The sequence $(A_n)_{n \in \mathbb{N}}$ is not necessarily decreasing. However, if we define
$$B_n := \bigcup_{k \geq n} A_k$$
then $B_n$ is decreasing since
$$B_{n+1} = \bigcup_{k \geq n+1} A_k \subset \bigcup_{k \geq n} A_k = B_n.$$
Since $B_n \downarrow B$ for $B:= \bigcap_{n \geq 1} B_k$, the continuity of the measure from above shows
$$\begin{align*} \mathbb{P} \left( \bigcap_{n \geq 1} \bigcup_{k \geq n} A_k \right) &= \mathbb{P}(B) \\\ &= \lim_{n \to \infty} \mathbb{P}(B_n) \\\ &= \lim_{n \to \infty} \mathbb{P} \left( \bigcup_{k \geq n} A_k \right). \end{align*}$$