= Solution
For the increasing family, define the <measurable sets>
$$
D_1=A_1,\qquad D_n=A_n\setminus A_{n-1}\quad(n\geq2).
$$
They are pairwise <disjoint>, with $A_N=\bigcup_{n=1}^ND_n$ and $\bigcup_nA_n=\bigcup_nD_n$. By <countable additivity>,
$$
\mu\left(\bigcup_{n=1}^{\infty}A_n\right)
=\sum_{n=1}^{\infty}\mu(D_n)
=\lim_{N\to\infty}\sum_{n=1}^N\mu(D_n)
=\lim_{N\to\infty}\mu(A_N).
$$
The nonnegative <series> is defined by its increasing partial-sum <limit>, so this proof also covers an infinite value. This is <continuity from below of a measure>.
For the decreasing family, put $C_n=B_1\setminus B_n$. These <measurable sets> increase, and
$$
\bigcup_nC_n=B_1\setminus\bigcap_nB_n.
$$
Since $\mu(B_1)<\infty$, <finite additivity> gives legitimate finite subtractions:
$$
\mu(C_n)=\mu(B_1)-\mu(B_n),\qquad
\mu\left(B_1\setminus\bigcap_nB_n\right)=\mu(B_1)-\mu\left(\bigcap_nB_n\right).
$$
Apply the first result to $C_n$ and subtract from $\mu(B_1)$ to obtain
$$
\boxed{\mu\left(\bigcap_{n=1}^{\infty}B_n\right)=\lim_{n\to\infty}\mu(B_n).}
$$
This is <continuity from above of a measure>. Its finiteness hypothesis matters: for <Lebesgue measure> and $B_n=[n,\infty)$, every $B_n$ has infinite <measure> but their <intersection> is empty. The proof must not subtract $\infty-\infty$.
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