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:

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

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

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

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

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


$A = B\cup C$ (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?

  • 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

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.

| cite | improve this answer | |
  • 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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