# $A_n\subseteq A: \mu(A_n)\xrightarrow{n}0 \Rightarrow \int_{A_n} f \xrightarrow{n} 0$ [duplicate]

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!

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 $$f$$.