Let $f:(X,\eta,\mu)\to [0,\infty]$ measurable and $E\in\eta$ with $\mu(E)=0$ ($\eta$ is a $\sigma$-algebra, $\mu$ a positive measure). Then $\displaystyle\int_E f\,d\mu=0$.
We have $$\int_E f\,d\mu=\sup\left\{\int_E s\,d\mu\ : \ s \text{ is a measurable simple function }, s\leq f \right\} \tag{*}.$$
But if $s$ is simple (and measurable) then $s=\displaystyle\sum_{i=1}^n \alpha_i\chi_{E_i}$, with $E_i\in\eta$.
Now
$$\int_E s\,d\mu=\sum_{i=1}^n \alpha_i\mu(E\cap E_i)=0,$$
because $E\cap E_i \subseteq E \Rightarrow \mu(E\cap E_i)\leq \mu(E)=0$.
So that, for all simple function $s\leq f$ , we have $\displaystyle\int_E s\,d\mu=0$, therefore $\displaystyle\int_E f\,d\mu=0$
Is this correct?
Thank you all.
