measure theory exercise: null integral implies null function

Let $\Omega \subset R^n$ a non empty open set and $f: \Omega \rightarrow R$ a nonnegative measurable function with $\int_{\Omega} f =0$. Then $f=0$ in $\Omega$ almost everywhere.

I have no idea of how to start this problem, someone could help me ?

Thanks in advance!

My try (I am not sure):

Let $E_n:= \{ x \in \Omega; f(x) > 1 / n\}, n \in N$ and define $E:= \{ x \in \Omega; f(x) > 0\} = \cup_{n \geq 1} E_n.$

Note that

$$ 0 = \int_{\Omega} f \geq \int_{E} f \geq \int_{E_n} f \geq \frac{|E_n|}{n} \geq 0.$$

Then $|E_n| = 0$ for all n, which implies $|E| = 0. $ Then $f=0 $ in $\Omega$ a.e

I am not sure because it seems that we can replace the set $\Omega$ by a measurable set with zero measure and if we consider a set like this the affirmation is not true.