Let $A_1,A_2,...$ be an infinite sequence of events which is monotonically increasing, $A_n \subset A_{n+1}$ for every $n$. Let $A=\cup_{n=1}^{\infty} A_n$. Show that $P(A)=\lim_{n \rightarrow \infty} P(A_n)$.
My approach : As the $A_i$'s are monotonically increasing, their respective probabilities are monotonically increasing as well. (Not sure if this holds, but my argument is each subsequent $A_i$ is a larger set, contains more or equal elements => greater probability) Therefore, $P(A_1) \leq \cdot \cdot \cdot \leq P(A_n) \leq \cdot \cdot$.
Claim : The sequence $P(A_i)$ is bounded above by $P(A)$.
Suppose P(A) is not the upper bound of the sequence $P(A_i)$, then at least $\exists n_1 \in \mathbb{N}$, such that $P(A_{n_1}) \geq P(A)$. But since $A=\cup_{n=1}^{\infty} A_n$, if there exists an event $A_{n_1}$, where $n_1 \in \mathbb{N}$ such that $P(A_{n_1}) \geq P(A)$ then $A \leq \cup_{n=1}^{\infty} A_n$. Contradiction.
Therefore, since the sequence $P(A_i), i \in \mathbb{N}$ is bounded above by $P(A)$, it converges to $P(A)$, which implies $P(A)= \lim_{n \rightarrow \infty} P(A_n)$.
Please correct me if I'm wrong, I'm just curious whether this works or not.
I know someone else has posted this question already, link :Increasing sequence of events and the probability of their limit.
This does not prove yet that the sequence "converges" to $P(A)$.
(It does not exclude that it converges to some value less than $P(A)$.)
For a real proof you should have a look at the sets $B_n=A_n-\bigcup_{k=1}^{n-1}A_k$.
They are disjoint and satisfy:$$\bigcup_{n=1}^{\infty} B_n=\bigcup_{n=1}^{\infty}A_n=A$$ Then consequently: $$P(A)=\sum_{k=1}^{\infty}P(B_k)=\lim_{n\to\infty}\sum_{k=1}^{n}P(B_k)=\lim_{n\to\infty}P(A_n)$$