Lebesgue Integral on a set of measure zero I need to show that if $f$ is an integrable function on $X$ and $\mu(E)=0 ,\ E\subset X$; then $\int _E f(x) d\mu(x)=0$ .
In my attempts I've showed that $\forall \epsilon > 0 \ \ \exists \delta>0 :$ if $\mu(E)<\delta,\ E\subset X$ then $\int _E |f(x)| d\mu(x)<\epsilon$ 
Then how can I conclude $\int _E f(x) d\mu(x)=0$ ?
 A: You've shown that for any $\epsilon>0$, you can find a delta such that $\mu(E)<\delta$ implies $\int_E|f(x)|d\mu<\epsilon$. You are given $\mu(E)=0$, so just show that you can take $\epsilon$ arbitrarily small.
A: Let $f:X\to\mathbb{R}$ be integrable, where $X$ is a measure space. By definition
$$
\int_Efd\mu=\int_Ef_+d\mu-\int_Ef_-d\mu
$$
consider $\int_Ef_+d\mu$. By definition
$$
\int_Ef_+d\mu=\int{}f_+\chi_Ed\mu=sup\int\phi{}d\mu
$$
where $\phi$ is a simple function satisfying $0\leq\phi\leq{}f\chi_E$ where $\chi_E=1$ if $x\in{}E$ and zero otherwise. The supremum is taken over all simple functions satisfying the our constraints. Recall that all such simple functions can be written by definition
$$
\phi=\sum_{n=1}^{N}a_n\chi_{E_n}
$$
It is trivial to notice that in order to satisfy the requirement that $0\leq\phi\leq{}f\chi_E$ all $E_n$ in the expansion gotta satisfy $E_n\subset{}E$. It is well known that if $E_n\subset{}E$ then $\mu(E_n)\leq\mu(E)$, but since the measure of $E$ is already zero we see that all $E_n$ gotta have zero measure. Recalling that an integral over a simple function is by definition
$$
\int\phi{}d\mu=\sum_{n=1}^{N}a_n\mu(\chi_{E_n})
$$
we have that for all simple function in the supremum
$$
\int\phi{}d\mu=\sum_{n=1}^{N}a_n\mu(\chi_{E_n})=0
$$
since the supremum of a set of zeros goes by the name of zero we get that 
$$
\int_Ef_+d\mu=\int{}f_+\chi_Ed\mu=sup\int\phi{}d\mu=0
$$
substitue $f_+$ with $f_-$ in the reasoning above and you conclude that
$$
\int_Ef_-d\mu=0
$$
Therefore
$$
\int_Efd\mu=0
$$
