A Finitely Additive Measure Is Countably Additive Exactly When It Is Continuous from Below

A finitely additive set function respects finite disjoint unions. Countable additivity asks for the same rule to hold for countably many disjoint sets. The extra condition that bridges the two is continuity from below: the measure of an increasing sequence of sets converges to the measure of their union.

Here is the precise equivalence, followed by proofs in both directions. The result applies to nonnegative set functions, which is the usual convention for measures.

Definitions and assumptions

Let $\mathcal{F}$ be a sigma-algebra and let $\mu:\mathcal{F}\to[0,\infty]$ satisfy $\mu(\varnothing)=0$. The function $\mu$ is finitely additive if, for disjoint $E,F\in\mathcal{F}$,

$$\mu(E\cup F)=\mu(E)+\mu(F).$$

It is countably additive if, for every pairwise disjoint sequence $E_1,E_2,\ldots$ in $\mathcal{F}$,

$$\mu\!\left(\bigcup_{n=1}^{\infty} E_n\right)=\sum_{n=1}^{\infty}\mu(E_n).$$

Continuity from below means that for every increasing sequence $A_1\subseteq A_2\subseteq\cdots$ in $\mathcal{F}$, with $A=\bigcup_{n=1}^{\infty}A_n$,

$$\lim_{n\to\infty}\mu(A_n)=\mu(A).$$

The values may be infinite; no assumption that the whole space has finite measure is needed.

Countable additivity implies continuity from below

Suppose $\mu$ is countably additive and $A_n$ increases to $A$. Define disjoint sets by

$$D_1=A_1,\qquad D_n=A_n\setminus A_{n-1}\quad(n\ge 2).$$

Each $D_n$ belongs to $\mathcal{F}$, the sets are pairwise disjoint, and their union is $A$. The first $n$ of them have union $A_n$. Countable additivity therefore gives

$$\mu(A)=\sum_{k=1}^{\infty}\mu(D_k), \qquad \mu(A_n)=\sum_{k=1}^{n}\mu(D_k).$$

Because the terms are nonnegative, the finite partial sums increase to the infinite series, including when the series is infinite. Thus $\mu(A_n)\to\mu(A)$.

Continuity from below implies countable additivity

Now suppose $\mu$ is finitely additive and continuous from below. Take any pairwise disjoint sequence $E_1,E_2,\ldots$ and form its finite partial unions:

$$B_n=\bigcup_{k=1}^{n}E_k.$$

The sequence $B_n$ increases to $B=\bigcup_{k=1}^{\infty}E_k$. By finite additivity, applied repeatedly,

$$\mu(B_n)=\sum_{k=1}^{n}\mu(E_k).$$

Continuity from below now yields

$$\mu(B)=\lim_{n\to\infty}\mu(B_n) =\lim_{n\to\infty}\sum_{k=1}^{n}\mu(E_k) =\sum_{k=1}^{\infty}\mu(E_k).$$

This is countable additivity. Together, the two arguments prove the equivalence.

A finitely additive example that fails the condition

Let the underlying set be $\mathbb{N}$, and consider the algebra consisting of all finite and cofinite subsets of $\mathbb{N}$. Define $\mu(E)=0$ for finite $E$ and $\mu(E)=1$ for cofinite $E$. This is finitely additive: disjoint finite sets have finite union, and a cofinite set cannot be disjoint from another nonempty cofinite set.

But let $A_n=\{1,2,\ldots,n\}$. Each $A_n$ is finite, so $\mu(A_n)=0$, while $\bigcup_n A_n=\mathbb{N}$ and $\mu(\mathbb{N})=1$. Hence $\mu(A_n)$ does not converge to $\mu(\bigcup_n A_n)$, and countable additivity fails: the singleton sets have measure zero, but their union has measure one.

This example is on an algebra rather than a sigma-algebra. It illustrates the same equivalence in the more general setting where the relevant unions belong to the domain. On a sigma-algebra, the proof above applies directly.

Common mistakes

• Using only finite additivity in the reverse direction. Finite additivity gives the measure of each partial union, but does not identify its limit with the measure of the infinite union. Continuity from below supplies that step.
• Assuming the total measure must be finite. The proof works for values in $[0,\infty]$ because it uses nonnegative sums and increasing partial sums.
• Confusing continuity from below with continuity from above. The theorem uses increasing sets. A continuity-from-above statement generally needs an additional finiteness condition, such as the first set having finite measure.
• Ignoring the domain of the set function. The union in each statement must belong to the domain. A sigma-algebra guarantees this for countable unions; on a general algebra, the equivalence is understood for sequences whose union is in the algebra.