Completeness of the Caratheodory Measure

Why Null Sets Control Completeness

An outer measure is defined on every subset of its underlying space, but its restriction to the Caratheodory measurable sets need not initially be assumed complete. Completeness asks whether subsets of measurable null sets are measurable too. The key point is that a set of outer measure zero contributes nothing to either side of any Caratheodory splitting test. This makes every subset of such a set measurable, even if the subset has no prior measurability description.

Let $X$ be a set, and let $\mu^*:\mathcal{P}(X)\to[0,\infty]$ be an outer measure. Thus $\mu^*(\varnothing)=0$, $\mu^*$ is monotone, and it is countably subadditive. In particular, $A\subseteq B$ implies $\mu^*(A)\le\mu^*(B)$, and $\mu^*(\bigcup_n A_n)\le\sum_n\mu^*(A_n)$. No finiteness or sigma-finiteness assumption is made.

Definition: A set $E\subseteq X$ is Caratheodory measurable if, for every $T\subseteq X$, $$ \mu^*(T)=\mu^*(T\cap E)+\mu^*(T\setminus E). $$ A set $N\subseteq X$ is outer-null if $\mu^*(N)=0$. Write $\mathcal{M}$ for the collection of Caratheodory measurable sets and $\mu=\mu^*|_{\mathcal{M}}$ for the restriction of the outer measure. The earlier results on Caratheodory splitting give that $\mathcal{M}$ is a sigma-algebra and $(X,\mathcal{M},\mu)$ is a measure space.

Outer-null is a property defined using $\mu^*$ on all subsets of $X$. A $\mu$-null set, by contrast, is a member of $\mathcal{M}$ whose restricted measure is zero. For measurable $E$, the restriction gives $\mu(E)=\mu^*(E)$, so these conditions agree for measurable sets. The result below also shows that an outer-null set is itself measurable, so the distinction disappears for sets of outer measure zero.

Every Subset of an Outer-Null Set Is Measurable

Proposition: Let $\mu^*$ be an outer measure on $X$. If $N\subseteq X$ satisfies $\mu^*(N)=0$, then every $A\subseteq N$ is Caratheodory measurable. Moreover, $\mu^*(A)=0$.

Proof. First, $A\subseteq N$ and monotonicity imply $0\le\mu^*(A)\le\mu^*(N)=0$, so $\mu^*(A)=0$. Fix an arbitrary $T\subseteq X$. Since $T=(T\cap A)\cup(T\setminus A)$, countable subadditivity (in its two-set case) gives $$ \mu^*(T)\le\mu^*(T\cap A)+\mu^*(T\setminus A)=\mu^*(T\setminus A). $$ Also $T\setminus A\subseteq T$, so monotonicity gives $\mu^*(T\setminus A)\le\mu^*(T)$. Therefore $\mu^*(T)=\mu^*(T\setminus A)$, including when these values are infinite. Since $\mu^*(T\cap A)=0$, this is exactly the Caratheodory equality for $A$. The choice of $T$ was arbitrary, so $A$ is measurable. $\square$

The argument does not need $N$ to be measurable in advance. Indeed, taking $A=N$ proves that every outer-null set is itself Caratheodory measurable. It also shows why the proof works for an arbitrary outer measure: its only ingredients are monotonicity, subadditivity, and the zero outer measure of $A$.

Worked Example: A Finite Outer-Null Set

Take $X=\mathbb{R}$ with Lebesgue outer measure $m^*$, and let $N=\{-1,0,1\}$. We first check directly that $N$ is outer-null. For any $\varepsilon>0$, cover each of the three points by an open interval of length $\varepsilon/6$ centered at that point. The total length is $3(\varepsilon/6)=\varepsilon/2$, so $0\le m^*(N)\le\varepsilon/2$ for every $\varepsilon>0$, and hence $m^*(N)=0$.

Let $A=\{-1,1\}$, so $A\subseteq N$. The proposition gives $m^*(A)=0$ and says that $A$ is Caratheodory measurable. To check the splitting on a particular set, take $T=[-2,2]$. Its Lebesgue measure is $4$. The split has $T\cap A=A$, of measure zero, and $T\setminus A=[-2,2]\setminus\{-1,1\}$. Since $A$ is measurable, the restricted measure is additive on this split, giving $m(T\setminus A)=4-0=4$.

Check: the two sides of the splitting equality are $m(T)=4$ and $m(T\cap A)+m(T\setminus A)=0+4=4$. The example uses the strict set membership $x\in A$: the two removed points are $-1$ and $1$, not the endpoints $-2$ and $2$.

Worked Example: Every Subset of a Countable Null Set

Again use $X=\mathbb{R}$ and Lebesgue outer measure $m^*$. Enumerate $\mathbb{Q}\cap[0,1]$ as $q_1,q_2,\ldots$. For any $\varepsilon>0$, cover $q_k$ by the open interval centered at $q_k$ of length $\varepsilon/2^{k+1}$. These intervals cover $\mathbb{Q}\cap[0,1]$, and their total length is $$ \sum_{k=1}^{\infty}\frac{\varepsilon}{2^{k+1}} =\varepsilon\left(\frac14+\frac18+\frac1{16}+\cdots\right) =\frac{\varepsilon}{2}. $$ Thus $0\le m^*(\mathbb{Q}\cap[0,1])\le\varepsilon/2$ for every $\varepsilon>0$, so $m^*(\mathbb{Q}\cap[0,1])=0$.

For instance, let $B=\{q_{2k}:k\ge1\}$, where membership means the enumeration index is even. Since $B\subseteq\mathbb{Q}\cap[0,1]$, the proposition says that $B$ is Caratheodory measurable and $m(B)=0$. The conclusion applies equally to any subset $B\subseteq\mathbb{Q}\cap[0,1]$, without requiring that the subset itself be given by an explicit enumeration or description.

Check: the cover bound is $\varepsilon/2$, not merely a bound for one chosen $\varepsilon$. Because it holds for every positive $\varepsilon$, the outer measure must be zero. Monotonicity then gives $m^*(B)=0$, and the proposition supplies measurability as well.

Worked Example: Subsets of the Cantor Set

Use $X=\mathbb{R}$ with Lebesgue outer measure $m^*$, and let $C$ be the middle-thirds Cantor set in $[0,1]$. At stage $n$, the set $C$ is covered by $2^n$ closed intervals, each of length $3^{-n}$. Their total length is $2^n3^{-n}=(2/3)^n$. Therefore $0\le m^*(C)\le(2/3)^n$ for every $n\ge1$. Since $(2/3)^n\to0$, it follows that $m^*(C)=0$.

For the particular subset $D=C\cap[0,1/4]$, we have $D\subseteq C$. Consequently $D$ is Caratheodory measurable and $m(D)=0$, whether or not one separately describes the points of $D$. More generally, every subset of $C$, including subsets not defined by a measurable construction, is measurable for this Caratheodory measure.

Check: at stage $n=2$, the cover consists of $2^2=4$ intervals of length $3^{-2}=1/9$, giving total length $4/9=(2/3)^2$. The bound tends to zero as the stage increases, which verifies the zero outer measure used in the proposition.

The Restricted Measure Is Complete

Definition: A measure space $(X,\mathcal{M},\mu)$ is complete if, whenever $N\in\mathcal{M}$ and $\mu(N)=0$, every subset $A\subseteq N$ belongs to $\mathcal{M}$. In that case $\mu(A)=0$ for each such subset.

Theorem: For any outer measure $\mu^*$ on $X$, the measure space $(X,\mathcal{M},\mu)$ obtained by restricting $\mu^*$ to its Caratheodory measurable sets is complete. No finiteness or sigma-finiteness hypothesis is required.

Proof. Let $N\in\mathcal{M}$ satisfy $\mu(N)=0$, and take any subset $A\subseteq N$. Since $\mu$ is the restriction of $\mu^*$, $\mu^*(N)=\mu(N)=0$. The proposition therefore implies that $A$ is Caratheodory measurable, so $A\in\mathcal{M}$, and that $\mu^*(A)=0$. Restricting once more gives $\mu(A)=\mu^*(A)=0$. This verifies both requirements in the definition of completeness. $\square$

The conclusion concerns completeness of the measure space, not finiteness of its measure. For example, the proof remains valid when $\mu(X)=\infty$: it never subtracts infinite quantities or assumes that a complement has finite measure. It also distinguishes this result from a claim about an arbitrary measure space. A measure defined on a smaller sigma-algebra may omit subsets of its null sets; the Caratheodory restriction does not, because the splitting criterion itself forces those subsets to be measurable.

A Common Pitfall: Assuming the Splitting Equality

A tempting but circular argument is: “If $A$ has outer measure zero, then $\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\setminus A)$, since $T$ is split into those two pieces.” That equality is precisely the Caratheodory condition being proved; merely writing $T$ as a union does not give equality for an outer measure. The direct fact available before measurability is subadditivity, which gives only $\mu^*(T)\le\mu^*(T\cap A)+\mu^*(T\setminus A)$.

For a concrete instance, take $X=\mathbb{R}$, $T=[0,1]$, and $A=\{0\}$. The desired equality has values $m^*([0,1])=1$ and $m^*(\{0\})+m^*((0,1])=0+1$. Those values agree, but using this equality as the reason that $A$ is measurable assumes the conclusion. Instead, the proof uses $m^*(T\cap A)=0$, subadditivity to obtain $m^*(T)\le m^*(T\setminus A)$, and monotonicity to obtain $m^*(T\setminus A)\le m^*(T)$. These two inequalities establish the equality without presupposing measurability.

Start with outer measure zero.
If $A\subseteq N$ and $\mu^*(N)=0$, monotonicity gives $\mu^*(A)=0$.

Test an arbitrary set.
For each $T\subseteq X$, apply subadditivity to $T=(T\cap A)\cup(T\setminus A)$.

Use monotonicity in the other direction.
Since $T\setminus A\subseteq T$, the two inequalities give $\mu^*(T)=\mu^*(T\setminus A)$, and the zero intersection supplies the splitting equality.

Conclude completeness.
Apply the result to every subset of a measurable $\mu$-null set in the restricted measure space.

Check Your Understanding

In each question, distinguish carefully between outer measure on all subsets and the restricted measure on Caratheodory measurable sets.

Which two outer-measure properties give the reverse inequalities needed to show $\mu^*(T)=\mu^*(T\setminus A)$ when $\mu^*(A)=0$?
Does the proposition require the outer-null set $N$ to be Caratheodory measurable beforehand? Explain why or why not.
If $N\in\mathcal{M}$ and $\mu(N)=0$, identify the equality that converts this restricted-measure condition into an outer-measure condition.
In the Cantor-set example, what is the total length of the stage-$n$ cover, and why does this imply outer measure zero?
Why is writing the Caratheodory splitting equality before proving measurability a circular argument?

Completeness of the Caratheodory Measure

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