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Let $f\in L^1(A)$ and $A_n\subseteq A: \mu(A_n)\xrightarrow{n\to\infty}0$, show that $$\int_{A_n} f \xrightarrow{n\to\infty}0 $$

What i tried:
We can write $f$ as following: $$f = f \cdot 1_{\{|f| >M\}} +f \cdot 1_{\{|f| \le M\}}, M>0. $$


Note: if $f\in L^1(A)$ and $g_n(x) = |f(x)| \cdot 1_{\{|f| \ge n\}}$ then $$\int_A g_n \xrightarrow{n\to\infty}0 \qquad (*)$$ because $|g_n(x)|\le |f(x)|\in L^1(A) $ and $\lim g_n = 0$ hence, from dominated convergence theorem we got $(*)$


$$\int_{A_n}f = \int_{A_n}f \cdot 1_{\{|f| >M\}} +\int_{A_n}f \cdot 1_{\{|f| \le M\}} $$ $$\int_{A_n}f \cdot 1_{\{|f| \le M\}} \le M\mu(A_n)\xrightarrow{n\to\infty}0$$ let $\epsilon >0$ , we can choose M large enough s.t: $$\int_{A_n}f \cdot 1_{\{|f| >M\}}\le \int_{A}f \cdot 1_{\{|f| >M\}} \le \epsilon.$$

is my approath correct?
Thank you!

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Your argument works fine if $f$ is non-negative but in general, $\lim \sup \int_{A_n}f \leq 0$ does not imply that $\int_{A_n}f \to 0$. Start with the inequality $|\int_{A_n}f| \leq \int_{A_n} |f|$ and then apply your argument with $|f|$ in place of $f$.

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