$\int_\Omega f d\mu = 0 $ if and only if $f(x)=0$ almost everywhere can someone give me a hint on what kind of theorem/definition I should make use of to solve this?

Let $(\Omega,\mathfrak A, \mu)$ be a measure space and $f:\Omega \to \mathbb R$ a non-negative measurable function.
 Show that
$$\int_\Omega f d\mu = 0 $$ if and only if $$f(x)=0$$ a.e.
Sketch: 

(1) Suppose not, then there is a positive measure subset $A$ where $f\not=0$.

(2) Let $A_n$ be the set of points where $f&gt;\frac{1}{n}$.

(3) Observe that $A_n\subseteq A_{n+1}$ and $\bigcup A_n=A$.

(4) Conclude that some $A_n$ has positive measure using the properties of measures.

(5) Conclude that $\int f d\mu\geq\frac{1}{n}\mu(A_n)&gt;0$.
If @$f(x) = 0$ a.e. is clear that $\int_{\Omega} fd\mu = 0$.

Now, suppose that $f$ is non-negative, $\int_{\Omega} fd\mu = 0$, but $f$ is not $0$ a.e. It means that there is $A$ such that $\mu(A) &gt;0$ and $f(x) \neq 0, \forall x \in A$. 

Then, 

$\int_{\Omega} f d\mu \geq \int_{A} f d\mu &gt; 0$. Contradiction!