If $A_1\subset A_2\subset A_3\subset\cdots$ is an increasing sequence of events with limit $A=\bigcup_{i=1}^\infty A_i$. Prove that $$\lim_{n\rightarrow\infty} P(A_n)=P(A)$$
My attempt so far:
Since $A_1\subset A_2\subset A_3\subset\cdots$ is increasing and $$\bigcup_{i=1}^\infty A_i=\lim_{n\rightarrow\infty} A_n \wedge A=\bigcup_{i=1}^\infty A_i$$ then $\lim_{n\rightarrow\infty}A_n=A$. I have that $$P\left(\lim_{n\rightarrow\infty}A_n\right)=P(A)\Rightarrow \lim_{n\rightarrow\infty}P(A_n)=P(A)$$
Best Answer
Let $B_1 = A_1$ and $B_{n+1} = A_{n+1}\setminus A_n$ for $n\ge 1$. Then $$ A_N = \bigcup_{n=1}^N B_n $$ and $$ \bigcup_{n=1}^\infty A_n = \bigcup_{n=1}^\infty B_n, $$ and $$ B_n \cap B_m = \varnothing \text{ for } n\ne m. $$ So \begin{align} P(A) & = P\left( \bigcup_{n=1}^\infty A_n \right) = P\left( \bigcup_{n=1}^\infty B_n \right) \\[10pt] & = \sum_{n=1}^\infty P(B_n) \qquad \left( \begin{array}{l} \text{by countable additivity of } P \text{ and} \\ \text{pairwise disjointness of } \{B_n\}_{n=1}^\infty \end{array} \right) \\[10pt] & = \lim_{N\to\infty} \sum_{n=1}^N P(B_n) = \lim_{N\to\infty} P\left( \bigcup_{n=1}^N B_n \right) = \lim_{N\to\infty} P(A_N). \end{align}