Take the 2-minute tour ×  If an integrable function $f(x)\ge0$
a.e., then $\int fd\mu\ge0$. Any hint is appreciated.
share|improve this question Suppose not ... –  Cantor Dec 10 '12
at 18:54 Sorry, I still cannot figure it out. –  Sam Dec 10 '12
at 19:14 Would be you be able to prove the statement if $f(x)
\geq 0$ instead of $f(x) \geq 0$ a.e. ? (This is such a
fundamental question, that we need to know exactly where you are
having trouble to provide useful comments) –  madprob Dec 10 '12
at 19:16 Use the definition of the integral, it will follow
directly. –  Amr Dec 10 '12 at 19:20 What is the definition of
$\int f d\mu$? –  Cantor Dec 10 '12 at 20:00 show 1 more comment
1 Answer  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\Righta