[proofplan]
For the increasing case, we write $\bigcup_{n=1}^\infty A_n$ as a disjoint union by defining the increments $B_n = A_n \setminus A_{n-1}$, apply countable additivity to obtain a series, and recognise its partial sums as $\mathbb{P}(A_n)$. The decreasing case follows by applying part (i) to the complements $A_1 \setminus A_n$, which form an increasing sequence.
[/proofplan]
[step:Disjointify the increasing sequence via increments $B_n = A_n \setminus A_{n-1}$]
Define $B_1 = A_1$ and $B_n = A_n \setminus A_{n-1}$ for $n \ge 2$.
We verify two properties. First, the $B_n$ are pairwise disjoint: if $i < j$, then $B_i \subset A_i \subset A_{j-1}$ (by the nesting $A_1 \subset A_2 \subset \cdots$), while $B_j = A_j \setminus A_{j-1}$ is disjoint from $A_{j-1}$, so $B_i \cap B_j = \varnothing$.
Second, $\bigcup_{n=1}^N B_n = A_N$ for every $N \ge 1$: this holds for $N = 1$ since $B_1 = A_1$, and if $\bigcup_{n=1}^{N-1} B_n = A_{N-1}$, then $\bigcup_{n=1}^N B_n = A_{N-1} \cup B_N = A_{N-1} \cup (A_N \setminus A_{N-1}) = A_N$. By induction, $\bigcup_{n=1}^N B_n = A_N$, and taking $N \to \infty$ gives $\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty A_n$.
[guided]
The central difficulty is that [countable](/page/Countable%20Set) additivity requires disjoint sets, but the $A_n$ are nested and overlapping. The standard trick is to "disjointify" the sequence by passing to the increments.
Define $B_1 = A_1$ and $B_n = A_n \setminus A_{n-1}$ for $n \ge 2$. Each $B_n$ captures the elements that appear in $A_n$ for the first time — the "new part" at stage $n$. We need to verify that this construction produces a disjoint sequence with the same union.
**Disjointness:** Suppose $i < j$. Then $B_i \subset A_i$ (since $B_i = A_i \setminus A_{i-1} \subset A_i$ for $i \ge 2$, and $B_1 = A_1$). By the nesting hypothesis $A_i \subset A_{j-1}$, so $B_i \subset A_{j-1}$. But $B_j = A_j \setminus A_{j-1}$, which is disjoint from $A_{j-1}$. Hence $B_i \cap B_j = \varnothing$.
**Same union:** We show $\bigcup_{n=1}^N B_n = A_N$ by induction. The base case $N = 1$ holds since $B_1 = A_1$. For the inductive step, $\bigcup_{n=1}^N B_n = A_{N-1} \cup B_N = A_{N-1} \cup (A_N \setminus A_{N-1}) = A_N$, where we used $A_{N-1} \cup (A_N \setminus A_{N-1}) = A_N$ (every element of $A_N$ either belongs to $A_{N-1}$ or to $A_N \setminus A_{N-1}$). Taking $N \to \infty$, $\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty A_n$.
[/guided]
[/step]
[step:Apply countable additivity and telescope the partial sums to prove part (i)]
Since the $B_n$ are pairwise disjoint with $\bigcup_{n=1}^\infty B_n = \bigcup_{n=1}^\infty A_n$, countable additivity gives
\begin{align*}
\mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n\right) = \sum_{n=1}^\infty \mathbb{P}(B_n).
\end{align*}
By definition of convergence of a series, this equals $\lim_{N \to \infty} \sum_{n=1}^N \mathbb{P}(B_n)$. For each finite $N$, since $B_1, \ldots, B_N$ are pairwise disjoint with $\bigcup_{n=1}^N B_n = A_N$, finite additivity gives
\begin{align*}
\sum_{n=1}^N \mathbb{P}(B_n) = \mathbb{P}\!\left(\bigcup_{n=1}^N B_n\right) = \mathbb{P}(A_N).
\end{align*}
Therefore $\mathbb{P}\!\left(\bigcup_{n=1}^\infty A_n\right) = \lim_{N \to \infty} \mathbb{P}(A_N)$.
[/step]
[step:Reduce the decreasing case to the increasing case via complements]
Now suppose $A_1 \supset A_2 \supset \cdots$. Define $C_n = A_1 \setminus A_n$ for $n \ge 1$. Since $A_n \supset A_{n+1}$, removing a larger set leaves a larger remainder: $C_n = A_1 \setminus A_n \subset A_1 \setminus A_{n+1} = C_{n+1}$. So $(C_n)_{n \ge 1}$ is an increasing sequence.
By part (i),
\begin{align*}
\mathbb{P}\!\left(\bigcup_{n=1}^\infty C_n\right) = \lim_{n \to \infty} \mathbb{P}(C_n) = \lim_{n \to \infty} \bigl(\mathbb{P}(A_1) - \mathbb{P}(A_n)\bigr),
\end{align*}
where we used $\mathbb{P}(C_n) = \mathbb{P}(A_1) - \mathbb{P}(A_n)$, which holds because $A_n \subset A_1$ and $C_n = A_1 \setminus A_n$ is the disjoint complement of $A_n$ within $A_1$.
The union $\bigcup_{n=1}^\infty C_n = \bigcup_{n=1}^\infty (A_1 \setminus A_n) = A_1 \setminus \bigcap_{n=1}^\infty A_n$ (an element of $A_1$ belongs to some $C_n$ if and only if it fails to belong to every $A_n$). Therefore
\begin{align*}
\mathbb{P}(A_1) - \mathbb{P}\!\left(\bigcap_{n=1}^\infty A_n\right) = \mathbb{P}(A_1) - \lim_{n \to \infty} \mathbb{P}(A_n).
\end{align*}
Subtracting $\mathbb{P}(A_1)$ from both sides gives $\mathbb{P}\!\left(\bigcap_{n=1}^\infty A_n\right) = \lim_{n \to \infty} \mathbb{P}(A_n)$.
[guided]
The strategy for the decreasing case is to convert it into an increasing problem. If the $A_n$ are shrinking, then the sets "removed so far" — $C_n = A_1 \setminus A_n$ — are growing.
**Why $C_n$ is increasing:** $A_n \supset A_{n+1}$ means $A_1 \setminus A_n \subset A_1 \setminus A_{n+1}$, since removing a larger set from $A_1$ leaves a larger remainder. So $C_n \subset C_{n+1}$.
**Applying part (i):** By the increasing case, $\mathbb{P}(\bigcup_{n=1}^\infty C_n) = \lim_{n \to \infty} \mathbb{P}(C_n)$.
**Computing $\mathbb{P}(C_n)$:** Since $A_n \subset A_1$, the set $A_1$ decomposes as the disjoint union $A_1 = A_n \cup C_n$. By finite additivity, $\mathbb{P}(A_1) = \mathbb{P}(A_n) + \mathbb{P}(C_n)$, giving $\mathbb{P}(C_n) = \mathbb{P}(A_1) - \mathbb{P}(A_n)$.
**Computing $\bigcup C_n$:** An element $\omega \in A_1$ belongs to $\bigcup_{n=1}^\infty C_n$ if and only if $\omega \in A_1 \setminus A_n$ for some $n$, which happens if and only if $\omega \notin A_n$ for some $n$, which happens if and only if $\omega \notin \bigcap_{n=1}^\infty A_n$. So $\bigcup_{n=1}^\infty C_n = A_1 \setminus \bigcap_{n=1}^\infty A_n$.
**Combining:** $\mathbb{P}(A_1 \setminus \bigcap A_n) = \lim_{n \to \infty} (\mathbb{P}(A_1) - \mathbb{P}(A_n))$. Since $\bigcap A_n \subset A_1$, the left side equals $\mathbb{P}(A_1) - \mathbb{P}(\bigcap A_n)$ by the same disjoint-union argument. Cancelling $\mathbb{P}(A_1)$ from both sides yields $\mathbb{P}(\bigcap_{n=1}^\infty A_n) = \lim_{n \to \infty} \mathbb{P}(A_n)$.
[/guided]
[/step]
