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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.2012-10-31
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    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)\})$2012-10-31

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