The Empty Set and Whole Space as Measurable Sets

The Splitting Test in Its Simplest Cases

The Caratheodory criterion tests whether a set $A$ divides every set $E$ into two pieces whose outer measures add back to the outer measure of $E$. The empty set and the whole space give the two extreme divisions: one piece is always empty, and the other is all of $E$. The verification is short, but it matters that outer measures may take the value $+\infty$. In particular, the argument must use addition of extended nonnegative values, not subtraction of possibly infinite quantities.

Let $X$ be a set and let $\mu^*$ be an outer measure on $X$. This means that $\mu^*$ is a function from the power set of $X$ to $[0,\infty]$, with $\mu^*(\varnothing)=0$, that is monotone and countably subadditive. No finiteness or sigma-finiteness assumption is imposed. The criterion below concerns every subset $E\subseteq X$, whether or not $E$ has already been shown measurable.

Definition (Caratheodory measurability): A set $A\subseteq X$ is $\mu^*$-measurable if, for every $E\subseteq X$, $$ \mu^*(E)=\mu^*(E\cap A)+\mu^*(E\setminus A). $$ The sum is taken in the extended nonnegative real numbers, so $+\infty+0=+\infty$.

The criterion is an equality, not merely the subadditivity inequality. Countable subadditivity already gives $\mu^*(E)\leq\mu^*(E\cap A)+\mu^*(E\setminus A)$, because the two pieces cover $E$. To pass the test, the reverse inequality must also hold. For the two extreme choices of $A$, the pieces can be identified exactly.

The Empty Set

Proposition: For every set $X$ with outer measure $\mu^*$, the empty set is $\mu^*$-measurable. This holds whether or not $\mu^*(X)$ is finite.

Proof. Fix an arbitrary set $E\subseteq X$. Since $E\cap\varnothing=\varnothing$ and $E\setminus\varnothing=E$, the right side of the splitting criterion is $$ \mu^*(E\cap\varnothing)+\mu^*(E\setminus\varnothing) =\mu^*(\varnothing)+\mu^*(E) =0+\mu^*(E) =\mu^*(E). $$ The last equality holds for every value in $[0,\infty]$, including $+\infty$. Since $E$ was arbitrary, the criterion holds for all $E\subseteq X$, which proves the proposition. $\square$

Nothing in this argument uses an assumption about the size of $E$, nor does it require $E$ to be measurable. The only outer-measure axiom used is $\mu^*(\varnothing)=0$, together with ordinary addition of nonnegative extended real numbers.

The Whole Space

Proposition: For every set $X$ with outer measure $\mu^*$, the whole space $X$ is $\mu^*$-measurable. This remains true when $\mu^*(X)=+\infty$.

Proof. Fix any $E\subseteq X$. Because $E\subseteq X$, we have $E\cap X=E$ and $E\setminus X=\varnothing$. Therefore $$ \mu^*(E\cap X)+\mu^*(E\setminus X) =\mu^*(E)+\mu^*(\varnothing) =\mu^*(E)+0 =\mu^*(E). $$ This equality also holds when $\mu^*(E)=+\infty$, since $+\infty+0=+\infty$. It holds for every subset $E$, so $X$ satisfies the Caratheodory criterion. $\square$

The two propositions do not assume that $X$ itself has finite outer measure. Indeed, the whole-space proof does not need to evaluate $\mu^*(X)$ at all: it works with the arbitrary test set $E$. Even if $E=X$ and $\mu^*(X)=+\infty$, the split is $+\infty+0=+\infty$, exactly as required.

Worked Example: Counting Outer Measure on an Infinite Space

Take $X=\mathbb{N}=\{1,2,\ldots\}$ and let $\mu^*(E)$ be the number of elements of $E$ if $E$ is finite, and $+\infty$ if $E$ is infinite. This is counting measure on the power set of $\mathbb{N}$, and is in particular an outer measure: the empty set has value zero, inclusion cannot decrease cardinality, and a countable union has cardinality at most the sum of the cardinalities of its members.

Test $A=\varnothing$ using first $E=\{2,5\}$. This test set has two elements, so $\mu^*(E)=2$. Its pieces are $E\cap A=\varnothing$, of outer measure $0$, and $E\setminus A=\{2,5\}$, of outer measure $2$. Thus the split sum is $0+2=2$, equal to $\mu^*(E)$. For $E=\mathbb{N}$, the outer measure is $+\infty$; the pieces have outer measures $0$ and $+\infty$, so their sum is $0+(+\infty)=+\infty$.

Now test $A=X=\mathbb{N}$. For $E=\{2,5\}$, the pieces are $E$ and $\varnothing$, with outer measures $2$ and $0$; the sum is $2+0=2=\mu^*(E)$. For $E=\mathbb{N}$, their outer measures are $+\infty$ and $0$, so the sum is $+\infty+0=+\infty=\mu^*(\mathbb{N})$. These calculations illustrate the criterion for the selected test sets; the propositions prove it for every $E\subseteq\mathbb{N}$.

Check: in each calculation, exactly one piece is empty and the other is the original test set. The computed split sum therefore agrees with the original outer measure, even for the infinite test set.

Worked Example: Lebesgue Outer Measure on the Real Line

Let $X=\mathbb{R}$ and let $\mu^*=m^*$, Lebesgue outer measure. Its total outer measure is $m^*(\mathbb{R})=+\infty$. For the whole-space test $A=\mathbb{R}$, choose $E=[0,1]$. The interval has outer measure $1$. The pieces are $E\cap A=[0,1]$, with outer measure $1$, and $E\setminus A=\varnothing$, with outer measure $0$. The split sum is $1+0=1=m^*([0,1])$.

For the same $A$, choose $E=\mathbb{R}$. Then $E\cap A=\mathbb{R}$ and $E\setminus A=\varnothing$. Their outer measures are $+\infty$ and $0$, so $$ m^*(E\cap A)+m^*(E\setminus A)=+\infty+0=+\infty=m^*(\mathbb{R}). $$ For the empty-set test $A=\varnothing$, with $E=[0,1]$, the pieces are $\varnothing$ and $[0,1]$, and the sum is $0+1=1$. With $E=\mathbb{R}$, the pieces are $\varnothing$ and $\mathbb{R}$, and the sum is $0+(+\infty)=+\infty$. Thus the finite and infinite test values agree with the propositions.

Check: the outer measure of $\mathbb{R}$ is infinite, but none of these computations subtracts it from another infinite quantity. Each calculation uses only a zero outer measure for the empty piece and the outer measure of the unchanged piece.

A Finite Example and the Extended-Value Rule

Worked Example: Weighted Outer Measure on Two Points

Let $X=\{a,b\}$, and define $$ \mu^*(E)=2\chi_E(a)+5\chi_E(b),\qquad E\subseteq X. $$ Here $\chi_E(x)$ equals $1$ if $x\in E$ and $0$ otherwise. This is a finite measure, hence an outer measure: it assigns zero to the empty set, is monotone, and is additive on disjoint sets, which implies countable subadditivity.

To check the splitting computations explicitly for $A=\varnothing$, list all possible test sets. For $E=\varnothing$, the value is $0$, and the pieces have values $0$ and $0$, so the split sum is $0+0=0$. For $E=\{a\}$, $\mu^*(E)=2$, and the pieces have values $0$ and $2$, so the sum is $0+2=2$. For $E=\{b\}$, $\mu^*(E)=5$, and the sum is $0+5=5$. For $E=X$, $\mu^*(E)=2+5=7$, and the split sum is $0+7=7$.

For $A=X$, the pieces are $E$ and $\varnothing$ in each of those four cases. Their split sums are respectively $0+0=0$, $2+0=2$, $5+0=5$, and $7+0=7$. Thus every possible test set passes for both choices of $A$. Check: the value for $X$ is $2+5=7$, and each split above reproduces the corresponding value of $E$.

The finite example makes the set identities visible by exhausting all test sets. For a general $X$, there may be too many subsets to list, so the proofs instead fix an arbitrary $E$ and identify its two pieces. This universal quantifier is essential: checking only $E=X$ would not by itself establish the criterion, whose definition requires equality for every subset $E$.

Takeaway: For any outer measure $\mu^*$ on any set $X$, both $\varnothing$ and $X$ are $\mu^*$-measurable. The proof uses $\mu^*(\varnothing)=0$ and the identities for intersections and set differences; it needs no finiteness, sigma-finiteness, or completeness assumption.

Common Pitfall: Subtracting Infinite Quantities

A tempting but invalid shortcut is to rewrite the splitting equation as a “defect” involving subtraction. For instance, with Lebesgue outer measure on $\mathbb{R}$, take $A=\mathbb{R}$ and $E=\mathbb{R}$. The correct split is $$ m^*(E\cap A)+m^*(E\setminus A)=m^*(\mathbb{R})+m^*(\varnothing)=+\infty+0=+\infty. $$ It equals $m^*(E)=+\infty$, so the criterion holds.

A wrong calculation might subtract $m^*(E)$ from the split sum and write $+\infty-(+\infty)=0$, then treat that as an ordinary arithmetic verification. In extended-real arithmetic, $+\infty-(+\infty)$ is undefined. The expression does not establish an equality or a defect value. The correction is to compare the two sides of the original criterion directly. Here both are $+\infty$, and the sum on the right is well-defined because it is $+\infty+0$. More generally, when applying this criterion, keep the nonnegative sum intact rather than subtracting outer measures that may be infinite.

Check Your Understanding

Use the splitting criterion directly. In each question, remember that it quantifies over every test set $E\subseteq X$.

For $A=\varnothing$, identify $E\cap A$ and $E\setminus A$ for an arbitrary $E\subseteq X$, and give the resulting split equation.
For $A=X$, explain why the proof remains valid if $\mu^*(X)=+\infty$.
On $\mathbb{N}$ with counting measure, compute the split sum for $A=X$ and $E=\{1,4,9\}$.
Why is the expression $+\infty-(+\infty)$ not an acceptable way to verify the criterion?
Does either extreme-set verification require $\mu^*$ to be finite or sigma-finite? Justify your answer from the proof.