X nonnegative and integral equal to zero implies measure of positive part zero I'm trying to show that if we have $X\geq0$ and $\int{X} d\mu=0$, then that implies $\mu([X>0])=0$.
I tried to do it using the definition $\int{X} d\mu = \sup\{\int{Y} d\mu\mid 0 \leq Y \leq X \text{ and $Y$ is a simple function}\}$ but keep getting stuck at the part where we use the definition of Lebesgue integral $\int{X} d\mu=\sum_{1}^n{x_i\mu(A_i)}$ then convert that back to the measure of positive part $\mu([X>0])=0$. Any insights? Thanks!

For a fixed integer $n$, consider the simple function 
$$f_n\colon x\mapsto \frac 1n\chi\{y, f(y)\geqslant 1/n\}.$$
This is a simple non-negative function which is smaller or equal of $f$, hence 
$\mu\{y, f(y)\geqslant 1/n\}=0$. Since $\{f\gt 0\}=\bigcup_{n\geqslant 1}\{y, f(y)\geqslant 1/n\}$, we obtain by $\sigma$-sub-additivity that $\mu\{f\gt 0\}=0$. 
