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Prove that if $f \in L^{1}(A)$ and {${A_{n}}$} is a sequence of measurable subsets of A with $\lim_{n \to \infty}m(A_{n})=0$ then $lim_{n \to \infty} \int_{A_n}f=0$

I know there exists a very similar post for tackling this particular problem but I cannot find the complete proof of it. I would be grateful if someone could help.

2 Answers 2

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$\lvert\int f1_{A_n}\,dm\rvert\leq\int\lvert f1_{A_n}\rvert\,dm\leq m(A_n)\int\lvert f\rvert\,dm\to0$ by Holder's inequality.

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    Thanks for the response. Will the same apply if we did not have that f was in $L^1$? For example if we had that f was integrable in $(R,Σ,m)$? – 2017-02-11
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    $f$ being integrable is the same as belonging to $L^1$. – 2017-02-12
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It's easy. For $n \in{} \mathbb{N}$, $\int_{A_n} f=\int_{A} f \mathbb{1}_{A_n}$. By Holder's inequallity, we have $\left |{\int_{A} f \mathbb{1}_{A_n}}\right |^2 \leq \int_{A} f^2 \int_{A} \mathbb{1}_{A_n}^2= \int_{A} f^2 \int_{A} \mathbb{1}_{A_n}=m(A_n)\int_{A} f^2 \xrightarrow{ n \to \infty } 0$ since $\lim_{ n \to \infty } m(A_n)=0$ and $f \in L^1(A)$. We conclude $\lim_{ n \to \infty } \int_{A} f \mathbb{1}_{A_n}=0$.

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    If $f \in L^1$, $f$ is not necessarily $L^2$... – 2017-02-11
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    You are right. I ignored that mistake. – 2017-02-20