[proofplan]
For continuity from below, we write the increasing union as a telescoping disjoint union $\bigcup A_k = A_1 \sqcup \bigsqcup_{k=2}^\infty (A_k \setminus A_{k-1})$, apply countable additivity, and recognize the partial sums as $\mu(A_n)$ via the telescoping identity. For continuity from above, we reduce to continuity from below by taking complements within $A_1$: the decreasing sequence $A_k$ becomes the increasing sequence $A_1 \setminus A_k$, and the finite-measure hypothesis $\mu(A_1) < \infty$ allows the subtraction needed to pass from $\mu(A_1 \setminus A_k)$ back to $\mu(A_k)$.
[/proofplan]
[step:Prove continuity from below]
Let $A_1 \subset A_2 \subset \cdots$ with $A_k \in \mathcal{F}$. Define $B_1 := A_1$ and $B_k := A_k \setminus A_{k-1}$ for $k \ge 2$. Each $B_k \in \mathcal{F}$, the $B_k$ are pairwise disjoint (since $B_k \subset A_k \setminus A_{k-1}$ and $B_j \subset A_j \subset A_{k-1}$ for $j < k$), and
\begin{align*}
\bigsqcup_{k=1}^\infty B_k = \bigcup_{k=1}^\infty A_k.
\end{align*}
By countable additivity:
\begin{align*}
\mu\!\left(\bigcup_{k=1}^\infty A_k\right) = \sum_{k=1}^\infty \mu(B_k) = \lim_{n \to \infty} \sum_{k=1}^n \mu(B_k).
\end{align*}
The finite partial sum telescopes: $\sum_{k=1}^n \mu(B_k) = \mu(B_1) + \sum_{k=2}^n \mu(A_k \setminus A_{k-1})$. Since $A_1 \subset A_2 \subset \cdots$, the sets $B_1, B_2, \ldots, B_n$ are disjoint with $\bigsqcup_{k=1}^n B_k = A_n$. By finite additivity (a special case of countable additivity):
\begin{align*}
\sum_{k=1}^n \mu(B_k) = \mu\!\left(\bigsqcup_{k=1}^n B_k\right) = \mu(A_n).
\end{align*}
Therefore $\mu(\bigcup_{k=1}^\infty A_k) = \lim_{n \to \infty} \mu(A_n)$.
[guided]
The idea is to decompose the increasing union into a disjoint union of "increments." Each $B_k = A_k \setminus A_{k-1}$ represents the new part added at stage $k$. The union $\bigsqcup_{k=1}^n B_k = A_n$ is a telescoping identity: start with $A_1$, add $A_2 \setminus A_1$, add $A_3 \setminus A_2$, etc. After $n$ steps, we recover $A_n$. Countable additivity converts the measure of the full disjoint union into a series, and the partial sums of that series are $\mu(A_n)$.
No finiteness assumption is needed: the series $\sum \mu(B_k)$ converges (possibly to $\infty$) in $[0, \infty]$, and the limit equals $\mu(\bigcup A_k)$.
[/guided]
[/step]
[step:Prove continuity from above]
Let $A_1 \supset A_2 \supset \cdots$ with $A_k \in \mathcal{F}$ and $\mu(A_1) < \infty$. Define $C_k := A_1 \setminus A_k$ for $k \ge 1$. Since $A_k \supset A_{k+1}$, $C_k \subset C_{k+1}$ (removing a smaller set leaves a larger complement). Hence $(C_k)_{k=1}^\infty$ is increasing, with
\begin{align*}
\bigcup_{k=1}^\infty C_k = A_1 \setminus \bigcap_{k=1}^\infty A_k.
\end{align*}
By continuity from below (applied to the increasing sequence $C_k$):
\begin{align*}
\mu\!\left(A_1 \setminus \bigcap_{k=1}^\infty A_k\right) = \lim_{k \to \infty} \mu(C_k) = \lim_{k \to \infty} \mu(A_1 \setminus A_k).
\end{align*}
Since $A_k \subset A_1$ and $\mu(A_1) < \infty$, [excision](/theorems/1081) (applied with $A_k \subset A_1$ and $\mu(A_k) \le \mu(A_1) < \infty$) gives $\mu(A_1 \setminus A_k) = \mu(A_1) - \mu(A_k)$. Similarly, $\bigcap_{k} A_k \subset A_1$ and $\mu(\bigcap_k A_k) \le \mu(A_1) < \infty$, so $\mu(A_1 \setminus \bigcap_k A_k) = \mu(A_1) - \mu(\bigcap_k A_k)$. Therefore
\begin{align*}
\mu(A_1) - \mu\!\left(\bigcap_{k=1}^\infty A_k\right) = \lim_{k \to \infty} \left(\mu(A_1) - \mu(A_k)\right) = \mu(A_1) - \lim_{k \to \infty} \mu(A_k).
\end{align*}
Since $\mu(A_1) < \infty$, we subtract it from both sides to obtain
\begin{align*}
\mu\!\left(\bigcap_{k=1}^\infty A_k\right) = \lim_{k \to \infty} \mu(A_k).
\end{align*}
[guided]
The idea is to convert continuity from above into continuity from below by complementing within $A_1$. The decreasing sequence $A_k$ becomes the increasing sequence $C_k = A_1 \setminus A_k$. Continuity from below (already proved) gives $\mu(\bigcup C_k) = \lim \mu(C_k)$. We then convert back using excision: $\mu(C_k) = \mu(A_1) - \mu(A_k)$.
The hypothesis $\mu(A_1) < \infty$ is essential for two reasons. First, excision requires $\mu(A_k) < \infty$ (which follows from $\mu(A_k) \le \mu(A_1) < \infty$). Second, the final step subtracts $\mu(A_1)$ from both sides, which requires $\mu(A_1) < \infty$ to avoid $\infty - \infty$.
Without $\mu(A_1) < \infty$, the conclusion can fail. For example, on $(\mathbb{N}, \mathcal{P}(\mathbb{N}), \#)$ (counting measure), $A_k := \{k, k+1, k+2, \ldots\}$ gives $A_k \downarrow \varnothing$ with $\mu(A_k) = \infty$ for all $k$ but $\mu(\varnothing) = 0 \ne \infty = \lim_k \mu(A_k)$.
[/guided]
[/step]
