Changing a Set on an Outer-Null Set
A measurable set can be altered on a set of measure zero without changing its measure. For a Caratheodory measure, this principle has a particularly strong form: the alteration need only be contained in an outer-null set, and the altered set is measurable even if the alteration itself was chosen arbitrarily. No finiteness or sigma-finiteness assumption is needed.
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. Write \(\mathcal{M}^*\) for its Caratheodory measurable sets: \(E\subseteq X\) belongs to \(\mathcal{M}^*\) when it splits every test set \(T\subseteq X\) according to the criterion below. As established in the earlier material on Caratheodory's theorem, \(\mathcal{M}^*\) is a sigma-algebra and \(\mu=\mu^*|_{\mathcal{M}^*}\) is a measure.
The key observation is that outer measure cannot distinguish two sets whose symmetric difference is contained in an outer-null set. This observation will let us compare the two pieces in the splitting criterion separately, without subtracting possibly infinite quantities.
Two Facts About Outer-Null Sets
Proof. Since \(C\subseteq D\cup Z\), monotonicity and countable subadditivity give \(\mu^*(C)\le\mu^*(D\cup Z)\le\mu^*(D)+\mu^*(Z)=\mu^*(D)\). The inclusion \(D\subseteq C\cup Z\) gives \(\mu^*(D)\le\mu^*(C)+\mu^*(Z)=\mu^*(C)\). The two inequalities prove equality, including when one or both values are infinite. \(\square\)
Proof. Monotonicity gives \(\mu^*(S)=0\), and also \(\mu^*(T\cap S)=0\) for every \(T\subseteq X\). Since \(T\subseteq (T\setminus S)\cup S\), countable subadditivity gives \(\mu^*(T)\le\mu^*(T\setminus S)+\mu^*(S)=\mu^*(T\setminus S)\). The reverse inequality follows from \(T\setminus S\subseteq T\) and monotonicity. Therefore \(\mu^*(T)=\mu^*(T\setminus S)\), and hence \(\mu^*(T)=\mu^*(T\cap S)+\mu^*(T\setminus S)\). This holds for every \(T\), so \(S\) is Caratheodory measurable. \(\square\)
The proposition includes the important case where \(Z\) itself is not initially known to be measurable: the splitting criterion proves that \(Z\), and indeed every subset of \(Z\), is measurable. This is stronger than merely saying that measurable null sets have measurable subsets.
The Measurable Modification Theorem
Proof. Fix an arbitrary test set \(T\subseteq X\). The sets \(T\cap A\) and \(T\cap B\) have symmetric difference contained in \(A\triangle B\), hence in \(Z\). The sets \(T\setminus A\) and \(T\setminus B\) also have symmetric difference contained in \(Z\). The lemma therefore gives \(\mu^*(T\cap A)=\mu^*(T\cap B)\) and \(\mu^*(T\setminus A)=\mu^*(T\setminus B)\). Because \(A\) is measurable, its splitting criterion says \(\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\setminus A)\). Substituting the two equalities yields \(\mu^*(T)=\mu^*(T\cap B)+\mu^*(T\setminus B)\). Since \(T\) was arbitrary, \(B\) is measurable. Applying the lemma to \(A,B\) themselves gives \(\mu^*(A)=\mu^*(B)\). For the stated special case, take \(B=A\triangle S\); then \(A\triangle B=S\subseteq Z\). \(\square\)
There are no hidden finiteness conditions in this proof. Each comparison is an equality of outer measures obtained from inclusions and subadditivity. In particular, the argument remains valid when a test set has infinite outer measure.
Worked Example: Removing a Countable Set from an Interval
Take \(X=\mathbb{R}\), \(\mu^*=m^*\), Lebesgue outer measure, and \(A=[0,1]\). Let \(S=\mathbb{Q}\cap[0,1]\), and set \(B=A\setminus S=A\triangle S\). We compute the outer measure of \(S\) first.
Enumerate \(S\) as \(q_1,q_2,\ldots\). Given any \(\varepsilon>0\), cover \(q_k\) by an open interval of length \(\varepsilon 2^{-k}\). These intervals cover \(S\), and their total length is \(\sum_{k=1}^{\infty}\varepsilon 2^{-k}=\varepsilon\sum_{k=1}^{\infty}2^{-k}=\varepsilon\). Thus \(m^*(S)\le\varepsilon\) for every \(\varepsilon>0\), so \(m^*(S)=0\).
The theorem now shows that \(B\) is Lebesgue measurable and \(m(B)=m(A)\). Since \(m([0,1])=1\), we obtain \(m(B)=1\). Check: \(S\) is countable and null, while \(A\) is the disjoint union of \(B\) and \(S\); countable additivity gives \(1=m(A)=m(B)+m(S)=m(B)+0\), again yielding \(m(B)=1\).
Worked Example: A Modification for a Dirac Outer Measure
Let \(X=\{p,q,r\}\) and define \(\mu^*(C)=1\) if \(p\in C\), and \(\mu^*(C)=0\) if \(p\notin C\), for every \(C\subseteq X\). This is an outer measure: it vanishes on the empty set and is monotone; for countable subadditivity, if a union contains \(p\), at least one set in the union contains \(p\), so the sum of their outer measures is at least \(1\), and if the union omits \(p\), its outer measure is \(0\).
Set \(A=\{p,q\}\), \(Z=\{q,r\}\), and \(S=\{r\}\). Since \(p\notin Z\), \(\mu^*(Z)=0\), and \(S\subseteq Z\). The modified set is \(B=A\triangle S=\{p,q,r\}\), because \(S\) is disjoint from \(A\). Direct computation gives \(\mu^*(A)=1\) and \(\mu^*(B)=1\). Both sets are measurable: for any \(T\subseteq X\), if \(p\in T\), exactly one of \(T\cap A\) and \(T\setminus A\) contains \(p\), so the two outer measures sum to \(1=\mu^*(T)\); if \(p\notin T\), both sides of the splitting criterion are \(0\). The same argument applies to \(B\). Check: \(A\triangle B=\{r\}\subseteq Z\), and both sets contain \(p\), so both have outer measure \(1\), as the theorem predicts.
Worked Example: Adding Null Points Between Two Intervals
Again use Lebesgue outer measure on \(\mathbb{R}\). Let \(A=[0,1]\cup[2,3]\) and \(S=\mathbb{Q}\cap[1,2]\). The same interval-cover argument as in the first example, applied to an enumeration of \(S\), gives \(m^*(S)=0\). The symmetric difference removes the rational endpoints \(1\) and \(2\) from \(A\), and adds the rational points in \((1,2)\). Thus \(B=A\triangle S=[0,1)\cup(\mathbb{Q}\cap(1,2))\cup(2,3]\).
The theorem gives measurability and \(m(B)=m(A)\). Direct computation checks the result: the three displayed pieces are disjoint measurable sets with measures \(1\), \(0\), and \(1\), respectively, so \(m(B)=1+0+1=2\). Also \(m(A)=m([0,1])+m([2,3])=1+1=2\). The endpoint changes do not affect the length, and the rational points added between the intervals have outer measure zero.
Completeness and a Common Pitfall
The same argument proves that the induced measure \((X,\mathcal{M}^*,\mu)\) is complete. Completeness means that whenever \(E\in\mathcal{M}^*\), \(\mu(E)=0\), and \(F\subseteq E\), then \(F\in\mathcal{M}^*\) and \(\mu(F)=0\). Indeed, \(\mu^*(E)=\mu(E)=0\), so the proposition applies to \(F\subseteq E\). It gives \(F\in\mathcal{M}^*\) and \(\mu(F)=\mu^*(F)=0\). This conclusion concerns subsets of measurable null sets, and the proposition gives the stronger fact that any subset of any outer-null set is measurable.
Common pitfall: subtracting infinite terms in the splitting criterion. For Lebesgue outer measure, take \(A=\mathbb{R}\) and \(B=\mathbb{R}\setminus\{0\}\). Both are measurable and differ only by a null set. With the test set \(T=\mathbb{R}\), the splitting identity for \(A\) reads \(m^*(\mathbb{R})=m^*(\mathbb{R})+m^*(\varnothing)=\infty+\!0\). It is invalid to subtract the first \(\infty\) from both sides: subtraction of \(\infty\) from \(\infty\) is undefined. The correction is to compare the corresponding terms in the two splitting identities. The lemma gives equality of the intersection terms and equality of the complement terms, even if a term is infinite; substituting those equalities proves the criterion for the modified set without any subtraction.
Check Your Understanding
Use the splitting criterion and the null-set lemma to justify each answer.
- If \(S\subseteq Z\) and \(\mu^*(Z)=0\), show directly that \(\mu^*(T\setminus S)=\mu^*(T)\) for every \(T\subseteq X\).
- Suppose \(A\in\mathcal{M}^*\) and \(A\triangle B\subseteq Z\), where \(\mu^*(Z)=0\). Which two pairs of sets must be compared to prove that \(B\) is measurable?
- For Lebesgue outer measure, why does covering the \(k\)-th point of an enumeration of \(\mathbb{Q}\cap[0,1]\) with an interval of length \(\varepsilon 2^{-k}\) prove that the set is outer-null?
- In the Dirac outer-measure example, compute the outer measures of \(\{q,r\}\) and \(\{p,r\}\). Which one is null, and why?
- Explain why the proof of measure invariance does not require subtracting terms from a Caratheodory splitting identity.