Completing the Extension with Null Subsets

Why Complete a Measure Space?

An extension of a premeasure gives a measure on a sigma-algebra, but that sigma-algebra need not contain every subset of its null sets. This can matter when a set differs from a measurable set only by points lying inside a null set: the difference may not belong to the original sigma-algebra at all. Completion repairs this omission by adjoining all such subsets, while assigning them measure zero. The construction does not require the original measure to be finite or sigma-finite.

Let $(X,\mathcal{M},\mu)$ be any measure space. A measurable null set is a set $Z\in\mathcal{M}$ with $\mu(Z)=0$. The word measurable is important: a subset of $Z$ need not itself belong to $\mathcal{M}$. A measure space is complete if every subset of every measurable null set belongs to its sigma-algebra.

Definition: The completion of $(X,\mathcal{M},\mu)$ is the measure space $(X,\overline{\mathcal{M}},\overline{\mu})$, where $$ \overline{\mathcal{M}} = \{E\subseteq X:\ E=A\cup N\text{ for some }A,Z\in\mathcal{M} \text{ with }\mu(Z)=0\text{ and }N\subseteq Z\}, $$ and $\overline{\mu}(E)=\mu(A)$ whenever $E=A\cup N$ is such a representation.

Equivalently, $E\in\overline{\mathcal{M}}$ if there are $A,B\in\mathcal{M}$ with $A\subseteq E\subseteq B$ and $\mu(B\setminus A)=0$. Indeed, a representation $E=A\cup N$, with $N\subseteq Z$ and $\mu(Z)=0$, gives $A\subseteq E\subseteq A\cup Z$, and $\mu((A\cup Z)\setminus A)=0$. Conversely, if $A\subseteq E\subseteq B$ and $\mu(B\setminus A)=0$, then $E=A\cup(E\setminus A)$, where $E\setminus A\subseteq B\setminus A$.

The Completion Theorem

Theorem: For any measure space $(X,\mathcal{M},\mu)$, the collection $\overline{\mathcal{M}}$ is a sigma-algebra containing $\mathcal{M}$. The formula $\overline{\mu}(A\cup N)=\mu(A)$ is well defined and makes $(X,\overline{\mathcal{M}},\overline{\mu})$ a complete measure space. Moreover, $\overline{\mu}$ agrees with $\mu$ on $\mathcal{M}$. No finiteness or sigma-finiteness assumption is needed.

Proof. First we show that $\overline{\mathcal{M}}$ is a sigma-algebra. The empty set belongs to it by taking $A=N=\varnothing$. Suppose $E=A\cup N$, where $A,Z\in\mathcal{M}$, $\mu(Z)=0$, and $N\subseteq Z$. Then $A\subseteq E\subseteq B=A\cup Z$, and $B\setminus A\subseteq Z$, so $\mu(B\setminus A)=0$. Taking complements gives $B^c\subseteq E^c\subseteq A^c$, with $A^c\setminus B^c=B\setminus A$. The equivalent sandwich description therefore shows that $E^c\in\overline{\mathcal{M}}$.

Now let $E_n\in\overline{\mathcal{M}}$ for $n\ge1$. Choose $A_n,Z_n\in\mathcal{M}$ such that $\mu(Z_n)=0$, $A_n\subseteq E_n$, and $E_n\setminus A_n\subseteq Z_n$. Put $A=\bigcup_{n=1}^{\infty}A_n$ and $Z=\bigcup_{n=1}^{\infty}Z_n$. Both sets are in $\mathcal{M}$, and countable subadditivity gives $\mu(Z)\le\sum_n\mu(Z_n)=0$, hence $\mu(Z)=0$. Also $A\subseteq\bigcup_nE_n$ and $(\bigcup_nE_n)\setminus A\subseteq Z$. Thus $\bigcup_nE_n\in\overline{\mathcal{M}}$. This proves closure under countable unions, and hence the sigma-algebra claim.

Next we check that the definition of $\overline{\mu}$ does not depend on the representation. Suppose $E=A\cup N=A'\cup N'$, where $A,A'\in\mathcal{M}$, $N\subseteq Z$, $N'\subseteq Z'$, and $\mu(Z)=\mu(Z')=0$. If $x\in A\setminus A'$, then $x\in E$ but $x\notin A'$, so $x\in N'\subseteq Z'$. Thus $A\setminus A'\subseteq Z'$; by the same reasoning, $A'\setminus A\subseteq Z$. Both differences are measurable null sets. The measurable sets $A$ and $A'$ therefore have the same measure: each differs from $A\cap A'$ by a null set, so $\mu(A)=\mu(A\cap A')=\mu(A')$, including when these values are infinite.

To check countable additivity, let $E_n\in\overline{\mathcal{M}}$ be pairwise disjoint and choose measurable $A_n\subseteq E_n$ with $E_n\setminus A_n$ contained in a measurable null set. The sets $A_n$ are pairwise disjoint because the $E_n$ are. Their union $A=\bigcup_n A_n$ is measurable, and $(\bigcup_n E_n)\setminus A$ is contained in a countable union of measurable null sets. Consequently, $$ \overline{\mu}\left(\bigcup_{n=1}^{\infty}E_n\right) =\mu\left(\bigcup_{n=1}^{\infty}A_n\right) =\sum_{n=1}^{\infty}\mu(A_n) =\sum_{n=1}^{\infty}\overline{\mu}(E_n). $$ Here the middle equality is countable additivity of the original measure $\mu$. Also $\overline{\mu}(\varnothing)=0$, so $\overline{\mu}$ is a measure. For $A\in\mathcal{M}$, the representation $A=A\cup\varnothing$ gives $\overline{\mu}(A)=\mu(A)$.

Finally, let $E\in\overline{\mathcal{M}}$ be a completed-measurable null set, represented as $E=A\cup N$ with $\overline{\mu}(E)=\mu(A)=0$. For a measurable null set $Z$ containing $N$, the set $A\cup Z$ is measurable and has measure zero. Every subset $F\subseteq E$ is a subset of $A\cup Z$, so it belongs to $\overline{\mathcal{M}}$ by the definition of completion and has completed measure zero. This proves completeness. $\square$

Computing What the Completion Adds

Worked Example: Completing a Space Without Splitting a Positive Block

Let $X=\{a,b,c,d\}$, let $N=\{a,b\}$, and let $P=\{c,d\}$. Take $\mathcal{M}=\{\varnothing,N,P,X\}$, and define $\mu(N)=0$ and $\mu(P)=1$. Compute the completion.

The only measurable null sets are $\varnothing$ and $N$: the other two nonempty measurable sets $P$ and $X$ have measures $1$ and $1$. Thus subsets of $N$ may be added, but the definition does not allow splitting $P$. Every completed set has the form $S\cup T$, where $S\subseteq N$ and $T\in\{\varnothing,P\}$. There are $2^2\cdot2=8$ such sets. Explicitly, they are $\varnothing,\{a\},\{b\},N,P,\{a\}\cup P,\{b\}\cup P,X$. The completed measure is zero on the four sets with $T=\varnothing$, and one on the four with $T=P$.

Check: $E=\{a\}\cup P=\{a,c,d\}$ is represented by $A=P$ and $N'=\{a\}\subseteq N$, so $\overline{\mu}(E)=\mu(P)=1$. Its complement is $\{b\}$, which is contained in the measurable null set $N$ and has completed measure zero. The two complement measures sum to $1=\overline{\mu}(X)$.

Worked Example: A Completion That Is the Power Set

Let $X=\{a,b,c\}$, $\mathcal{M}=\{\varnothing,\{a,b\},\{c\},X\}$, with $\mu(\{a,b\})=0$ and $\mu(\{c\})=2$. Find the completion and its measure.

The measurable null sets are $\varnothing$ and $\{a,b\}$. Every subset of $\{a,b\}$ can therefore be adjoined. The other measurable block is the singleton $\{c\}$, whose subsets, $\varnothing$ and $\{c\}$, are already measurable. Hence every subset of $X$ is in $\overline{\mathcal{M}}$, so $\overline{\mathcal{M}}=\mathcal{P}(X)$. For any $E\subseteq X$, a representation is $E=A\cup S$, where $A=\{c\}$ if $c\in E$, and $A=\varnothing$ otherwise, while $S=E\cap\{a,b\}$. It follows that $\overline{\mu}(E)=2$ if $c\in E$, and $0$ if $c\notin E$.

Check: The disjoint sets $\{a\},\{b\},\{c\}$ have completed measures $0,0,2$, respectively. Their union is $X$, whose measure is $0+0+2=2$, agreeing with the original value $\mu(X)=\mu(\{a,b\})+\mu(\{c\})=0+2=2$.

Worked Example: When Completion Adds No Sets

Let $X=\{a,b,c\}$, $\mathcal{M}=\{\varnothing,\{a,b\},\{c\},X\}$, and define $\mu(E)$ to be the number of points in $E$ for $E\in\mathcal{M}$. Thus the four values are $0,2,1,3$. Determine whether completion enlarges $\mathcal{M}$.

Each nonempty member of $\mathcal{M}$ has positive measure: the values are $2$ for $\{a,b\}$, $1$ for $\{c\}$, and $3$ for $X$. So the only measurable null set is $\varnothing$, whose only subset is itself. The definition of completion adds no sets: $\overline{\mathcal{M}}=\mathcal{M}$, and $\overline{\mu}=\mu$.

Check: $\{a\}$ is a subset of the measurable set $\{a,b\}$, but it is not in $\mathcal{M}$. It is not added, because $\{a,b\}$ has measure $2$, not zero. Thus having no nonempty measurable null sets makes this space complete, without making its sigma-algebra the power set.

A Common Mistake: Adding Too Much

Completion means adjoining subsets of measurable null sets, not arbitrary subsets of measurable sets. In the first example, a tempting but incorrect argument is: “$\{c\}\subseteq P$, so $\{c\}$ belongs to the completion.” The premise that $\{c\}$ is a subset of $P$ is true, but $P$ has measure $1$, not zero. To check directly, the only members of the original sigma-algebra contained in $\{c\}$ are $\varnothing$; and $\{c\}$ is not contained in any measurable null set, since the only such sets are $\varnothing$ and $N$. Therefore $\{c\}\notin\overline{\mathcal{M}}$. By contrast, $\{a\}\subseteq N$ and $\mu(N)=0$, so $\{a\}$ is in the completion.

The theorem also shows why the assigned measure is forced: adjoining $N\subseteq Z$ with $\mu(Z)=0$ cannot change the measure of the measurable set $A$ in a representation $E=A\cup N$. The well-definedness argument ensures that choosing another representation cannot give a different value. Completion is therefore a controlled enlargement of the domain, not a change to the original measure.

Check Your Understanding

For each question, identify the measurable null set or sets involved before deciding whether completion adds a set.

Suppose $E=A\cup N$, where $A,Z\in\mathcal{M}$, $\mu(Z)=0$, and $N\subseteq Z$. Explain why $A\subseteq E\subseteq A\cup Z$ and why the two bounding sets differ by a null set.
In the first worked example, why is $\{c\}$ not in the completion, even though it is a subset of the measurable set $P$?
Give the definition of a complete measure space and use it to explain why the completion in the second worked example is complete.
In the third worked example, what property of the measurable null sets implies that the completion adds no sets?
In the countable-additivity proof, why are the measurable sets $A_n\subseteq E_n$ pairwise disjoint when the completed-measurable sets $E_n$ are pairwise disjoint?

Completing the Extension with Null Subsets

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