\begin{proof} Since $A_{n}\subseteq A_{n+1}$ and $A=\bigcup_{n}A_{n}$ then we define $B_{n}$ such that: - $B_{1}=A_{1}$ - $B_{2}=A_{2}\setminus A_{1}$ - $B_{3}=A_{3}\setminus A_{2}$ - … Then $$A=\bigcup_{j} B_{j}$$and hence $$\mathbb{P}(A)=\mathbb{P}\left( \bigcup_{j}B_{j} \right)=\sum_{j}\mathbb{P}(B_{j})=\lim_{ n \to \infty } \sum_{j=1}^{n}\mathbb{P}(B_{j})=\lim_{ n \to \infty } \mathbb{P}(A_{n})$$same logic holds for decreasing events but with complements involved. \end{proof} ## Remark Simply a by-product of Measure.