# An inequality involving $\limsup$ to prove continuity of probability

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 $$\lim$$?

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 $$\limsup_{n \to \infty} P\left(\bigcup_{k = n}^{\infty} A_k\right) = \lim_{n \to \infty}\left[\sup_{m \geq n} P\left(\bigcup_{k = m}^{\infty} A_k\right)\right] = \lim_{n \to \infty} P\left(\bigcup_{k = n}^{\infty} A_k\right)$$
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: $$\bigcup_{k=1}^{\infty} A_k \supseteq \bigcup_{k=2}^{\infty} A_k \supseteq \bigcup_{k=3}^{\infty} A_k \supseteq \dots \bigcup_{k=n}^{\infty} A_k \supseteq \bigcup_{k = n+1}^{\infty} A_k \supseteq \dots$$