Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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.

share|cite|improve this answer

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 $$ \int_Efd\mu=0 $$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.