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Problem:

$f$ is integrable on $[0,2]$, for any measurable set $E \subseteq [0,2]$, when $m(E) = 1$, $\int_E fdm = 0$ prove $f = 0$ $a.e.$ on $[0,2]$

I think the following theorem might be of help, but I'm not sure.

Theorem. If $f$ is integrable on $[a,b]$ and $\int_a^x f(t)dt = 0$ $\forall x \in [a,b]$, then $f = 0$ $a.e.$ on $[a,b]$.

The proof is in this link.

Especially Lime
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Guanfei
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1 Answers1

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Let $E_- = \{f < 0\}$ and $E_+ = \{f > 0\}$ and $A = [0,2] \setminus E_- \setminus E_+ = \{f = 0\}$. Suppose that $m(E_-) \geq 1$ or $m(E_+) \geq 1$. In either case we get a contradiction when integrating over the larger set. Thus $m(E_-),m(E_+) < 1$. Now we integrate over $E_- \cup A$ and $E_+ \cup A$ to find that in fact $m(E_-) = m(E_+) = 0$.

pg_star
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  • How to get the contradiction when integrating over the larger set? – Guanfei Jun 25 '20 at 09:27
  • Say $m(E_-) \geq 1$. Take a subset $E_-^* \subseteq E_-$ with $m(E_-^) = 1$. Then the integral over $E_-^$ is zero. But $f < 0$ on $E_-^*$, a contradiction. – pg_star Jun 25 '20 at 09:30
  • I see. But can we always find a subset $E_-^*$ $\subseteq$ $E_-$ with measure 1? – Guanfei Jun 25 '20 at 09:41