Proof that when $n \to \infty$, then $P(A_{n}\cap B_{n})\to p$

Question

Show that $A_1$, $A_2$, ... and $B_1$, $B_2$ ... are all aleatory events of the same probability space such that $P(A_{n}) \to 1$ and $P (B_{n})\to p$, when $n \to \infty$ then $P(A_{n}\cap B_{n})\to p$.

I know there is a theorem that says:

If the sequence $(A_{n})_{n>1}$, where $A_{n} \in \mathbb{A}$, decrease to the empty set, $A_{n} \supset A_{n + 1}$ for all n, then $P (A_{n}) \to 0$ when $n \to \infty$.

I am almost sure that I have to use this Theorem but I am not able to do this.

Any help?

Answer 1

Unfortunately there is no assumption of non-increasingness /decreasingness hence we cannot use directly this result. However, notice that $$ \mathbb P\left(A_n\cap B_n\right)+\mathbb P\left(A_n^c\cap B_n\right)=\mathbb P\left(B_n\right) $$ hence $$ \mathbb P\left(A_n\cap B_n\right)=\mathbb P\left(B_n\right)-\mathbb P\left(A_n^c\cap B_n\right). $$ The only remaining question is: what is $\lim_{n\to +\infty}\mathbb P\left(A_n^c\cap B_n\right)$?