Measure of set equal to zero when $n$ goes to infinity

Let $(X,A,\mu)$ a measure space and $f:X\to\mathbb R$ measurable. For each $n\in\mathbb N$ let $E_n=\{x\in X:\vert f(x)\vert\gt\frac{1}{n}\}$. Prove that each $E_n$ is measurable and if $\lim\mu(E_n)=0$ then $f=0$ a.e.

I suppose there is not much dificulty proving $E_n$ measurable since if we split $\vert f(x)\vert\gt\frac{1}{n}$ then we get $f^{-1}(-\infty,-1/n)$ and $f^{-1}(1/n,\infty)$ for all $n$. As $f$ is measurable, both preimages are measurable.

I am not sure how to procced on the second part

if $\lim\mu(E_n)=0$ then $f=0$ a.e.

I can see that when $n\to\infty$ then $f=0$

Taking the measure $\lim\mu(E_n)=\mu\{x\in X:f(x)=0\}=0$ ? this clearly does not fit the definition of a.e

Answer

Since $E_1 \subset E_2 \subset E_3 \subset \cdots$ you have $\mu(E_1) \le \mu(E_2) \le \mu(E_3) \le \cdots$

The fact that $\lim_{n \to \infty} \mu(E_n) = 0$ means that $\mu(E_n) = 0$ for all $n$.

Since $\cup_n E_n = \{x : |f(x)| > 0\}$ you get

$$\mu(\{x : f(x) \not= 0\}) = \mu(\{x : |f(x)| > 0\}) \le \sum_n \mu(E_n) = 0.$$