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Let $I\subset \Bbb R$ be an interval and let $g\in L^1(I)$ integrable over $I$ such that $$\int_a^b g(x) dx=0\qquad \forall\,\, a,b\in I. $$

I would like to prove that $g=0$ a.e. without using density argument. My idea is to consider the set $$A=\{x\in I\,:\, |g(x)|>0\}= \bigcup_{n\geq1} A_n $$ where $$A_n=\{x\in I\,:\, |g(x)|>\frac{1}{n}\}.$$

How to show that $|A_n|=0$, i.e., $A_n$ is of measure zero?

In this way we can conclude that $|A|=0$ and hence $g=0$ almost everywhere.

Guy Fsone
  • 23,903
  • One strategy is to se how that the subsets of $[a,b]$ over which the integral is $0$ forms a Dynkin system. The conclusion follows by Dynkin’s monotone theorem – Mittens Nov 23 '23 at 17:55
  • Check this: https://math.stackexchange.com/q/16244/42969 – Martin R Nov 23 '23 at 17:57

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