How to prove this problem about integrable function?

Question

If an integrable function $f(x)\ge0$ a.e., then $\int fd\mu\ge0$. Any hint is appreciated.

Answer 1

Thanks to @Cantor, @madprob, @Amr and @Davide, I post an answer for my own question.

$A=\{x;f(x)\ge0\}=\cup\{x;f(x)\ge\frac1n\}=\cup A_n$. $$ \int_{A}fd\mu\ge\int_{A_n}fd\mu\ge\int_{A_n}\frac1nd\mu\ge\frac1n\mu(A_n)\ge0. $$

$B=\{x;f(x)<0\}=\cup\{x;f(x)<-\frac1n\}=\cup B_n$. $$ \int_{B}fd\mu=-\int_{B}f^-d\mu. $$ $f^-\ge0\Rightarrow \exists$ a sequence of nonnegative simple functions $\varphi_n\nearrow f^-$ and $\varphi_n$ can be represented as a linear combination of characteristic functions. $$ 0\le\int_{B}\varphi_nd\mu=\int_{B}\sum_{k=1}^ma_k\mathbf{1}_{E_k}d\mu=\sum_{k=1}^ma_k\mu(E_k\cap B)\le\sum_{k=1}^ma_k\mu(B)=0. $$ $\Rightarrow \int_{B}\varphi_nd\mu=0\Rightarrow \int_{B}f^-d\mu=0$ as $n\to\infty\Rightarrow\int_{B}fd\mu=0\Rightarrow \int fd\mu\ge0$.