Proof $\mu(E)=0 \rightarrow \int_E f d\mu=0$ 1

Question

I want to show: $$\mu(E)=0 \rightarrow \int_E f d\mu=0$$

$f:X \rightarrow [0, \infty] $ is measurable and $E \in \mathcal{A} $

Consider a step function $s=\sum_i a_i \chi_{A_i}$

Then I get: $\int_E s d \mu = \sum_i a_i \mu(A_i\cap E) $

Can I conclude that $ \mu(A_i \cap E) \leq \mu(E)=0 $?

Answer 1

Yes, by $\sigma$-additivity, $\mu(E)=\mu(A_i \cap E)+\mu(E\setminus A_i)$ and, being a measure, $\mu(E\setminus A_i)\geq 0$.

Answer 2

Yes, surely. $0 \leq \mu (A_i \cap E) \leq \mu (E)$ implies that $\mu (A_i \cap E) =0$.