The limit of the integral when the set is decreasing in probability to zero. This is an exercise problem(#2 in section 3.2) from 'A course in probability theory'.
If $E(\vert X \vert ) < \infty$ and $\lim_{n \to \infty} P(A_n) = 0,$ then $\lim_{n \to \infty} \int_{A_n} X \ dP = 0.$
I tried using the modulus inequality of integrals but I cannot get the result. Thank you.
 A: The answer given by @d.k.o. is almost correct: Let us assume that $\int_{A_n} X \, dP \not \to 0$. Then there is some $\epsilon > 0$ and a subsequence $A_{n_k}$ with $|\int_{A_{n_k}} X \, dP | \geq \epsilon$ for all $k$.
Now, since $P(A_n) \to 0$, we can choose a further subsequence $A_{n_{k_\ell}}$ (which we call $B_\ell$ for brevity) with $P(B_\ell) \leq 2^{-\ell}$. In particular, $\sum_\ell P(B_\ell) <\infty$, so that the Borel Cantelli lemma implies $P(B_\ell \text{ i.o.}) = 0$ ($B_\ell$ infinitely often). This means $\chi_{B_\ell} \to 0$ almost surely.
Now we can use the dominated convergence theorem to conclude
$$
\int_{B_\ell} X \, dP \to 0,
$$
which is in contradiction to $|\int_{A_{n_k}} X \, dP|\geq \epsilon$ for all $k$.
A: The function $$\mu (A) =\int_A |X(\omega ) | dP(\omega ) $$ is a finite measure since $$E (|X|) <\infty .$$
Since $P(A) =0$ implies that $\mu (A) =0$ the measure $\mu $ is absolutely continuous with respect to the measure $P.$ Hence the condition $P(A_n ) \to 0 $ implies $\mu (A_n ) \to 0 $  and therefore if $P(A_n ) \to 0 $ we have $$\left |\int_{A_n} X(\omega )dP(\omega )\right|\leqslant \int_{A_n } |X(\omega )| dP(\omega ) =\mu (A_n ) \to 0.$$
A: Since $X1_{A_n}\xrightarrow{p}0$ and $|X1_{A_n}|\le |X|$ with $\mathbb{E}|X|<\infty$ the dominated convergence theorem (for sequences converging in prob.) implies that $\mathbb{E}X1_{A_n}\rightarrow 0$.
