Show that $\int_E f d\mu=0$ if $\mu(E)=0$.

Question

Suppose $(X,S,\mu)$ is a measure space, $f$ is a non-negative measurable function, $E\in S$, and $\mu(E)=0$.

I want to show that $\int_E f d\mu=0$.

$\int_E f d\mu:=\int_X f\chi_E d\mu = sup\{\int_Xu\chi_E d\mu\mid 0\leq u\leq f$ and $u$ is a simple function $\}$.

It suffices to show that $\int_X u\chi_Ed\mu=0$ for each simple function $u$, which seems obvious but I'm not sure how to show it.

Answer 1

For any $0\leq u\leq f$ for $u=\displaystyle\sum_{i}a_{i}\chi_{E_{i}}$, then $\displaystyle\int_{E}u=\displaystyle\sum_{i}a_{i}\int_{E}\chi_{E_{i}}=\sum_{i}a_{i}\mu(E_{i}\cap E)=0$.

Now find a sequence $(u_{n})$ of simple functions such that $0\leq u_{n}\leq f$, and $\displaystyle\int_{E}u_{n}\rightarrow\int_{E}f$, then $\displaystyle\int f=0$.