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Proof that if $f$ is integrable and $\int_E {f}\,{d\mu} = 0$ for all $E\in\Sigma$ then $f=0$ almost everywhere

My attempt:

I called:

$A = \{ x\in X : f(x) \ne 0 \}$

$B_n = \{ x\in X: f(x) \gt \frac{1}{n} \}$

$C_n = \{ x\in X:f(x) \lt -\frac{1}{n} \}$

$B = \bigcup_{n=1}^{\infty}{B_n}$

$C = \bigcup_{n=1}^{\infty}{C_n}$

Then:

$A = B\cup C$ (with $B\cap C=\emptyset$), $B_{n} \subset B_{n+1}$ and $C_{n} \subset C_{n+1}$.

Since $\mu$ is continuous we have:

$\mu(B_n)\to \mu(B)$ and $\mu(C_n)\to \mu(C)$

Let's suppose that $\mu(A)>0$. Since $\mu(A) = \mu(B) + \mu(C)$, we have two possibilities:

a) $\mu(B)>0$

Since $\mu(B_n)\to \mu(B)$, there is an $k$ such that $\mu(B_{k}) \gt 0$.

Thus $$\int_{B_k} {f}\,{d\mu} \geq \int_{B_k} {\frac{1}{k}}\,{d\mu} = \frac{\mu(B_k)}{k} \gt 0$$ which is a contradiction.

b) $\mu(C)>0$

Since $\mu(C_n)\to \mu(C)$, there is an $k$ such that $\mu(C_{k}) \gt 0$.

Thus $$\int_{C_k} {(-f)}\,{d\mu} \geq \int_{C_k} {\frac{1}{k}}\,{d\mu} = \frac{\mu(C_k)}{k} \gt 0$$ which is a contradiction.

Thus $\mu(A)=0$.

Am I right? Is there an easier way to solve this problem?

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  • 5
    $\begingroup$ Excellent job. There is (as far as I know) no proof that is essentially different. $\endgroup$ – drhab Oct 12 '15 at 20:15
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Your proof is right, and here is my attempt to the proof. If not, then there exist a set $E_0$ such that $\mu(E_0) >0$ and $f(x) \neq 0 ~ \text{for any} ~ x \in E_0$, without loss of generality, we may assume that $$f(x) > 0 ~\text{for any} ~ x \in E_0.$$ So we get $\int_{E_0} f(x) d \mu (x) > 0$ since $\mu(E_0) >0$ and $f(x) > 0 $ for any $x \in E_0$, which is contradictory to the condition.

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  • 1
    $\begingroup$ The fact that $\mu(E_0) >0$ and $f(x) > 0 $ for any $x \in E_0$ implies $\int_{E_0} f(x) d \mu (x) > 0$ is (true but) at least as elaborate as the result to be shown hence taking the former for granted to prove the latter is odd. $\endgroup$ – Did Sep 14 '16 at 8:38
  • $\begingroup$ Since $\mu (E_0) >0$ and $f(x) >0$ for any $x \in E_0$, so $\inf_{x \in E} f(x) \geq 0.$ Then we have $$\int_{E_0} f(x) d \mu (x) > \int_{E_0} \inf_{x\in E_0} f(x) d \mu(x) \geq \int_{E_0} 0 d \mu(x) = 0.$$ $\endgroup$ – Haipeng Chen Sep 14 '16 at 12:23
  • $\begingroup$ Why the strict inequality? That $\int_{E_0}d(x)d\mu(x)\geqslant0$ is direct, but one needs $>0$. $\endgroup$ – Did Sep 14 '16 at 13:16
  • $\begingroup$ If $\int_{E_0} f(x) d\mu(x) =0$, since $\mu(E_0)>0$ and $f(x)\geq 0$, then we see $\int_{E_0} f(x) d\mu(x) =0$ if and only if $f(x)=0$ for any $x \in E_0$. $\endgroup$ – Haipeng Chen Sep 15 '16 at 14:54
  • $\begingroup$ "We see" Do we? Since you are again taking the result for granted, your comment does not hold water either. Sorry. $\endgroup$ – Did Sep 15 '16 at 15:34

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