In Billingsley, Theorem 10.2 (ii),
Theorem 10.2 Let $\mu$ be a measure on the field $\mathscr{F}$.
(i) Continuity from below: …
(ii) Continuity from above: If $A_n$ and $A$ lie in $\mathscr{F}$ and $A_n \downarrow A$, and if $
\mu(A_1) < \infty$, then $\mu(A_n) \downarrow \mu(A)$.(ii) Countable Subadditivity: …
What is the significance of $A_1$?
Suppose we say $\mu(A_5) < \infty$. By monotonicity, $A_1 \supset \cdots \supset A_5 \supset \cdots$, so as long as some $\mu(A_i) < \infty$, $i \geq n$ holds then (ii) also holds.
So to prove, all one would need to say is that for some $\mu(A_k) < \infty$, $k \geq n$, by monotonicity, all subsequent sequences are also finite.
Then proceed as usual,
$$A_n \downarrow A \Rightarrow A_k \cap A_n^c \uparrow A_k \cap A^c$$ such that
$$\mu(A_k \cap A_n^c) \uparrow \mu(A_k \cap A^c).$$
But,
$$\mu(A_k) = \mu(A_k \cap A_n^c) + \mu(A_n) = \mu(A_k \cap A^c) + \mu(A),$$
which implies that $\mu(A_n) \downarrow \mu(A)$.
Best Answer
If I recall correctly, you are right: in fact, since $(A_n)_{n \geq 1}$ decreases to $A$, we have that for any $k$,
$$ \bigcap_{n \geq k}A_n = \bigcap_{n \geq 1}A_n $$
and so we can ask for one of the terms to be of finite measure, say $n_0$, because in that case we can apply the reasoning to $(A_{n+n_0})_{n \geq 1}$ and note that $\lim_{n \to \infty}\mu(A_n) = \lim_{n \to \infty}\mu(A_{n+n_0})$. In any case, there are some versions of the proof in which it is useful to assume $n_0 = 1$, such as the following: since $$ A_1 = A \stackrel{d}{\cup} \coprod_{n \geq 1}A_{n}\setminus A_{n+1} $$
we get $\mu(A_1) = \mu(A) + \sum_{n \geq 1}\mu(A_{n}) - \mu(A_{n+1}) = \mu(A) + \mu(A_1) - \lim_n\mu(A_n)$, and now because $A_1$ has finite measure, we can subtract $\mu(A_1)$ to both sides, proving the claim.