Finite Splitting by Several Measurable Sets

From One Split to Finitely Many

Caratheodory measurability is defined by an exact splitting identity: a measurable set cuts every test set into two parts whose outer measures add to the outer measure of the whole. To split a test set by several measurable sets, the key is to apply this identity successively to the part that remains, not repeatedly to the original test set. When the measurable sets are pairwise disjoint, this procedure accounts for each piece once and leaves a final common remainder.

Let $X$ be a set and let $\mu^*$ be an outer measure on the power set $\mathcal{P}(X)$. Thus $\mu^*:\mathcal{P}(X)\to[0,\infty]$, $\mu^*(\varnothing)=0$, $\mu^*$ is monotone, and it is countably subadditive. No finiteness or sigma-finiteness assumption is imposed. All sets considered below are subsets of $X$.

Definition: A set $E\subseteq X$ is Caratheodory measurable with respect to $\mu^*$ if, for every $A\subseteq X$, $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E). $$ This is a condition on the outer measure of every subset $A$, not only on sets already known to be measurable.

The values in this identity may be infinite. Sums below are sums in $[0,\infty]$; in particular, no subtraction of outer measures will be used. The goal is a finite identity for disjoint measurable sets, with the remainder defined as what is left after removing all of them.

The Finite Splitting Identity

Theorem: Let $\mu^*$ be an outer measure on $\mathcal{P}(X)$, let $n\geq 0$ be an integer, and let $E_1,\ldots,E_n\subseteq X$ be pairwise disjoint Caratheodory measurable sets. For every $A\subseteq X$, $$ \mu^*(A)=\sum_{i=1}^{n}\mu^*(A\cap E_i)+ \mu^*\left(A\setminus\bigcup_{i=1}^{n}E_i\right). $$ For $n=0$, the sum is the empty sum $0$, the union is empty, and the identity reads $\mu^*(A)=\mu^*(A)$.

Proof. Define $R_0=A$, and for $1\leq k\leq n$ define $R_k=A\setminus\bigcup_{i=1}^{k}E_i$. Thus $R_k$ is the part of $A$ left after the first $k$ sets have been removed. For each $k\geq1$, apply the defining criterion for $E_k$ to the test set $R_{k-1}$. It gives $\mu^*(R_{k-1})=\mu^*(R_{k-1}\cap E_k)+\mu^*(R_{k-1}\setminus E_k)$. Because the $E_i$ are pairwise disjoint, $E_k$ is disjoint from every $E_i$ with $i<k$. Hence $R_{k-1}\cap E_k=A\cap E_k$, while $R_{k-1}\setminus E_k=R_k$. Therefore, for every $k=1,\ldots,n$, $$ \mu^*(R_{k-1})=\mu^*(A\cap E_k)+\mu^*(R_k). $$ Substitute the identity for $k=2$ into the one for $k=1$, then the identity for $k=3$, and continue through $k=n$. This yields $\mu^*(R_0)=\sum_{k=1}^n\mu^*(A\cap E_k)+\mu^*(R_n)$. Since $R_0=A$ and $R_n=A\setminus\bigcup_{i=1}^{n}E_i$, this is the asserted formula. If $n=0$, there are no identities to substitute and the stated empty-family formula holds directly. Every substitution adds nonnegative extended-real values, so it remains valid when some terms equal $+\infty$. $\square$

The proof uses only the individual measurability of each $E_k$ and the fact that the sets are pairwise disjoint. In particular, the criterion for $E_k$ is applied to the residual set $R_{k-1}$, which is an allowed test set whether or not that residual set has already been shown to be measurable.

Corollary: Under the hypotheses of the theorem, for every $A\subseteq X$, $$ \sum_{i=1}^{n}\mu^*(A\cap E_i)\leq \mu^*(A). $$ This inequality is understood in the extended nonnegative reals; if $\mu^*(A)=+\infty$, it does not assert that the sum is finite.

Proof. The finite splitting identity writes $\mu^*(A)$ as the sum of the displayed finite sum and the nonnegative quantity $\mu^*(A\setminus\bigcup_{i=1}^{n}E_i)$. Addition of a nonnegative extended-real number cannot decrease a value, so the finite sum is at most $\mu^*(A)$. The conclusion also holds for $n=0$, when its left side is $0$. $\square$

Worked Splittings

Worked Example: A Finite Counting Outer Measure

Let $X=\{a,b,c,d,e\}$ and define $\mu^*(S)=|S|$ for every $S\subseteq X$, where $|S|$ is the number of elements of $S$. This is an outer measure: the empty set has cardinality zero, inclusion cannot decrease cardinality, and the cardinality of a countable union is at most the sum of the cardinalities of its members. Every $E\subseteq X$ is Caratheodory measurable, because $A\cap E$ and $A\setminus E$ are disjoint and have union $A$, so $|A|=|A\cap E|+|A\setminus E|$.

Take $E_1=\{a,c\}$, $E_2=\{b,d\}$, and $A=\{a,b,c,e\}$. The two measurable sets are disjoint. The first piece is $A\cap E_1=\{a,c\}$, so its outer measure is $2$. The second piece is $A\cap E_2=\{b\}$, so its outer measure is $1$. Their union is $\{a,b,c,d\}$; after removing it from $A$, the remainder is $\{e\}$, of outer measure $1$. Thus the right side of the theorem is $2+1+1=4$. The left side is $\mu^*(A)=|\{a,b,c,e\}|=4$. Check: the three pieces $\{a,c\},\{b\},\{e\}$ are disjoint and their union is exactly $A$, so their total cardinality is $2+1+1=4$.

Worked Example: Splitting for a Dirac Outer Measure

On $X=\mathbb{R}$, fix the point $0$ and define $\mu^*(S)=1$ if $0\in S$, and $\mu^*(S)=0$ if $0\notin S$. This is an outer measure. It vanishes on the empty set and is monotone. For countable subadditivity, if a union contains $0$, at least one set in the union contains $0$, so the sum of their outer measures is at least $1$; if the union does not contain $0$, its outer measure is $0$.

Every $E\subseteq\mathbb{R}$ is Caratheodory measurable. Indeed, if a test set $A$ contains $0$, exactly one of $A\cap E$ and $A\setminus E$ contains $0$, so the two outer measures add to $1=\mu^*(A)$. If $A$ does not contain $0$, neither part contains $0$, so both sides of the criterion are $0$.

Set $A=[-2,3]$, $E_1=(-\infty,0)$, and $E_2=\{0\}$. The measurable sets are disjoint. Since $0\notin A\cap E_1=[-2,0)$, the first piece has outer measure $0$. Since $A\cap E_2=\{0\}$, the second has outer measure $1$. The union $E_1\cup E_2=(-\infty,0]$, so the remainder is $A\setminus(E_1\cup E_2)=(0,3]$, which has outer measure $0$. The sum is $0+1+0=1$, and $\mu^*(A)=1$ because $0\in A$. Check: the only point relevant to this outer measure is $0$, and it occurs in exactly the second piece.

Why the Residual Set and the Hypotheses Matter

A common mistake is to apply the criterion to the original $A$ for every measurable set and then combine all the resulting equations. That does not produce the finite splitting identity: each individual equation divides $A$ into two parts, but it does not divide $A$ into all the pieces simultaneously.

Worked Example: Avoiding Double Counting

Use the counting outer measure on $X=\{a,b,c\}$, take $A=X$, $E_1=\{a\}$, and $E_2=\{b\}$. Each set is measurable, and the two sets are disjoint. Applying the criterion to $E_1$ and $A$ gives $\mu^*(A)=\mu^*(\{a\})+\mu^*(\{b,c\})=1+2=3$. Applying it to $E_2$ and the same original $A$ gives $\mu^*(A)=\mu^*(\{b\})+\mu^*(\{a,c\})=1+2=3$. Adding those equations would count the original set twice and does not give a valid splitting formula.

The correct second application uses the remainder $R_1=A\setminus E_1=\{b,c\}$. Splitting $R_1$ by $E_2$ gives $\mu^*(R_1)=\mu^*(\{b\})+\mu^*(\{c\})=1+1=2$. Substituting into the first equation yields $\mu^*(A)=1+1+1=3$: the piece in $E_1$, the piece in $E_2$, and the common remainder are $\{a\}$, $\{b\}$, and $\{c\}$, respectively. Check: their cardinalities are $1,1,1$, and their disjoint union is $X$, whose cardinality is $3$.

Disjointness is also essential to the formula in its stated form. In the same counting outer measure, let $A=X=\{a,b,c\}$, $E_1=\{a,b\}$, and $E_2=\{b,c\}$. Both sets are measurable, but they are not disjoint. The proposed sum of piece measures plus the remainder would be $\mu^*(A\cap E_1)+\mu^*(A\cap E_2)+\mu^*(A\setminus(E_1\cup E_2))=2+2+0=4$, whereas $\mu^*(A)=3$. The point $b$ is counted in both pieces. Thus pairwise disjointness is what makes the successive pieces $A\cap E_k$ the part of the current remainder removed at stage $k$.

Finally, the criterion cannot be applied to a set that is not Caratheodory measurable. To see this with an outer measure, let $X=\{a,b\}$ and define $\rho(S)=0$ when $S=\varnothing$ and $\rho(S)=1$ when $S\ne\varnothing$. This is an outer measure: it is monotone, and a nonempty union has at least one nonempty member, so countable subadditivity holds. But $E=\{a\}$ is not measurable, since for the test set $A=X$, $\rho(A)=1$, while $\rho(A\cap E)+\rho(A\setminus E)=\rho(\{a\})+\rho(\{b\})=1+1=2$. The exact splitting identity fails. The finite formula is therefore a consequence of applying the Caratheodory criterion to each measurable set; it is not a general additivity rule for arbitrary pieces of an outer-measure space.

Check Your Understanding

For each question, identify the test set to which the Caratheodory criterion is applied at each stage.

Write the residual sets $R_1,R_2,R_3$ when splitting $A$ by three pairwise disjoint measurable sets $E_1,E_2,E_3$.
Why does $R_{k-1}\cap E_k=A\cap E_k$ in the proof of the finite splitting theorem?
In the finite counting example, replace $A$ by $\{a,d,e\}$. Compute the two pieces and remainder and verify the identity.
For the Dirac outer measure at $0$, what is the outer measure of a set that does not contain $0$, even if that set is nonempty?
Using counting outer measure on $\{a,b,c\}$, explain why $E_1=\{a,b\}$ and $E_2=\{b,c\}$ cannot be used in the disjoint finite splitting formula.

Finite Splitting by Several Measurable Sets

What You'll Learn