Complement Closure via the Splitting Criterion

The Splitting Test and Its Symmetry

Caratheodory measurability is defined by testing how a set divides every possible test set. The complement question is therefore especially direct: replacing a set by its complement reverses the two pieces, but does not change their sum. The proof works for every outer measure, without assuming that its values are finite.

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 $\mu^*$ is monotone, and that it is countably subadditive. Here $\mathcal{P}(X)$ is the collection of all subsets of $X$; an outer measure is not assumed to be a measure on that collection. In particular, its values are allowed to be $+\infty$.

Definition: A set $E\subseteq X$ is Caratheodory measurable with respect to $\mu^*$ if, for every $A\subseteq X$, $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c). $$ The equality is in $[0,\infty]$, and $E^c=X\setminus E$. The set $A$ is arbitrary: it need not be measurable.

The two intersections partition $A$, since $A\cap E$ and $A\cap E^c$ are disjoint and their union is $A$. The criterion asks for exact equality of outer measure with the sum of the outer measures of these pieces. Countable subadditivity always gives the inequality $\mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\cap E^c)$; measurability requires equality for every $A$.

Complement Closure

Theorem (Complement Closure): For an outer measure $\mu^*$ on $X$ and any $E\subseteq X$, $E$ is Caratheodory measurable if and only if $E^c$ is Caratheodory measurable.

Proof. Suppose first that $E$ is measurable. To show that $E^c$ is measurable, take an arbitrary $A\subseteq X$. The criterion for $E$ gives $$ \mu^*(A)=\mu^*(A\cap E)+\mu^*(A\cap E^c). $$ Addition in $[0,\infty]$ is commutative, including when one or both summands are $+\infty$. Thus $$ \mu^*(A)=\mu^*(A\cap E^c)+\mu^*(A\cap E). $$ This is precisely the criterion for $E^c$, because $(E^c)^c=E$. Since $A$ was arbitrary, $E^c$ is measurable.

Conversely, suppose that $E^c$ is measurable. Apply its criterion to an arbitrary $A\subseteq X$: $$ \mu^*(A)=\mu^*(A\cap E^c)+\mu^*(A\cap (E^c)^c) =\mu^*(A\cap E^c)+\mu^*(A\cap E). $$ Commutativity of addition gives the splitting identity for $E$. It holds for every $A$, so $E$ is measurable. Both implications are proved. $\square$

The proof does not subtract one outer-measure value from another. Such subtraction would be unsafe if values could be infinite. Instead, it uses only the identity in the definition and swaps its two summands.

Corollary: A set and its complement always have the same measurability status. In particular, if one of them fails the splitting criterion for even one test set $A$, then the other fails for that same $A$.

Proof. The theorem gives both implications. For the final statement, the two right-hand sides of the splitting tests are the same pair of terms in opposite orders: $$ \mu^*(A\cap E)+\mu^*(A\cap E^c) = \mu^*(A\cap E^c)+\mu^*(A\cap E). $$ Therefore, if this value is not $\mu^*(A)$ for one of the tests, it is not $\mu^*(A)$ for the other set either. $\square$

Worked Checks of the Criterion

Worked Example: Counting Outer Measure on the Natural Numbers

Let $X=\mathbb{N}=\{1,2,\ldots\}$, and define $\mu^*(A)$ to be the number of elements of $A$ when $A$ is finite, and $+\infty$ when $A$ is infinite. This is an outer measure: it vanishes on the empty set, is monotone, and is countably subadditive. For subadditivity, if a countable union is finite, its elements are covered by the sets in the union, so its size is at most the sum of their sizes. If the union is infinite, either at least one covering set is infinite or the sum of the finite sizes is unbounded; in either case the sum of the outer measures is $+\infty$.

Take $E=\{2,4\}$, so $E^c=\mathbb{N}\setminus\{2,4\}$. We check the splitting identity for every $A\subseteq\mathbb{N}$. If $A$ is finite, its two disjoint pieces have cardinalities adding to the cardinality of $A$, including if either piece is empty. If $A$ is infinite, removing at most two elements leaves $A\cap E^c$ infinite. Hence $\mu^*(A)=+\infty$, $\mu^*(A\cap E^c)=+\infty$, and the sum on the right is $+\infty$. In both cases the identity holds, so $E$ is measurable.

For a numerical check, set $A=\{1,2,3,4,5\}$. Then $\mu^*(A)=5$, $A\cap E=\{2,4\}$ has outer measure $2$, and $A\cap E^c=\{1,3,5\}$ has outer measure $3$. Thus $5=2+3$. Interchanging the pieces gives $5=3+2$, as required for $E^c$.

Check: for the infinite test set $A=\mathbb{N}$, the two pieces have outer measures $2$ and $+\infty$, so their sum is $+\infty=\mu^*(\mathbb{N})$. This also checks the extended-value case in the identity.

Worked Example: An Outer Measure That Counts Blocks

Let $X=\{a,b,c,d\}$, partitioned into blocks $B_1=\{a,b\}$ and $B_2=\{c,d\}$. Define $\mu^*(A)$ to be the number of these two blocks that intersect $A$. This is an outer measure. It is zero on the empty set and monotone. For countable subadditivity, every block intersecting a union $\bigcup_n A_n$ intersects at least one $A_n$; the number of blocks touched by the union is therefore at most the sum of the numbers touched by the individual sets.

Take $E=\{a,b\}=B_1$. For any $A\subseteq X$, each block that intersects $A$ lies entirely in either $E$ or $E^c$. Thus a block contributes one to $\mu^*(A)$ exactly when it contributes one to one of $\mu^*(A\cap E)$ and $\mu^*(A\cap E^c)$, and contributes zero to the other. Counting the two blocks gives the splitting identity for every $A$.

For example, if $A=\{a,c,d\}$, then $\mu^*(A)=2$, $\mu^*(A\cap E)=\mu^*(\{a\})=1$, and $\mu^*(A\cap E^c)=\mu^*(\{c,d\})=1$. Hence $2=1+1$. For the complement $E^c=\{c,d\}$, the same test reads $2=1+1$, with the pieces reversed.

Now consider a set that cuts a block, $F=\{a\}$. Use $A=B_1=\{a,b\}$ as a test. The whole block has outer measure $1$, while $A\cap F=\{a\}$ and $A\cap F^c=\{b\}$ each have outer measure $1$. The proposed equality would be $1=1+1=2$, which is false. The complement $F^c=\{b,c,d\}$ fails on this same test: its two pieces are $\{b\}$ and $\{a\}$, again giving $1\ne2$.

Worked Example: A Concrete Failure Shared by Complements

On any nonempty set $X$, define $$ \mu^*(A)= \begin{cases} 0,&A=\varnothing,\\ 1,&A\ne\varnothing. \end{cases} $$ This is an outer measure. Monotonicity follows because a subset of the empty set is empty, and every nonempty set has value $1$. For countable subadditivity, if the union is empty, every set in the union is empty and both sides are zero. If the union is nonempty, at least one set in the union is nonempty, so the sum on the right is at least $1$, while the left side is $1$.

The empty set satisfies the criterion: for every $A$, the two pieces are $\varnothing$ and $A$, giving $0+\mu^*(A)=\mu^*(A)$. The whole space also satisfies it, since its two pieces are $A$ and $\varnothing$. But if $F$ is nonempty and $F^c$ is nonempty, take $A=X$. Then $\mu^*(X)=1$, while $\mu^*(X\cap F)=1$ and $\mu^*(X\cap F^c)=1$. The splitting test fails: $1\ne1+1=2$. For the complement, the two intersection values are still $1$ and $1$, so the same test fails.

Check: the failure is a strict inequality, $2>1$, in the direction already guaranteed by subadditivity. It is not a failure caused by subtracting or manipulating infinite values.

A Common Mistake

A tempting but incorrect argument is to think that complementing a set changes the splitting sum. For example, for the block outer measure and $A=\{a,c,d\}$, one might claim that $E=\{a,b\}$ gives $2=1+1$, but $E^c$ gives $2=0+2$, because $A\cap E^c$ has two elements. This calculation confuses cardinality with the block outer measure. The set $\{c,d\}$ touches one block, so its outer measure is $1$, not $2$. The correct complement calculation is $2=1+1$. More generally, the terms are always $\mu^*(A\cap E)$ and $\mu^*(A\cap E^c)$; complementing $E$ merely exchanges their order.

Another potential error is to write a difference such as $\mu^*(A)-\mu^*(A\cap E)$ and identify it with $\mu^*(A\cap E^c)$. The splitting criterion does not license that subtraction, especially if either term is infinite. The complement proof needs no such step: it preserves the original equality and uses commutativity alone.

Complement closure is a local structural fact about the splitting criterion. It does not by itself establish closure under unions or countable unions; those require additional arguments. Its precise content is that every test set $A$ gives the same two outer-measure terms for $E$ and $E^c$, in reversed order.

Check Your Understanding

Use the splitting identity directly in each question. Remember that all test sets range over subsets of $X$, and outer-measure values may be infinite.

Write the Caratheodory criterion for $E^c$, and identify how its two terms compare with the terms for $E$.
If the criterion holds for $E^c$ for every $A\subseteq X$, prove that it holds for $E$.
For the block outer measure on $\{a,b,c,d\}$, use $A=\{a,b\}$ to test whether $F=\{a\}$ is measurable. Give both outer-measure values on the right-hand side.
For the outer measure that assigns value $1$ to every nonempty set, test a nonempty proper set $F$ using $A=X$. Does $F^c$ pass the same test?
Why is it unsafe to derive the complement identity by subtracting one outer-measure value from another when values in $[0,\infty]$ are allowed?

Complement Closure via the Splitting Criterion

What You'll Learn