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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 $\int _E f(x) d\mu(x)=0$ ?

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Show that you can bound the integral by $1/n$ for all $n$, or something else that goes to $0$ as $n\to\infty$. For each $n$, you will get a $\delta_n>0$. But $\mu(E) = 0$ implies that $\mu(E) < \delta_n$ for all $n$. You should be careful with this, but this is in the same direction as what Alex suggests below. – John Martin Oct 31 '12 at 3:48
    
Another way is by using that for $f$ nonnegative $\int_E f=\mu(\{(x,y):x\in E\text{ and } 0\leq y\leq f(x)\})$ – leo Oct 31 '12 at 5:10

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.

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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 $$ \int_Efd\mu=0 $$

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