Continuity of measure from below and above

Remarks

For part 1, put $B_1 = A_1$ and $B_n = A_n \setminus A_{n-1}$ for $n \ge 2$. The sets $B_n$ are pairwise disjoint, their union is $\bigcup_n A_n$, and $B_1 \cup \cdots \cup B_n = A_n$. Countable additivity gives $\mu(\bigcup_n A_n) = \sum_n \mu(B_n)$, and the partial sums of this series are the numbers $\mu(A_n)$. For part 2, apply part 1 to the increasing sets $A_1 \setminus A_n$, and subtract from the finite number $\mu(A_1)$.

For Lebesgue measure $\lambda$ on $\R$ and a point $x$, the sets $[x - 1/n, x + 1/n]$ decrease to $\set{x}$ and have measure $2/n$. Part 2 gives $\lambda(\set{x}) = 0$.

The hypothesis $\mu(A_1) < \infty$ in part 2 is needed. For Lebesgue measure, the sets $A_n = [n, \infty)$ decrease to $\varnothing$, but $\lambda(A_n) = \infty$ for every $n$.

Going deeper

For a student

For a finitely additive measure on a σ-algebra, continuity from below is equivalent to countable additivity. So it is a useful way to check that a set function is a measure. It is the set version of the monotone convergence theorem. For probability measures both parts hold without extra hypothesis, since every set has finite measure. Continuity of measures underlies the Borel–Cantelli lemmas and the right-continuity of distribution functions.