Show $\{x:f(x) \neq 0\}$ has measure $0$

Question

If $f:A \to \mathbb{R}$ is non-negative and $\displaystyle \int_{A} f(x)dx = 0$, show that $\{x:f(x)≠ 0\}$ has measure $0$.

I know that to do this I should prove that $\{x:f(x)>\frac{1}{n}\}$ has content $0$. How do I do this?

Answer 1

If $S_n := \{x : f(x) > 1/n\}$, then

$$|S_n| = \int_A 1_{S_n}(x)\, dx = n \int_A \frac{1}{n}1_{S_n}(x)\, dx \le n \int_A f(x)1_{S_n}(x)\, dx \le n\int_A f(x)\, dx = 0.$$

Since the set $S := \{x : f(x) \neq 0\}$ is the union of the sets $S_n$ and each $S_n$ has measure zero, $S$ has measure zero.