Caratheodory Measurability by Exact Splitting

Outer Measure and Exact Splitting

An outer measure assigns a size to every subset of a space, including sets that may not belong to any chosen sigma-algebra. Its countable subadditivity guarantees that splitting a set into two pieces cannot make the sum of their outer measures smaller than the outer measure of the original set. Caratheodory measurability asks for more: a set is measurable when this inequality is an equality for every set being split.

Let $X$ be a set and let $\mu^*:\mathcal{P}(X)\to[0,\infty]$ be an outer measure. This means that $\mu^*(\varnothing)=0$, that $\mu^*(B)\le\mu^*(C)$ whenever $B\subseteq C\subseteq X$, and that for every sequence $(B_n)$ of subsets of $X$, $\mu^*(\bigcup_n B_n)\le\sum_n\mu^*(B_n)$. The values may be infinite. No measure space $(X,\mathcal{M},\mu)$ is assumed here: $\mu^*$ is defined on all subsets, and has not yet been asserted to be a measure on any sigma-algebra.

Definition: A set $E\subseteq X$ is Caratheodory measurable with respect to $\mu^*$ if, for every $A\subseteq X$, it satisfies the exact-splitting identity $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E). $$ The two pieces are disjoint and their union is $A$. The condition quantifies over every subset $A$ of $X$, not merely over measurable sets from some previously selected sigma-algebra.

The identity is the defining test. It is important that the test concerns the outer measure of the two pieces separately: knowing only $\mu^*(E)$, or checking the identity for one particular $A$, does not settle whether $E$ is Caratheodory measurable.

What Subadditivity Already Gives

For any $E,A\subseteq X$, the sets $A\cap E$ and $A\setminus E$ cover $A$. Therefore outer-measure subadditivity gives one direction of the desired equality without any measurability assumption.

Proposition: For every outer measure $\mu^*$ on $X$ and all $E,A\subseteq X$, $$ \mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E). $$ Consequently, $E$ is Caratheodory measurable if and only if the reverse inequality holds for every $A\subseteq X$.

Proof. Since $A=(A\cap E)\cup(A\setminus E)$, countable subadditivity, applied to these two sets (or to the sequence formed by them and then empty sets), gives $\mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E)$. If $E$ is Caratheodory measurable, its defining identity immediately gives the reverse inequality for every $A$. Conversely, if the reverse inequality holds for every $A$, it combines with the displayed subadditivity inequality to give equality for every $A$. That equality is precisely the definition. $\square$

This proposition isolates the substance of the test: the direction “outer measure of the whole is at most the sum” is automatic, while measurability requires that the split pieces do not have a sum strictly larger than the whole. The two sides are interpreted in $[0,\infty]$, so an equality with value $+\infty$ is allowed.

Worked Example: A Two-Point Outer Measure That Fails the Test

Let $X=\{p,q\}$ and define $\mu^*(\varnothing)=0$, while $\mu^*(B)=1$ for every nonempty $B\subseteq X$. This is an outer measure: it is monotone, and if a union is nonempty then at least one set in the union is nonempty, so the sum of the outer measures of the sets in any countable cover is at least $1$. Test $E=\{p\}$ using $A=X$.

The two pieces are $A\cap E=\{p\}$ and $A\setminus E=\{q\}$. Thus $\mu^*(A)=\mu^*(\{p,q\})=1$, whereas $\mu^*(A\cap E)+\mu^*(A\setminus E)=\mu^*(\{p\})+\mu^*(\{q\})=1+1=2$. The exact-splitting identity fails: $1\ne2$. Check: the subadditivity direction still holds, since $1\le2$; the failure is exactly in the reverse direction required for measurability.

It Is Enough to Test Finite Outer Measure

Although the definition quantifies over all subsets $A\subseteq X$, sets with infinite outer measure cannot cause a failure of the identity. This observation gives an equivalent version of the test that can be more convenient in calculations.

Theorem: Let $\mu^*$ be an outer measure on $X$, and let $E\subseteq X$. Then $E$ is Caratheodory measurable if and only if $$ \mu^*(A)\ge\mu^*(A\cap E)+\mu^*(A\setminus E) $$ for every $A\subseteq X$ with $\mu^*(A)<\infty$.

Proof. If $E$ is Caratheodory measurable, its defining equality implies the stated inequality for each $A$ of finite outer measure. For the converse, assume the inequality holds for every $A$ with $\mu^*(A)<\infty$. Fix any $A\subseteq X$. If $\mu^*(A)<\infty$, the assumed inequality and the subadditivity inequality proved above give equality. If $\mu^*(A)=\infty$, subadditivity gives $\infty=\mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E)$. A sum of two nonnegative extended-real numbers that is at least $+\infty$ must be $+\infty$. Hence the exact-splitting identity also holds for this $A$. The two cases cover every subset $A$, so $E$ is Caratheodory measurable. $\square$

The finite-test version does not mean that the original definition has changed: it is an equivalent way to verify it. In particular, one must not infer measurability just by checking a few convenient finite sets. The inequality must hold for every $A$ of finite outer measure.

Worked Example: Splitting Sets Under Counting Outer Measure

Take $X=\mathbb{N}=\{1,2,3,\ldots\}$ and let $\mu^*(B)$ be the number of elements of $B$ when $B$ is finite, and $+\infty$ when $B$ is infinite. Fix $E=\{1,2,3\}$. For the finite test set $A=\{2,4,5\}$, determine both pieces and check the identity.

The intersection is $A\cap E=\{2\}$, containing one element. The difference is $A\setminus E=\{4,5\}$, containing two elements. Therefore $\mu^*(A)=3$, $\mu^*(A\cap E)=1$, and $\mu^*(A\setminus E)=2$, so the split reads $3=1+2$.

Now take $A=\mathbb{N}$, which has infinite counting outer measure. For the same fixed $E=\{1,2,3\}$, the pieces are $\mathbb{N}\cap E=\{1,2,3\}$ and $\mathbb{N}\setminus E=\{4,5,6,\ldots\}$. Their outer measures are $3$ and $+\infty$, respectively, so the split reads $+\infty=3+\infty$. Check: these pieces are disjoint and their union is $\mathbb{N}$; their outer measures add to $+\infty$, as required.

In fact, every $E\subseteq\mathbb{N}$ satisfies the exact-splitting identity for this outer measure. For any $A\subseteq\mathbb{N}$, the pieces $A\cap E$ and $A\setminus E$ are disjoint and partition $A$. If $A$ is finite, its number of elements is the sum of the numbers in the two pieces. If $A$ is infinite, at least one piece must be infinite: if both were finite, their union $A$ would be finite. Thus the sum of the two piece counts is $+\infty=\mu^*(A)$. This verifies the identity also in the infinite case.

Splitting for Lebesgue Outer Measure

On $\mathbb{R}$, Lebesgue outer measure is denoted $m^*$. It is an outer measure on all subsets of $\mathbb{R}$, whether or not they are Lebesgue measurable. For intervals, its value is their length; in particular, endpoints do not change the value. When checking a proposed split, compute the outer measure of each piece and the original set separately rather than assuming additivity for arbitrary subsets.

Worked Example: Two Exact Splits for Lebesgue Outer Measure

Take $X=\mathbb{R}$, $\mu^*=m^*$, and $E=[0,1]$. First choose $A=[-1,2]$. Compute the two pieces and their outer measures.

The intersection is $A\cap E=[0,1]$, whose outer measure is $1-0=1$. The difference is $A\setminus E=[-1,0)\cup(1,2]$. These two intervals have lengths $0-(-1)=1$ and $2-1=1$; their disjoint union has outer measure $1+1=2$. The original interval has outer measure $2-(-1)=3$. Hence the identity reads $3=1+2$.

For another check, take $A=[-2,0]$. Then $A\cap E=\{0\}$, whose outer measure is $0$, and $A\setminus E=[-2,0)$, whose outer measure is $0-(-2)=2$. Also $m^*(A)=0-(-2)=2$, so this split reads $2=0+2$. Check: in each case the intersection and difference are disjoint and together recover the specified $A$. These calculations illustrate the criterion for these particular test sets; by themselves, they are not a verification for every $A\subseteq\mathbb{R}$.

When Splitting Is Already Additive

There is a useful setting in which the exact identity follows directly from the definition of a measure. This explains why the splitting test is needed when starting from an outer measure: an outer measure has subadditivity, but not necessarily additivity on every pair of disjoint subsets.

Proposition: Suppose $\nu:\mathcal{P}(X)\to[0,\infty]$ is a countably additive measure on the power set of $X$. Then every $E\subseteq X$ satisfies $$ \nu(A)=\nu(A\cap E)+\nu(A\setminus E) $$ for every $A\subseteq X$.

Proof. Fix $E,A\subseteq X$. The sets $A\cap E$ and $A\setminus E$ are disjoint: an element in the first belongs to $E$, while an element in the second does not. Their union is $A$, since each element of $A$ either belongs to $E$ or does not. Countable additivity of $\nu$, applied to this disjoint pair (and empty sets for any further terms), gives $\nu(A)=\nu(A\cap E)+\nu(A\setminus E)$. This holds for every $A$, including sets on which the value is infinite, so every $E$ satisfies the exact-splitting condition. $\square$

Counting measure on $\mathbb{N}$ is one example of a measure defined on the entire power set, which is why every subset passed the test in the counting example. By contrast, the two-point outer measure above was only required to be subadditive; it assigned value $1$ to each nonempty singleton and value $1$ to their union, so its split into two singletons exposed the failure of additivity.

A Common Pitfall

A common error is to reverse the direction supplied by subadditivity or to assume that the two pieces must have additive outer measure. For every outer measure and every $A,E\subseteq X$, the guaranteed inequality is $\mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E)$. In the two-point example, the incorrect calculation would be to declare $1=1+1$ because the two pieces partition $X$. The correction is to evaluate each quantity from the definition: the whole has outer measure $1$, each singleton has outer measure $1$, and the sum is $2$. That proposed split fails, so $E=\{p\}$ is not Caratheodory measurable there.

The criterion therefore has a precise role: it identifies the subsets that split every test set without creating extra outer measure. The definition itself makes no claim here about how the collection of all such subsets behaves under complements or unions; those are separate structural questions.

Check Your Understanding

For each question, keep the underlying space, outer measure, and test set distinct, and check which inequality is automatic.

State the exact-splitting definition of Caratheodory measurability for an outer measure $\mu^*$ on $X$.
Why does outer-measure subadditivity always give $\mu^*(A)\le\mu^*(A\cap E)+\mu^*(A\setminus E)$?
For counting outer measure on $\mathbb{N}$, take $E=\{1,2,3\}$ and $A=\{1,3,6,7\}$. Write both pieces and verify the identity.
Explain why a set $A$ with $\mu^*(A)=+\infty$ cannot violate exact splitting.
For the two-point outer measure with value $1$ on every nonempty set, test $E=\{p\}$ using $A=\{q\}$. Does this test alone establish measurability?