Countable Additivity from Caratheodory Splitting

From Exact Splitting to Countable Additivity

An outer measure is countably subadditive, but it need not be additive on arbitrary disjoint sets. Caratheodory measurability supplies the missing direction: a measurable set splits the outer measure of every test set exactly. Applying that finite splitting repeatedly gives a lower bound by each finite partial sum. Outer subadditivity gives the upper bound for the full countable union. Together they prove countable additivity on the measurable sets, with no finiteness or sigma-finiteness assumption.

Let $X$ be a set and let $\mu^*: \mathcal{P}(X)\to[0,\infty]$ be an outer measure: $\mu^*(\varnothing)=0$, $\mu^*$ is monotone, and it is countably subadditive. Here $\mathcal{P}(X)$ is the collection of all subsets of $X$. A set $E\subseteq X$ is Caratheodory measurable if, for every $A\subseteq X$,

\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c).

Write $\mathcal{M}^*$ for the Caratheodory-measurable sets, and define the restricted set function $\mu$ on $\mathcal{M}^*$ by $\mu(E)=\mu^*(E)$. The earlier results on Caratheodory measurability establish that $\mathcal{M}^*$ is a sigma-algebra. Thus the union of a countable sequence of its members is again in $\mathcal{M}^*$. The aim here is to prove that this restriction is countably additive: for pairwise disjoint $E_n\in\mathcal{M}^*$, its value on their union equals the sum of their values. All sums and equalities below are in $[0,\infty]$, so $+\infty$ is allowed.

The Finite Splitting Inequality

The decisive lower bound is finite. For pairwise disjoint measurable sets $E_1,\ldots,E_N$, splitting a set $A\subseteq X$ successively shows that the outer measure of $A$ is at least the sum of the outer measures of its pieces $A\cap E_i$.

Finite splitting lemma: If $E_1,\ldots,E_N\in\mathcal{M}^*$ are pairwise disjoint and $A\subseteq X$, then $$ \mu^*(A)\ge\sum_{i=1}^{N}\mu^*(A\cap E_i). $$

Proof. We prove the inequality by induction on $N$. For $N=1$, monotonicity gives $\mu^*(A)\ge\mu^*(A\cap E_1)$. Suppose the inequality holds for $N-1$. Since $E_N$ is measurable, the splitting criterion, applied to $A$, gives $\mu^*(A)=\mu^*(A\cap E_N)+\mu^*(A\cap E_N^c)$. For each $i<N$, disjointness implies $E_i\subseteq E_N^c$, and hence $A\cap E_i\subseteq A\cap E_N^c$. The induction hypothesis applied to $A\cap E_N^c$ and $E_1,\ldots,E_{N-1}$ therefore gives $\mu^*(A\cap E_N^c)\ge\sum_{i=1}^{N-1}\mu^*(A\cap E_i)$. Adding $\mu^*(A\cap E_N)$ proves the claim. These inequalities and sums are valid for extended nonnegative values as well: if a term is infinite, the corresponding lower bound forces the left side to be infinite. $\square$

Notice that the lemma uses the exact splitting criterion for each $E_i$, not additivity of the outer measure. Its conclusion is only a lower bound for finitely many disjoint measurable pieces. We now apply it to finite initial segments of a countable disjoint family.

Countable Additivity on the Measurable Sets

Theorem: Let $\mu^*$ be an outer measure on $X$, let $\mathcal{M}^*$ be its Caratheodory sigma-algebra, and let $\mu=\mu^*|_{\mathcal{M}^*}$. For every pairwise disjoint sequence $(E_n)_{n\ge1}$ in $\mathcal{M}^*$, $$ \mu\left(\bigcup_{n=1}^{\infty}E_n\right)=\sum_{n=1}^{\infty}\mu(E_n). $$ No finiteness or sigma-finiteness assumption is required.

Proof. Put $E=\bigcup_{n=1}^{\infty}E_n$. Since $\mathcal{M}^*$ is a sigma-algebra, $E\in\mathcal{M}^*$, and so $\mu(E)=\mu^*(E)$. For each positive integer $N$, apply the finite splitting lemma with $A=E$ and $E_1,\ldots,E_N$. Because $E_i\subseteq E$, we have $E\cap E_i=E_i$. Consequently, $$ \mu(E)=\mu^*(E)\ge\sum_{i=1}^{N}\mu^*(E_i)=\sum_{i=1}^{N}\mu(E_i). $$ The partial sums on the right are nondecreasing. By definition, their limit in $[0,\infty]$ is the series $\sum_{i=1}^{\infty}\mu(E_i)$. Since every partial sum is at most $\mu(E)$, their limit is also at most $\mu(E)$. This proves $\sum_{i=1}^{\infty}\mu(E_i)\le\mu(E)$, including the case where the series is infinite.

For the reverse inequality, countable subadditivity of the outer measure gives $$ \mu(E)=\mu^*(E) \le\sum_{i=1}^{\infty}\mu^*(E_i) =\sum_{i=1}^{\infty}\mu(E_i). $$ Combining the two inequalities proves the asserted equality. If the series on the right is infinite, the second inequality forces $\mu(E)=\infty$. If the series is finite, both inequalities give equality at that finite value. Empty members of the sequence contribute $\mu(\varnothing)=0$, so they require no separate exception. $\square$

The proof has two distinct directions. Finite exact splitting controls every finite partial sum from below by the measure of the whole union; taking the supremum over those partial sums gives the countable lower bound. Countable subadditivity controls the union from above by the entire series. Neither direction alone proves equality.

Worked Example: A Countable Partition of the Natural Numbers

Let $X=\mathbb{N}=\{1,2,\ldots\}$ and define $\mu^*(A)$ to be the number of elements of $A$ if $A$ is finite, and $+\infty$ if $A$ is infinite. This is an outer measure, and every subset $E\subseteq X$ is Caratheodory measurable: for any $A$, the disjoint pieces $A\cap E$ and $A\cap E^c$ partition $A$, so their cardinalities add when $A$ is finite; if $A$ is infinite, at least one piece is infinite, and the sum of the two outer measures is $+\infty$.

Take $E_n=\{2n-1,2n\}$ for $n\ge1$. These sets are pairwise disjoint and their union is $\mathbb{N}$. Each contains exactly two elements, so $\mu(E_n)=2$. For a finite initial union, there are $2N$ elements, giving $\mu(\bigcup_{n=1}^N E_n)=2N=\sum_{n=1}^N\mu(E_n)$. As $N$ tends to infinity, these values tend to $+\infty$. The full union is infinite, so $\mu(\mathbb{N})=+\infty=\sum_{n=1}^{\infty}2$.

Check: the finite partial sums are $2,4,6,\ldots$, which are unbounded, and the measure of the union is $+\infty$, as the theorem predicts.

Worked Example: A Finite Total Split into Odd and Even Integers

On $X=\mathbb{N}$, define $\mu^*(A)=\sum_{k\in A}2^{-k}$, with the sum of nonnegative terms interpreted as the supremum of its finite partial subsums. This is an outer measure. For any $E,A\subseteq\mathbb{N}$, the terms indexed by $A$ split into those indexed by $A\cap E$ and $A\cap E^c$; hence $\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c)$, and every subset is measurable.

Partition $X$ into $E_1=\{1,3,5,\ldots\}$ and $E_2=\{2,4,6,\ldots\}$. The geometric series give $$ \mu(E_1)=\sum_{j=1}^{\infty}2^{-(2j-1)} =\frac{1/2}{1-1/4}=\frac23, \qquad \mu(E_2)=\sum_{j=1}^{\infty}2^{-2j} =\frac{1/4}{1-1/4}=\frac13. $$ Thus $\mu(E_1)+\mu(E_2)=\frac23+\frac13=1$. The whole-space value is $\mu(X)=\sum_{k=1}^{\infty}2^{-k}=(1/2)/(1-1/2)=1$, so the two sides agree. Check: the even and odd sets are disjoint and together contain every positive integer; their computed weights add to the weight of all of $X$.

Worked Example: An Infinite Lebesgue Measure Sum

For Lebesgue outer measure $m^*$ on $\mathbb{R}$, intervals are Caratheodory measurable and $m([a,b))=b-a$. Consider $E_n=[n,n+1)$ for $n\ge1$. The intervals are pairwise disjoint: the right endpoint $n+1$ is excluded from $E_n$ and included in $E_{n+1}$. Each has measure $(n+1)-n=1$. Their union is $[1,\infty)$, a measurable set of infinite Lebesgue measure.

For the first $N$ intervals, their union is $[1,N+1)$ and has measure $(N+1)-1=N$. The corresponding partial sum is $\sum_{n=1}^{N}1=N$. Both quantities tend to $+\infty$, and therefore $\displaystyle m([1,\infty))=\sum_{n=1}^{\infty}m(E_n)=+\infty$. This verifies the theorem in an example where neither side is finite. The calculation uses half-open intervals so that no endpoint is counted in two members of the partition.

Why Subadditivity Is Not Enough

A common error is to use outer subadditivity as if it were an equality. The direction supplied by subadditivity is only $\mu^*(A\cup B)\le\mu^*(A)+\mu^*(B)$; the reverse inequality does not hold for arbitrary disjoint sets.

Worked Example: A Disjoint Pair That Is Not Measurable

Let $X=\{a,b\}$ and define $\mu^*(\varnothing)=0$ and $\mu^*(A)=1$ for every nonempty $A\subseteq X$. This is an outer measure: it is monotone, and a countable union has outer measure $0$ if every member is empty, and otherwise has outer measure $1$, which is at most the sum of the outer measures of the covering sets.

The disjoint sets $\{a\}$ and $\{b\}$ have outer measure $1$ each, while $\mu^*(X)=1$. Thus $\mu^*(\{a\}\cup\{b\})=1<2=\mu^*(\{a\})+\mu^*(\{b\})$. In fact, $\{a\}$ is not Caratheodory measurable: testing the splitting criterion with $A=X$ would require $\mu^*(X)=\mu^*(\{a\})+\mu^*(\{b\})$, or $1=2$, which is false. The correction is to apply the theorem only to measurable pieces. Their exact splitting gives the finite lower bound; subadditivity then supplies the other direction.

Countable additivity is what makes the restricted outer measure a measure on $(X,\mathcal{M}^*)$. The argument depends on measurability for the lower bound and on the defining outer-measure inequality for the upper bound. It does not require a finite total mass, and it remains valid when individual terms, partial sums, or the total measure are infinite.

Check Your Understanding

Use the finite splitting lemma and countable subadditivity to justify each conclusion.

For a disjoint sequence of measurable sets, why is each finite partial sum bounded above by the measure of the full union?
Where in the countable additivity proof is the sigma-algebra property of $\mathcal{M}^*$ used?
If $\sum_n\mu(E_n)=+\infty$, which inequality forces $\mu(\bigcup_nE_n)=+\infty$?
For the weighted measure on $\mathbb{N}$ above, compute the measure of the set of integers divisible by $3$, and compare it with the total measure.
Why does outer subadditivity alone not prove the reverse inequality for two disjoint sets?