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Suppose $f$ belongs to $L^1[0,1]$ and given any pair of rationals $0\leq p<q\leq 1$ in the case $$\int_p^q fd\mu =0$$ Prove that $f=0$ a.e on [0,1].

I ma preparing for an exam and I came across this question, I need some help. This is what I am thinking.

I have $\int_0^1 |f|d\mu <\infty$ and so $f$ is finite a.e. If I decompose $f$ as $f=f^+-f^-$ where $f^+=max(f(x),0)$ and $f^-=max(-f(x),0)$ and so $f^+$ and $f^-$ are both finite a.e

and let $p=0$, $q=1$.

That means $\int_p^qfd\mu=\int_0^1( f^+-f^-)d\mu=\int_o^1f^+d\mu-\int_0^1f^-d\mu$

but $\int_p^qfd\mu=0$ implies $ \int_0^1f^+d\mu=\int_0^1f^-d\mu$

Hence $f^+=f^-$ a.e and this will mean $f^+=0=f^-$ according to the definition of $f^+$ and $f^-$. To conclude, I will say by monotonicity of the integral, this holds for any other pair of $p,q$.

Does this make sense? if not can someone show me how? Thank you

Cnine
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  • Why would $\int_0^1 f^+ d\mu = \int_0^1 f^- d\mu$ imply that $f^+ = f^-$ a.e.? Consider e.g. $f(x) = \sin(2\pi x)$. –  Sep 08 '17 at 18:01
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    It would be helpful to know what theorems are available. One line of argument: the condition gives us that $\int_0^q f d\mu = 0$ for any rational $q \in [0,1]$. Since $f \in L^1$, the function $x \mapsto \int_0^x f d\mu$ is continuous on $[0,1]$, so this forces $\int_0^x f d\mu = 0$ for any real $x \in [0,1]$. Now the Lebesgue differentiation theorem immediately implies that $f = 0$ a.e. –  Sep 08 '17 at 18:12
  • @Bungo, My assumption may be wrong. Any idea how I should approach it? Because since $f^+ and f^-$ are both positive functions I though that should hold. – Cnine Sep 08 '17 at 18:19
  • Consider the example $f(x) = \sin(2\pi x)$. Then $f^+$ and $f^-$ have the same area (they are shifted copies of each other). But they are not equal a.e.; in fact, $f^+$ is nonzero on $(0,1/2)$, whereas $f^-$ is nonzero on $(1/2,1)$. –  Sep 08 '17 at 18:38

1 Answers1

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If $f,g\in L^1(\Omega,\mu)$, then $\int_{\Omega}fd\mu=\int_{\Omega}gd\mu\not\Rightarrow f=g\;a.e.$ A counterexample is given in the comments. However, what is true is that if $\int_{A}fd\mu=\int_{A}gd\mu$ for all $A$ measurable, then $f=g\;a.e.$ (See Folland's Real Analysis proposition $2.23(b)$ for a proof).

With this in mind, I think your idea of a proof can be saved. First, as $f^+$ and $f^-$ are positive functions, the functions $\mu(E)=\int_Ef^+d\mu$ and $\nu(E)=\int_Ef^-d\mu$ are measures on $[0,1]$. Therefore, it suffices to prove that if two measures, $\mu,\nu$ agree on a generating set of a sigma-algebra, then they agree on the entire sigma-algebra (this follows because the set of intervals with rational endpoints generates the Borel sigma-algebra).

A proof of this fact can be found using the Monotone Class Theorem. See, for example, here.

Moya
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  • Thanks, I looked the book up, I understand that , for what I did to work, I need to show that , $\int_p^qf^+=\int_p^qf^-$ holds for all measurable set. So what I am thinking now is to try and show that any measurable set $E$ can be written as a combination ( maybe countable union or intersection ) intervals with rational endpoints. $E=\cup_n(a_n,b_n)$ where $a_n,b_n \in \mathbb{Q}$. Can you help me do that, if is a good approach? – Cnine Sep 22 '17 at 05:57
  • See here: https://math.stackexchange.com/a/166089/192336 – Moya Sep 23 '17 at 20:14