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Let $f \in L_1([0, 1])$ and $A_n \subset [0, 1]$ be measurable such that $\lambda(A_n) \to 0$ where $\lambda$ is the Lebesgue measure.

Can we prove that $\int_{A_n} f \mathrm d \lambda \to 0$ as $n \to +\infty$? If not, is it true if we impose more that $A_{n} \supset A_{n+1}$?

My attempt: If $A_{n} \supset A_{n+1}$ then $$ a_n :=\int_{A_n} |f| \mathrm d \lambda $$ decreases as $n \to \infty$. However, I could not prove that $a_n \to 0$.

Akira
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  • Integral over $A_n$ is equivalent to multiplying $f$ by the indicator of $A_n$. Since $f$ is $L^1$ you can apply DCT. – Leonid Jul 12 '22 at 23:33
  • @Leonid It seems we need $A_{n} \supset A_{n+1}$ to apply DCT, right? – Akira Jul 13 '22 at 00:01
  • No, why do you say that? The sequence $1_{A_n}f$ is dominated by $|f|$ and its pointwise limit is the $0$ function a.e. (since $f$ is $L^1$ it is finite a.e. hence the pointwise limit exists a.e. and is zero) – Leonid Jul 13 '22 at 00:05
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    Actually never-mind the comment above. After having considered more carefully the pointwise convergence of that sequence, it seems that the above assumption is indeed required. Since intuitively one can construct a sequence which is shrinking but is "oscillating" through the points. – Leonid Jul 13 '22 at 00:22

1 Answers1

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Recall the absolute continuity of Lebesgue integral: For each $\epsilon>0$, there is $\delta>0$ such that for all measurable $E\subset[0,1]$ with $\lambda(E)<\delta$, we have $$\int_E |f|\,d\lambda<\epsilon.$$

Now for this $\delta>0$, we can find $N>0$ such that $\lambda(A_n)<\delta$ for all $n>N$, so $$\left|\int_{A_n}f\,d\lambda\right|\leq\int_{A_n}|f|\,d\lambda<\epsilon,\quad n>N.$$

We don’t need $A_{n+1}\subset A_n$.

Feng
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