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 $