Caratheodory Measurability of Ring Sets

From Ring Covers to the Splitting Criterion

In the covering construction, an outer measure is defined on every subset of $X$, even though the sets used to cover a subset come only from a ring $\mathcal{R}$. The key question here is whether a set $A$ from that original ring separates every outer-measure calculation into its part inside $A$ and its part outside $A$. The answer follows by splitting each covering set using the ring operations.

Let $\mathcal{R}$ be a ring of subsets of $X$: it contains $\varnothing$, and is closed under finite unions and differences. Let $\mu_0:\mathcal{R}\to[0,\infty]$ be a premeasure, meaning that $\mu_0(\varnothing)=0$ and that it is countably additive whenever a disjoint sequence in $\mathcal{R}$ has its union in $\mathcal{R}$. In particular, $\mu_0$ is finitely additive on disjoint ring sets. No finiteness or sigma-finiteness assumption is needed below.

Definition: For any $E\subseteq X$, define the induced outer measure by $$ \mu^*(E)=\inf\left\{\sum_{n=1}^{\infty}\mu_0(C_n):\ C_n\in\mathcal{R},\ E\subseteq\bigcup_{n=1}^{\infty}C_n\right\}. $$ The infimum of an empty collection is $+\infty$. A set $A\subseteq X$ satisfies the Caratheodory splitting criterion for $\mu^*$ if, for every $E\subseteq X$, $$ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A). $$ Here $\mu^*$ is an outer measure on $\mathcal{P}(X)$, not yet a measure on that entire power set.

The criterion asks for an equality for every subset $E$, not just for $E\in\mathcal{R}$. The two inequalities have different sources. The inequality from $E$ to its two pieces follows from outer-measure subadditivity. For the reverse inequality, take a ring cover of $E$, split each covering set at $A$, and use finite additivity of $\mu_0$.

Why Every Ring Set Satisfies the Criterion

Theorem: For the outer measure $\mu^*$ induced by the premeasure $\mu_0$ on the ring $\mathcal{R}$, every $A\in\mathcal{R}$ satisfies $$ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A) $$ for every $E\subseteq X$. The equality is in $[0,\infty]$, so it includes infinite values.

Proof. Fix $A\in\mathcal{R}$ and $E\subseteq X$. First, $E=(E\cap A)\cup(E\setminus A)$. Outer-measure subadditivity therefore gives $$ \mu^*(E)\le \mu^*(E\cap A)+\mu^*(E\setminus A). $$ For completeness, this instance of subadditivity follows directly from the covering infimum: if either term on the right is infinite, the inequality holds immediately. Otherwise, for any $\varepsilon>0$, cover $E\cap A$ and $E\setminus A$ by ring sets with total costs less than their respective outer measures plus $\varepsilon/2$. Combining those covers covers $E$, so $\mu^*(E)\le\mu^*(E\cap A)+\mu^*(E\setminus A)+\varepsilon$. Letting $\varepsilon$ decrease to zero proves the inequality.

For the other direction, suppose first that $E$ has a cover $E\subseteq\bigcup_{n=1}^{\infty}C_n$, where each $C_n\in\mathcal{R}$. Because $\mathcal{R}$ is closed under intersections and differences, both $C_n\cap A$ and $C_n\setminus A$ belong to $\mathcal{R}$. The first sequence covers $E\cap A$, and the second covers $E\setminus A$. These two ring sets are disjoint and have union $C_n$. Finite additivity of the premeasure gives $$ \mu_0(C_n)=\mu_0(C_n\cap A)+\mu_0(C_n\setminus A). $$ Consequently, $$ \begin{aligned} \mu^*(E\cap A)+\mu^*(E\setminus A) &\le \sum_{n=1}^{\infty}\mu_0(C_n\cap A) +\sum_{n=1}^{\infty}\mu_0(C_n\setminus A)\\ &=\sum_{n=1}^{\infty}\mu_0(C_n). \end{aligned} $$ This holds for every ring cover of $E$. Taking the infimum over those covers proves $\mu^*(E\cap A)+\mu^*(E\setminus A)\le\mu^*(E)$. If $E$ has no ring cover, then $\mu^*(E)=+\infty$, and this inequality holds as well. Combining the two inequalities proves the required equality, including when one or more terms are infinite. If $E=\varnothing$, all three outer measures are zero, so that case is included. $\square$

The essential step is not that $A$ itself covers part of $E$. Rather, every member $C_n$ of any cover is replaced by the two ring members $C_n\cap A$ and $C_n\setminus A$. The proof therefore applies to arbitrary subsets $E$, whether or not they belong to the original ring.

Worked Example: A Weighted Finite Ring

Let $X=\{a,b,c,d\}$, let $\mathcal{R}=\mathcal{P}(X)$, and assign weights $2,3,5,7$ to $a,b,c,d$, respectively. Define $\mu_0(S)$ to be the sum of the weights of the points in $S$. This is a premeasure: for disjoint sets, each point in their union appears in exactly one set, so the sum of the weights over the union is the sum of the individual weights.

Take $A=\{a,b\}$ and $E=\{a,c,d\}$. Since $E$ itself is an allowed cover, its outer measure is at most $\mu_0(E)=2+5+7=14$. Conversely, any cover of $E$ counts each point of $E$ at least once in the total cost, so every cover has cost at least $14$. Thus $\mu^*(E)=14$. The two pieces have outer measures $\mu^*(E\cap A)=\mu^*(\{a\})=2$ and $\mu^*(E\setminus A)=\mu^*(\{c,d\})=5+7=12$. Hence $2+12=14=\mu^*(E)$. Check: the cover $E$ has cost $2+5+7=14$, matching the outer measure calculation.

Worked Example: An Infinite Outer Measure

Let $X=\mathbb{N}=\{1,2,\ldots\}$, let $\mathcal{R}$ be the ring of finite subsets of $\mathbb{N}$, and let $\mu_0(F)=|F|$. This is a premeasure: whenever a disjoint sequence of finite sets has a union that belongs to $\mathcal{R}$, the union is finite and its cardinality is the sum of the cardinalities.

For $E=\mathbb{N}$, the singleton sets $\{n\}$ give a ring cover whose total cost is $\sum_{n=1}^{\infty}1=+\infty$. Every countable cover of $\mathbb{N}$ by finite sets also has infinite total cost: if its total cardinality sum were finite, only finitely many points could lie in its union. Therefore $\mu^*(\mathbb{N})=+\infty$. Let $A=\{1,2\}$. Then $\mu^*(E\cap A)=2$, while $E\setminus A=\{3,4,\ldots\}$ has infinite outer measure by the same argument. Thus $$ \mu^*(E\cap A)+\mu^*(E\setminus A)=2+(+\infty)=+\infty=\mu^*(E). $$ Check: the cover of $\mathbb{N}$ by its singletons has cost $\sum_{n=1}^{\infty}1=+\infty$; the equality uses extended nonnegative values, not a subtraction of infinite quantities.

Reading the Splitting Argument in an Interval Cover

Worked Example: Splitting an Interval Cover

Let $X=[0,3)$, and let $\mathcal{R}$ consist of unions of the three half-open intervals $$ I_0=[0,1),\qquad I_1=[1,2),\qquad I_2=[2,3). $$ Set $\mu_0(S)$ equal to the number of these unit intervals in $S$. This is a premeasure on this finite ring: disjoint ring sets cannot share a unit interval, so the number of intervals in a disjoint union is the sum of their numbers. Consider $A=I_0\cup I_2$ and $E=[1/4,5/2)$.

A ring cover of $E$ must include $I_0$, because $1/4\in E\cap I_0$; any ring set containing $1/4$ contains the whole interval $I_0$. It must also include $I_1$, because $3/2\in E\cap I_1$, and $I_2$, because $5/2\in E\cap I_2$. Thus every cover costs at least $1+1+1=3$. The cover $I_0\cup I_1\cup I_2=X$ costs $3$, so $\mu^*(E)=3$.

Now $E\cap A=[1/4,1)\cup[2,5/2)$, which meets $I_0$ and $I_2$, so its outer measure is $2$: both intervals are needed in a cover, and together they suffice. Also $E\setminus A=[1,2)$ has outer measure $1$. Therefore $\mu^*(E\cap A)+\mu^*(E\setminus A)=2+1=3=\mu^*(E)$.

The splitting can be seen directly in a cover. Use $C_1=I_0\cup I_1$ and $C_2=I_2$; their total cost is $2+1=3$. Then $C_1\cap A=I_0$, $C_1\setminus A=I_1$, $C_2\cap A=I_2$, and $C_2\setminus A=\varnothing$. The split costs on the $A$-side are $1+1=2$, and on the other side they are $1+0=1$. Check: the split total $2+1=3$ equals the original cover cost $2+1=3$, as finite additivity requires.

A Common Error and Why the Result Matters

Common mistake: For a cover member $C_n$, it is incorrect to use $C_n\cap A$ as a cover of all of $C_n$, and then conclude that its cost accounts for $\mu_0(C_n)$. In the interval example, $C_1=I_0\cup I_1$ has cost $2$, but $C_1\cap A=I_0$ has cost $1$; the part $I_1$ has been lost. The correction is to retain both pieces: $$ C_n=(C_n\cap A)\,\dot\cup\,(C_n\setminus A), \qquad \mu_0(C_n)=\mu_0(C_n\cap A)+\mu_0(C_n\setminus A). $$ The split cover on each side then gives the lower bound for the sum of the two outer measures.

Another important distinction is that this theorem establishes the criterion for the original ring sets; it does not, by itself, assert that every subset of $X$ satisfies the criterion. The covering construction starts with an outer measure on all subsets, while the criterion identifies particular sets whose intersections and complements split outer measure exactly. This is the step that makes the original ring usable in the extension argument. Neither finite total mass nor sigma-finiteness is used in the proof.

Check Your Understanding

For each question, keep the ring-cover definition of $\mu^*$ separate from the criterion being tested.

For $A\in\mathcal{R}$ and $C\in\mathcal{R}$, why do both $C\cap A$ and $C\setminus A$ belong to $\mathcal{R}$?
In the interval example, why must every ring cover of $E=[1/4,5/2)$ include all three unit intervals?
In the counting-measure example, explain why no countable cover of $\mathbb{N}$ by finite sets can have finite total cost.
Which inequality in the theorem follows from splitting each member of an arbitrary ring cover?
Why must the proof retain both $C\cap A$ and $C\setminus A$ when it splits a cover member $C$?

Caratheodory Measurability of Ring Sets

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