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.

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 $$\left |\int_{A_n} X(\omega )dP(\omega )\right|\leqslant \int_{A_n } |X(\omega )| dP(\omega ) =\mu (A_n ) \to 0.$$
