If $f \in L^1$ $\mu(A_n)\to 0$ then $\int_{A_n}f\to 0$ In this quesiton, what happens if $p=1$? Does the statement remain true? That is, if $(X,\Sigma,\mu)$ is a measure space, $f\in L^1$ and if $\{A_n\}$ be a sequence in $\Sigma$ such that $\mu(A_n)\to 0$, is it true that $\int_{A_n}fd\mu\to 0$?
Following the answer of that question, we'd get
$$\left|\int_{A_n} f\, d\mu\right| \le \int_X \chi_{A_n}|f|\, d\mu \le \|\chi_{A_n}\|_{\infty} \|f\|_1.$$
Which is true, but what's next? Do we have that $\|\chi_{A_n}\|_{\infty} \to 0$ when $\mu(A_n)\to 0$? How to prove that?
 A: You can always do the following -  let $k \ge 1$ and observe $$\left| \int_{A_n} f \, d\mu \right| \le \int_{A_n \cap \{|f| \le k\}} |f| \, d\mu + \int_{A_n \cap \{|f| > k\}} |f| \, d\mu \le k \mu(A_n) + \int_{\{|f| > k\}} |f| \, d\mu.$$ Let $n \to \infty$ to find $$\tag{*}\limsup_{n \to \infty} \left| \int_{A_n} f \, d\mu \right| \le \int_{\{|f| > k\}} |f| \, d\mu = \int_X \chi_{\{|f| > k\}} |f| \, d\mu.$$ But $\chi_{\{|f| > k\}} |f| \le |f|$ and $\chi_{\{|f| > k\}} |f| \to 0$ almost everywhere, so 
$$ \lim_{k \to \infty} \int_X \chi_{\{|f| > k\}} |f| \, d\mu = 0$$ by dominated convergence. Thus the right hand side of (*) can be made arbitrarily small for large $k$, giving you
$$\limsup_{n \to \infty} \left| \int_{A_n} f \, d\mu \right| = 0.$$
A: If $\mu(A_n)\xrightarrow{n\rightarrow\infty}0$ then $\mathbb{1}_{A_n}\xrightarrow{n\rightarrow\infty}0$ in $L_1$. This implies that every subsequence of $\{\mathbb{1}_{A_n}\}$ has a subsequence along which $\mathbb{1}_{A_n}$ converges to $0$ $\mu$-a.s. 
Combining these with  Lebesgue dominated convergence should imply that $\int_{A_n}f\,d\mu\xrightarrow{n\rightarrow\infty}0$.
AS for your second question, Umberto is right. 
