The theorem states: for a function $f:X\rightarrow [0,\infty]$ that is measurable, if $$\int_E f\,\,d\mu=0$$Then, $f=0$ for almost everywhere on $E$. (Here $E\in\mathfrak M$, where $\mathfrak M$ is a $\sigma$-algebra, also $\mu$ is a positive measure).
The proof states: If $A_n=\{x\in E: f(x)>1/n\}$, where $n=1,2,3...$, then$$\frac{1}{n}\mu(A_n)\leq \int_{A_n} f\,d\mu\leq \int_{E} f\,d\mu=0$$ so that $\mu(A_n)$.
My question: I don't understand the motivation behind the equality $$\frac{1}{n}\mu(A_n)\leq \int_{A_n} f\,d\mu$$
The integral here is the Lebesgue integral, for the middle integral, $$\frac{1}{n}\mu(A_n)\leq \int_{A_n} f\,d\mu\equiv sup \int_{A_n}s\,d\mu$$
Here $s$ are simple measurable functions, and the supremum is taken over all simple measurable functions. The integral of $s$ is also defined as $$\int_{A_n}s\,d\mu=\sum_{i=1}^{n}\alpha_i\mu(B_i\cap A_n)$$
Here $\alpha_i$ are the distinct values of $s$ and $B_i=\{x:s(x)=\alpha_i\}$.
$f(x)\geq\frac{1}{n}1_{A_n}(x)$ is true for each $x$.
Here $1_{A_n}$ denotes the characteristic function of set $A_n$.
(It sends $x$ to $1$ if $x\in A_n$ and to $0$ otherwise.)
Consequently $0=\int fd\mu\geq \int\frac{1}{n}1_{A_n}d\mu=\frac{1}{n}\mu(A_n)$, hence $\mu(A_n)=0$.