Chebyshev's inequality, measure theory proof? Can someone lead me to to the answer (that means you don't post the answer).


  Let $f$ be measurable with $f &gt; 0$ almost everywhere. If $\int_E f = 0$ for some measurable set $E$, then $m(E) = 0$.


So I think by Chebyshev's inequality, we get for each $a \geq 0$,

$$\int_E f \geq a m(x \in E : f \geq a).$$

Select $a = 1/n$, then

$$0 = \int_E f \geq (1/n) m(x \in E : f \geq 1/n).$$

So $m(x \in E : f \geq 1/n) = 0 \implies m(\cup_{n\geq 1}E_n) = 0$

with $E_n = (x \in E : f \geq 1/n)$, but $ m(\cup_{n\geq 1}E_n) = m(E) = 0$ by assumption that $m(f &gt; 0) = 0$
Assume that domain of f is a measurable set A.Since f>$0$ almost everywere then the set

B={x$\in$ A :f(x)&lt;=$0$} has measure m(B)=$0$.Then
f>$0$ in A\B.Also E $\subseteq$ A and f>$0$ in E\B  .

So we apply Chebyshev.For $\frac{1}{n}$ we get that,
$\frac{1}{n}$*$m(E_{n})$ $\leq\ \int_E f\ $ with

$E_{n}$ = {x $\in$ E\B : f>= $\frac{1}{n}$ }.

Hence,$m(E_{n})$=$0$ and since the sequence of sets $(E_n)_{n=1}^\infty$
is increasing so  $\lim_{n\to\infty}(E_n) $ = m$(\cup_{n\geq 1}E_n)$=m(E/B)=$0$

We Conclude that m(E\B)=$0$ and m(B)=$0$.Therefore m(E)=$0$.
Assume that domain of f is a measurable set A.Since f>$0$ almost everywere then the set

B={x$\in$ A :f(x)&lt;=$0$} has measure m(B)=$0$.Then
f>$0$ in A\B.Also E $\subseteq$ A and f>$0$ in E\B  .

So we apply Chebyshev.For $\frac{1}{n}$ we get that,
$\frac{1}{n}$*$m(E_{n})$ $\leq\ \int_E f\ $ with

$E_{n}$ = {x $\in$ E\B : f>= $\frac{1}{n}$ }.

Hence,$m(E_{n})$=$0$ and since the sequence of sets $(E_n)_{n=1}^\infty$
is increasing so  $\lim_{n\to\infty}(E_n) $ = m$(\cup_{n\geq 1}E_n)$=m(E/B)=$0$

We Conclude that m(E\B)=$0$ and m(B)=$0$.Therefore m(E)=$0$.