= Solution
A <sigma-algebra> $\mathcal F$ on $\Omega$ contains $\Omega$, is closed under complements relative to $\Omega$, and is closed under countable unions. Consequently it contains the empty set and is closed under countable intersections. A <measure space> is a triple $(\Omega,\mathcal F,\mu)$ with a map $\mu:\mathcal F\to[0,\infty]$ satisfying $\mu(\varnothing)=0$ and <countable additivity> on pairwise disjoint measurable sets.
For the increasing sequence, set $C_1=A_1$ and $C_n=A_n\setminus A_{n-1}$ for $n\geq2$. These measurable sets are disjoint, their union is $A$, and $A_N=\bigcup_{n=1}^NC_n$. Thus <countable additivity> and finite additivity give
$$
\boxed{\mu(A)=\sum_{n=1}^{\infty}\mu(C_n)
=\lim_{N\to\infty}\sum_{n=1}^N\mu(C_n)
=\lim_{N\to\infty}\mu(A_N).}
$$
The identity permits the value infinity; it uses no subtraction of infinite measures. This is <continuity from below of a measure>.
For a decreasing sequence, the analogous assertion is \b[not always true]. On $\Omega=\mathbb N$ with the <counting measure>, take $B_n=\{n,n+1,\ldots\}$. Then $\bigcap_nB_n=\varnothing$, so $\mu(B)=0$, whereas $\mu(B_n)=\infty$ for every $n$ and their limit is infinity.
A sufficient extra condition for <continuity from above of a measure> is $\mu(B_1)<\infty$. Indeed $B_1\setminus B_n$ increases to $B_1\setminus B$, and the increasing-sequence result gives
$$
\mu(B_n)=\mu(B_1)-\mu(B_1\setminus B_n)\longrightarrow\mu(B).
$$
More generally it suffices that one of the $B_n$ has finite measure. The missing finiteness assumption is exactly what the counterexample violates.
Back to article page
