Proof that $P(A_{n} \cap B)\rightarrow P(B)$ given $A_{n} \nearrow A$ and $P(A)=1$

Question

Let $A$, $B$ and $A_{1},A_{2},\ldots,A_{n}$ aleatory events in a sample space, with $A_{n} \nearrow A$, and $P(A)=1$.

Proof that $P(A_{n} \cap B)\rightarrow P(B)$

This is how I am starting the answer:

Let $B = \cup (A_{n}\cap B)$

Is it the right way?

Any help. I am really stuck in here.

Answer 1

Elaborating on the answer above, note that since $A_n \nearrow A$, then $(A_n \cap B) \nearrow (A \cap B)$. Proving this should be straightforward (e.g. from elementary set theory).

Furthermore, since $P(A) = 1$, $P(A^c) = 0$. Therefore, because $P(A^c \cap B) \leq P(A^c) = 0$, we see that \begin{align}P(B) &= P(A \cap B) + P(A^c \cap B) \\&= P(A \cap B).\end{align} Putting it all together, we see that \begin{align} P\bigg( \bigcup_i A_i \cap B \bigg) &= \lim_i P(A_i \cap B) \\ &= P( A \cap B)\\&= P(B).\end{align}

Answer 2

Just show $(A_n \cap B) \nearrow (A \cap B)$ and note $P(A \cap B) = P(B)$.