Continuity (from above) theorem Let $\mathcal{A}_1 \supseteq \mathcal{A}_2 \supseteq \mathcal{A}_3 \supseteq \cdots$ be a sequence of non-increasing sets with elements in the sample space $\mathcal{S}$.   Define the union of the sequence as the set:
$$\mathcal{A} \equiv \bigcap_{k=1}^\infty \mathcal{A}_k.$$
Show that $\mathbb{P}(\mathcal{A}) = \lim_{k \rightarrow \infty} \mathbb{P}(\mathcal{A}_k)$.  Can someone please provide the proof and explanation as to why this is the case?  I am confused on how I would show this to be true.
 A: Since you've now offered an explanation of your attempted working (in the comments) I'm going to provide a solution.  This property of the probability measure is often referred to as "continuity from above", and it follows as a consequence of countable additivity.  The property is usually established via the corresponding property of "continuity from below", but here I will fold that result in to give a proof that only uses the properties of sets and the axioms of probability.
(I note from your comments that you are trying to prove the result by invoking continuity from below.  That is not particularly helpful without also proving continuity from below, since these are essentially just variations on the same substantive result.)
Since $\mathcal{A}_1 \supseteq \mathcal{A}_2 \supseteq \mathcal{A}_3 \supseteq \cdots$ we have $\overline{\mathcal{A}}_1 \subseteq \overline{\mathcal{A}}_2 \subseteq \overline{\mathcal{A}}_3 \subseteq ...$, which means we can form disjoint sets of the form $\overline{\mathcal{A}}_k - \overline{\mathcal{A}}_{k-1}$ for $k \geqslant 2$.  Using DeMorgan's law and the countable additivity property, we have:
$$\begin{equation} \begin{aligned}
\mathbb{P}(\mathcal{A}) 
= \mathbb{P} \Bigg( \bigcap_{k=1}^\infty \mathcal{A}_{k} \Bigg) 
&= 1 - \mathbb{P} \Bigg( \ \overline{\bigcap_{k=1}^\infty \mathcal{A}_{k}} \ \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( \bigcup_{k=1}^\infty \overline{\mathcal{A}}_k \ \Bigg) \\[6pt]
&= 1 - \mathbb{P} \Bigg( \overline{\mathcal{A}}_1 \cup \bigcup_{k=2}^\infty (\overline{\mathcal{A}}_k - \overline{\mathcal{A}}_{k-1}) \Bigg) \\[6pt]
&= 1 - \mathbb{P} ( \overline{\mathcal{A}}_1) - \sum_{k=2}^\infty \mathbb{P} (\overline{\mathcal{A}}_k - \overline{\mathcal{A}}_{k-1}) \\[6pt]
&= 1 - \mathbb{P} ( \overline{\mathcal{A}}_1) - \sum_{k=2}^\infty [ \mathbb{P} (\overline{\mathcal{A}}_k) - \mathbb{P}(\overline{\mathcal{A}}_{k-1}) ] \\[6pt]
&= 1 - \mathbb{P} ( \overline{\mathcal{A}}_1) - \lim_{n \rightarrow \infty}  \sum_{k=2}^n [ \mathbb{P} (\overline{\mathcal{A}}_k) - \mathbb{P}(\overline{\mathcal{A}}_{k-1}) ] \\[6pt]
&= \lim_{n \rightarrow \infty} \Bigg( 1 - \mathbb{P} ( \overline{\mathcal{A}}_1) - \sum_{k=2}^n [ \mathbb{P} (\overline{\mathcal{A}}_k) - \mathbb{P}(\overline{\mathcal{A}}_{k-1}) ] \Bigg) \\[6pt]
&= \lim_{n \rightarrow \infty} \Big( 1 - \mathbb{P} ( \overline{\mathcal{A}}_n) \Big) \\[6pt]
&= \lim_{n \rightarrow \infty} \mathbb{P} ( \mathcal{A}_n). \\[6pt]
\end{aligned} \end{equation}$$
(The penultimate step in this equation involves recognition of cancelling terms in the sum.)
A: For a decreasing sequences of sets $\{\mathcal A_k\}_{k=1}^\infty$ we note that $\{\mathcal A_k^c\}_{k=1}^\infty$ is an increasing sequence. Further (as you have mentioned already)
$$\mathbb P\left(\bigcup_{k=1}^\infty \mathcal B_k\right) = \lim_{k\rightarrow\infty} \mathbb P(\mathcal B_k)$$
when $\{\mathcal B_k\}_{k=1}^\infty$ is increasing. It follows that
$$\begin{aligned}
\mathbb P\left(\bigcap_{k=1}^\infty \mathcal A_k\right) &= 1 - \mathbb P\left(\left[\bigcap_{k=1}^\infty \mathcal A_k\right]^c\right) = 1 - \mathbb P\left(\bigcup_{k=1}^\infty \mathcal A_k^c\right)  \\
& = 1 - \lim_{k\rightarrow\infty} \mathbb P(\mathcal A_k^c) = \lim_{k\rightarrow\infty}[1 - \mathbb P(\mathcal A_k^c)] \\&= \lim_{k\rightarrow\infty}\mathbb P(\mathcal A_k).
\end{aligned}$$
