Convergence of probability implies convergence of integral Given $A$ is an integrable random variable and $P(A_n)\rightarrow 0$ as $n\rightarrow \infty$. Prove that $\int_{A_n} A\ dP\rightarrow 0$ as $n\rightarrow \infty$.
My Solution By using SLNN result, which gives $P(A_n)\rightarrow 0$ as $n\rightarrow \infty$ implies $1_{A_n}\rightarrow 0$ as $n\rightarrow \infty$ almost surely. Since $A$ is a random variable, $A\cdot 1_{A_n}\rightarrow 0.A= 0$ almost surely. Since $A$ is integrable, $\int_{\omega} A\cdot 1_{A_n} dP\rightarrow \int_{\omega} 0\ dP$ as $\ n\rightarrow \infty$, or $\int_{A_n} A\ dP = \int_{\omega} A\cdot 1_{A_n} dP\rightarrow 0$ as $n\rightarrow \infty$.
My Question Is the solution above correct? If not, please help me correct it.

Your argument doesn't look right, as pointed out in the comments. Here's an alternative proof.
Suppose the conclusion is false.  Then there exists an $\epsilon > 0$ and a sequence $(A_n)$ of events such that $P(A_n) < 2^{-n}$ and $\int_{A_{n}} A dP > \epsilon '$. By the Borel-Cantelli Lemma, $P(\limsup A_n) = 0$. But, letting $F = \limsup A_{n}$, (Reverse) Fatou's Lemma implies that $\int_F A dP \geq \limsup \int_{A_n} A dP > \epsilon$. This is a contradiction.
