I am trying to understand a step in a proof about continuity of probability.
Consider $\{A_n:n\in\mathbb{N}\}$ a sequence of events and $A$ an event such that $\lim_{n\to\infty}A_n = A$, where we say a sequence of events converge to a certain $A$ event iff
$$ \liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k = \bigcap_{n=1}^{\infty}\bigcup_{k=n}^{\infty}A_k= \limsup_{n\to\infty}A_n $$
The proof uses the following inequality:
$$ \limsup_{n\to\infty}\mathbb{P}(A_n)\leq \mathbb{P}(\limsup_{n\to\infty}A_n) $$
And the proof goes:
Notice that $\forall n \in \mathbb{N}$ $A_n\subseteq\bigcup_{k=n}^{\infty}{A_k}$. Then $\mathbb{P}(A_n)\leq\mathbb{P}(\bigcup_{k=n}^{\infty}A_k)$. Then,
$$ \limsup_{n\to\infty}\mathbb{P}\left(A_n\right)\leq \limsup_{n\to\infty}\mathbb{P}\left(\bigcup_{k=n}^{\infty}A_k\right) = \lim_{n\to\infty}\mathbb{P}\left(\bigcup_{k=n}^{\infty}A_k\right) = \mathbb{P}\left(\lim_{n\to\infty}\bigcup_{k=n}^{\infty}A_k\right) $$
The proof continues and I get it. However, my question is: how do we get the last equality between $\limsup$ and $\lim$ and the last equality between each $\lim$?
For the second to last equality first
By definition of $\limsup$ for the real numbers $$ \limsup_{n \to \infty} x_n = \lim_{n \to \infty} \left(\sup_{m \geq n} x_m\right) $$ It should be clear that $\forall m, n \in \mathbb{N}$, $$ \bigcup_{k = m}^{\infty} A_k \subseteq \bigcup_{k = n}^{\infty} A_k \implies P\left(\bigcup_{k = m}^{\infty} A_k\right) \leq P\left(\bigcup_{k = n}^{\infty} A_k\right) $$ when $m \geq n$. Thus it follows $$ \sup_{m \geq n} P\left(\bigcup_{k = m}^{\infty} A_k\right) = P\left(\bigcup_{k = n}^{\infty} A_k\right) $$ and $$ \limsup_{n \to \infty} P\left(\bigcup_{k = n}^{\infty} A_k\right) = \lim_{n \to \infty}\left[\sup_{m \geq n} P\left(\bigcup_{k = m}^{\infty} A_k\right)\right] = \lim_{n \to \infty} P\left(\bigcup_{k = n}^{\infty} A_k\right) $$
For the last equality
You cannot generally interchange the limit and probability. However, since you have a nested monotonic sequence within the probability, you can. Explicitly, each $n + 1^{\text{th}}$ term is a subset of the $n^{\text{th}}$ term: $$ \bigcup_{k=1}^{\infty} A_k \supseteq \bigcup_{k=2}^{\infty} A_k \supseteq \bigcup_{k=3}^{\infty} A_k \supseteq \dots \bigcup_{k=n}^{\infty} A_k \supseteq \bigcup_{k = n+1}^{\infty} A_k \supseteq \dots $$
You can find the discussion on why you can interchange the limit and probability here or here (in many textbooks, blogs, etc.)