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Let $f$ be Lebesgue-integrable on $[0,1]$. Suppose $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$. Show $\int_Af(x)\,dx=0$ for every measurable subset $A$ of $[0,1]$.

*Let $A$ be a measurable subset of $[0,1]$. Then $A$ can be written as the union of disjoint, countable? intervals. Since $\int_a^bf(x)\,dx=0$ for all $0\leq a\leq b\leq 1$, each integral of $f$ over each interval is $0$ so $\int_Af(x)\,dx=0$.

I'm not sure if I did it right...

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  • $\begingroup$ You can make $[a,b]$ arbitrarily small like a Cantor set whose Lebesgue measure is 0. $\endgroup$ – Pedro Gomes Aug 11 '17 at 20:35
  • $\begingroup$ My initial inclination without thinking about this much is to show that if the hypothesis holds then $f$ is $0$ almost everywhere, and then go on from there. $\endgroup$ – Michael Hardy Aug 11 '17 at 20:41
  • $\begingroup$ The set of rational numbers is measurable but is not a union countably many disjoint intervals. $\endgroup$ – Michael Hardy Aug 11 '17 at 20:41
  • $\begingroup$ @MichaelHardy if $f\geq 0$ or $f\leq 0$, then it would be much easier to show $f=0$, but here I'm not sure how to show that $f=0$. It does fell a little off to say that "countable..." I'm not quite sure how to prove this problem... $\endgroup$ – 2ndaccount Aug 11 '17 at 20:45
  • $\begingroup$ You could argue as I do here $\endgroup$ – David Mitra Aug 11 '17 at 20:52
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The Lebesgue measure is regular, and in particular outer regular. Therefore there exists a sequence $(O_n)$ of open measurable sets containing $A$ such that $\lambda(A)=\inf \lambda(O_n)$.

Then $\displaystyle \int_A f(x)\, \mathrm{d}x=\int_{O_n} f(x)\, \mathrm{d}x - \int_{O_n - A} f(x)\, \mathrm{d}x$.

Since $O_n$ is an open set of $\mathbb{R}$, it is a countable union of open intervals and $\int_{O_n} f(x)\, \mathrm{d}x=0$. By dominated convergence you can show that $\int_{O_n - A} f(x)\, \mathrm{d}x \to 0$: use the sequence of functions $f1_{O_n-A}$ dominated by $|f|$.

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  • $\begingroup$ what do you mean regular? $\endgroup$ – 2ndaccount Aug 11 '17 at 20:45
  • $\begingroup$ en.wikipedia.org/wiki/Regular_measure $\endgroup$ – fonfonx Aug 11 '17 at 20:47
  • $\begingroup$ What are the sequence of function bounded by? (in DCT) $\endgroup$ – 2ndaccount Aug 11 '17 at 20:48
  • $\begingroup$ $|f|$, I edited my answer $\endgroup$ – fonfonx Aug 11 '17 at 20:50
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This is a standard application of Dynkin's $\pi$-$\lambda$ Theorem.

Define $\mathcal{P}=\{(a,b]\mid0\leq a<b\leq1\}\cup\{\emptyset\}$ and $\mathcal{L}=\{A\in\mathcal{B}([0,1])\mid\int_{A}f(x)\,dx=0\}$. It is routine to verify that $\mathcal{P}$ is a $\pi$-class and $\mathcal{L}$ is a $\lambda$-class. That is:

  • For any $A,B\in\mathcal{P}$, we have $A\cap\mathcal{B\in\mathcal{P}}$,
  • $\emptyset\in\mathcal{L}$,
  • $A^{c}\in\mathcal{L}$ whenever $A\in\mathcal{L}$, where the complement is taken with respect to $[0,1]$,
  • $\cup_{i=1}^{\infty}A_{i}\in\mathcal{L}$ whenever $A_{i}\in\mathcal{L}$ and $A_{1},A_{2}\ldots$ are pairwisely disjoint.

By the given condition, we have $\mathcal{P}\subseteq\mathcal{L}$. Therefore, by Dynkin's $\pi$-$\lambda$ Theorem, we have $\sigma(\mathcal{P})\subseteq\mathcal{L}$. Note that $\sigma(\mathcal{P})=\mathcal{B}([0,1])$. Therefore, $\mathcal{L=B}([0,1])$. Q.E.D.

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  • $\begingroup$ I forget to say, for every Lebesgue measurable set $A$, there exists a Borel set $B$ such that $m(A\Delta B)=0$. (In case if you want to prove that the proposition holds for all Lebesgue measurable subsets of [0,1] rather than Borel measurable subsets) $\endgroup$ – Danny Pak-Keung Chan Aug 11 '17 at 22:43

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