = 1}^{i = n} v(U_i) < \varepsilon$, where $v(U_i)$ is the volume of $U_i$.
3) The oscillation of $f$ at $x$ is defined by $o(f,x) := \lim_{\delta \rightarrow 0} \left[ \sup f \left( B(x, \delta \right) - \inf f \left( B(x, \delta) \right) \right]$.

 A: Let $k \in \mathbb{N}$. We wish to show that $C_k = \left\{x \in A : f(x) \ge \frac1k\right\}$ has content $0$.
Let $\varepsilon > 0$. Set $A = [a_1, b_1] \times [a_2, b_2] \times \cdots
 \times[a_n, b_n]$.
Since $\displaystyle\int_A f = 0$, there exists a partition $P$
$$a_1 = x_0^{(1)} < x_1^{(1)} < \cdots < x_{r_1-1}^{(1)} < x_{r_1}^{(1)} = b_1$$
$$a_2 = x_0^{(2)} < x_1^{(2)} < \cdots < x_{r_2-1}^{(2)} < x_{r_2}^{(2)} = b_2$$
$$\vdots$$
$$a_n = x_0^{(n)} < x_1^{(n)} < \cdots < x_{r_n-1}^{(n)} < x_{r_n}^{(n)} = b_n$$
of $A$ such that the upper Darboux sum of $f$ with respect to $P$ is $U_{f, P} < \frac{\varepsilon}{k}$.
Denote $$A_{i_1, \ldots, i_n} = \left[x^{(1)}_{i_1-1}, x^{(1)}_{i_1}\right] \times \left[x^{(2)}_{i_2-1}, x^{(2)}_{i_2}\right] \times\cdots \times \left[x^{(n)}_{i_n-1}, x^{(n)}_{i_n}\right]$$ for $i_j = 1, \ldots, r_j$; $j = 1, \ldots, n$ and $M_{i_1, \ldots, i_n} = \displaystyle\sup_{x \in A_{i_1, \ldots, i_n}}|f(x)|$.
If $A_{i_1, \ldots, i_n}$ intersects $C_k$, then certainly $M_{i_1, \ldots, i_n} \ge \frac{1}k$.
We have:
\begin{align}\frac{1}{k}\sum_{A_{i_1, \ldots, i_n} \cap C_k\ne \emptyset} \nu(A_{i_1, \ldots, i_n}) &= \sum_{A_{i_1, \ldots, i_n} \cap C_k \ne \emptyset} \frac{1}{k}\nu(A_{i_1, \ldots, i_n})\\
 &\le \sum_{A_{i_1, \ldots, i_n} \cap C_k \ne \emptyset} M_{i_1, \ldots, i_n}\nu(A_{i_1, \ldots, i_n}) \\
&\le \sum_{i_1, \ldots, i_n} M_{i_1, \ldots, i_n}\nu(A_{i_1, \ldots, i_n}) \\
&= U_{f, P}\\
&< \frac{\varepsilon}k
\end{align}
Hence
$$\displaystyle\sum_{A_{i_1, \ldots, i_n} \cap C_k\ne \emptyset} \nu(A_{i_1, \ldots, i_n}) < \varepsilon$$
so $$\big\{A_{i_1, \ldots, i_n} : A_{i_1, \ldots, i_n} \cap C_k\ne \emptyset\big\}$$ is a finite family of rectangles which cover $C_k$ and sum of their volumes is $< \varepsilon$. We conclude that $C_k$ has content $0$.
