In probability space $(\Omega,\mathcal{F},P)$, we have random variable $X$ that maps $\Omega$ into $R$. Assume $\mathbb{E}{|X|}<\infty$ and let $A_n$ be a sequence in $\mathcal{F}$ satysfying $\lim_{n \to \infty} P(A_n)=0$. Prove that $\lim_{n \to \infty} \int_{A_n} X dP =0$.
I have proved it when we know $\lim_{n \to \infty} A_n$ exists:
$\lim_{n \to \infty} \int_{A_n} X dP = \lim_{n \to \infty} \int_{\Omega} 1_{A_n}X dP $
Since $|X 1_{A_n}| \leq |X|$ , we can use Dominated convergence theorem. Therefore,
$\lim_{n \to \infty} \int_{A_n} X dP = \int_{\Omega} \lim_{n \to \infty} 1_{A_n}X dP = \int_{\Omega} 1_{ \lim_{n \to \infty} A_n}X dP = \int_{\lim_{n \to \infty} A_n} X dP $
Then, by absolute continuity, I have claimed that $\lim_{n \to \infty} P(A_n)=0$ yields $\lim_{n \to \infty} \int_{A_n} X dP =0$.
But, how we can prove this when we don't know $\lim_{n \to \infty} A_n$ exists?
We do not know that $\lim_n A_n$ exists, nor do we need it for the entire sequence. Suppose for contradiction that $\limsup_n \int_{A_n} X dP > \epsilon > 0$. Then after passing to a subsequence we may assume that $ \int_{A_n} X dP > \epsilon$ for all $n$. Since $P(A_n) \to 0$ by passing to a further subsequence we may assume $\sum_n P(A_n) < \infty$. By a Borel-Cantelli lemma we know that the probability that infinitely many $A_n$ occurs is zero, in other words $1_{A_n} \to 0$ a.s. Now we can apply dominated convergence to get $0<\epsilon<\lim_n \int_{A_n} X dP = 0$, a contradiction.