If $\mathbb{P}(A_n)\to 0$, prove that $\int\limits_{A_n}X\mathrm{d}\mathbb{P}\to 0.$ 
Let $(\varOmega,\mathcal{F},\mathbb{P})$ be a probability space. If $X:\varOmega\to \mathbb{R}$ is integrable and $(A_n) \subset\mathcal{F}$ such that $\mathbb{P}(A_n)\to 0$, then prove that $\int\limits_{A_n}X\mathrm{d}\mathbb{P}\to 0.$

Attempt. If $X=1_A$ for some $A\in \mathcal{F}$, then $$\int\limits_{A_n}X\mathrm{d}\mathbb{P}=\int1_{A_n}1_A\mathrm{d}\mathbb{P}=\int1_{AA_n}\mathrm{d}\mathbb{P}=\mathbb{P}(AA_n)\to 0,~~n\to \infty$$
since $\mathbb{P}(A_n)\to 0$. If $X=\sum_{k=1}^{m}a_k1_{A_k}$ is simple, then from linearity of integral:
$$\int\limits_{A_n}X\mathrm{d}\mathbb{P}=\sum_{k=1}^{m}a_k\int\limits_{A_n}1_{A_k}\mathrm{d}\mathbb{P}\to \sum_{k=1}^{m}a_k\cdot 0=0,~~n\to \infty.$$
If $X$ is non negative, then for some increasing sequence of simple rv's,
say $(X_k)$, $X_k\nearrow X$.
$$\lim_{n\to \infty}\int\limits_{A_n}X\mathrm{d}\mathbb{P}=\lim_{n\to \infty}\int\limits_{A_n} \lim_{k\to \infty} X_k\mathrm{d}\mathbb{P}=\lim_{n\to \infty}\int  \lim_{k\to \infty} X_k1_{A_n}\mathrm{d}\mathbb{P}\overset{!}{=}$$
$$  \lim_{n\to \infty}\lim_{k\to \infty} \int   X_k1_{A_n}\mathrm{d}\mathbb{P}\overset{?}{=}  \lim_{k\to \infty}\lim_{n\to \infty} \int   X_k1_{A_n}\mathrm{d}\mathbb{P}=\lim_{k\to \infty}0=0,$$
where the $!$ holds by the MCT. I am not sure about the $?$ equality, how the change of limits is justified.
(the case where $X$ takes values on $(-\infty,\infty)$ follows easily from the above, since $X=X^+-X^-$).
Thanks in advance for the help.

EDIT. According to the comment by @TheSilverDoe, we may use DCT. Indeed, $|X1_{A_n}|\leqslant |X|$ and $X$ is integrable. Also, is it true that $X1_{A_n}\to 0$ with prob. $1$, since $\mathbb{P}(A_n)\to 0$?
 A: This is consequence of the following lemma which holds for general measure spaces:
Lemma: If $X\in L_1(\Omega,\mathscr{F},\mu)$, then for any $\varepsilon>0$ there is $\delta>0$ such that $\mu(A)<\delta$ implies $\int_A|X|\,d\mu<\varepsilon$
Here is a short proof
Let $X_n=|X|\wedge n$. Then $X_n$ is nondecreasing (in $n$) and $X_n\nearrow |X|$ as $n\rightarrow\infty$. By monotone convergence $\int X_n\,d\mu\xrightarrow{n\rightarrow\infty}\int|X|\,d\mu$. Thus, given $\varepsilon>0$, there is $N_\varepsilon$ such that if $n\geq N_\varepsilon$,
$\int(|X|-X_n)\,d\mu<\frac{\varepsilon}{2}$.
Let $\delta=\frac{\varepsilon}{2N_\varepsilon}$. Then, for any $A\in\mathscr{F}$ with $\mu(A)<\delta$ we have
$$
\begin{align}
\int_A|X|\,d\mu&=\int_A(|X|-X_{N_\varepsilon})\,d\mu+\int_AX_{N_\varepsilon}\,d\mu\\
&\leq \int_\Omega(|X|-X_{N_\varepsilon})\,d\mu +N_\varepsilon\mu(A)< \frac{\varepsilon}{2}+N_\varepsilon\delta=\varepsilon
\end{align}
$$

Back to the OP: set $\mu=\mathbb{P}$. Given $\varepsilon>0$, let $\delta>0$ be as in the statement of the Lemma above. Since $\lim_n\mathbb{P}[A_n]=0$, there is $N$ such that $n\geq N$ implies $\mathbb{P}[A_n]<\delta$. Hence
$$\big|\int_{A_n}X\,d\mathbb{P}\big|\leq\int_{A_n}|X|\,d\mathbb{P}<\varepsilon$$
for all $n\geq N$. This shows that $\lim_n\mathbb{E}\big[\mathbb{1}_{A_n}X\big]=0$.
A: There is a pretty straightforward approach to this:
Split the integral into
$$\int\limits_{A_{n}} X dP = \int\limits_{A_{n} \cap \{|X| \le K\}} X dP + \int \limits_{A_{n} \cap \{|X|>K\}} X dP \le K \cdot P(A_{n}) + \int\limits_{|X|>K} |X| dP$$
Now note that $$\int\limits_{|X|>K} |X| dP \rightarrow 0$$ as $K \rightarrow \infty$ because of the dominated convergence theorem: $|X| \mathbb{1}_{|X|>K}$ is dominated by $|X|$.
So let $\epsilon > 0$ be arbitrary.
Choose $K$ large enough such that $$\int\limits_{|X|>K} |X| dP < \frac{1}{2}\epsilon$$
Then choose $n$ large enough such that $K \cdot P(A_{n}) < \frac{1}{2}\epsilon$.
Done!
A: Here another approach:
Let $X_n = X1_{A_n}$.
Note the following:

*

*$(X_n)$ is uniformly integrable family, since $|X_n| \le |X|$ which is integrable random variable.


*$X_n \to 0$ in probability. Indeed $0 \le \mathbb P(|X_n| > \varepsilon) = \mathbb P( \{ |X| > \varepsilon \} \cap A_n) \le \mathbb P(A_n) \to 0$
Now, it is known that $Y_n \to Y$ in probability + $(Y_n)$ uniformly integrable $\iff$ $Y_n \to Y$ in $L_1$
Hence $X_n \to 0$ in $L_1$ which means $\mathbb E[|X_n|] = \int_{A_n}|X|d\mathbb P \to 0$
A: A different approach.
What do we know? $X$ is integrable, meaning $\mathbb{E}[\vert X\vert]=\int_\Omega \vert X\vert d\mathbb{P} < \infty$.
Now, why would we expect $\int\limits_{A_n}X\mathrm{d}\mathbb{P}\to 0 \;$? When $\; \mathbb{P}(A_n)\to 0$ we are limiting our "area" of sampling, essentially, by multiplying or r.v. $X$ with the indicator $1_{A_n}$, And we get the new Integrabe function $Y_n=X\cdot1_{A_n}$
Now let's be original, for any such series of sets we can produce a worse one! We produce new sets $B_n$, such that $\mathbb{P}(B_n) \leq \text{min}(\mathbb{P}(A_n),\mathbb{P}(B_{n-1}))$ such that for any set S of measure $\mathbb{P}(S)\leq\mathbb{P}(B_n)$
$$\int\limits_{S}X\mathrm{d}\mathbb{P}\leq \int\limits_{B_n}X\mathrm{d}\mathbb{P}$$
The sets we constructed also have the property $B_n\subseteq B_{n-1}$ and more importantly, there is a monotone increasing function $\phi:\mathbb{N}\to\mathbb{N}$ such that $\mathbb{P}(A_{\phi (n)}) \leq \mathbb{P}(B_n)\to0$ and more importantly:
$$\int\limits_{A_{\phi (n)}}X\mathrm{d}\mathbb{P}\leq \int\limits_{B_n}X\mathrm{d}\mathbb{P}$$
We can now define the monotone decreasing integrable functions  $Y_n=X\cdot1_{B_n}$ with the point limit $0$ and achieve the desired result.
The proof of the existence of $B_n$ as such is a cute measure-theoretic exercise.
