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Outer Measures and Caratheodory's Theorem · Tutorial 57 of 798

Finite Unions of Caratheodory Measurable Sets

Learn how repeated Caratheodory splitting proves finite-union closure without assuming that the whole space has finite outer measure.

Advanced 10 min read

What You'll Learn

  • State the Caratheodory splitting criterion for an outer measure.
  • Split a test set successively by two measurable sets.
  • Handle extended-real equalities without subtracting infinities.
  • Prove closure under arbitrary finite unions by induction.
  • Identify double-counting and infinity-subtraction errors.

Why Repeated Splitting Proves Union Closure

Caratheodory measurability is defined by how a set splits every test set: the outer measure before splitting must equal the sum of the outer measures on the two sides. To show that the union of two measurable sets is measurable, the natural strategy is to split first by one set and then split the part outside it by the other. The resulting pieces keep the overlap from being counted twice.

The argument must work for extended nonnegative values, including \(+\infty\). In particular, it cannot rely on subtracting one outer measure from another. We will establish the required equality using only successive splitting, subadditivity, and monotonicity, so no assumption that the outer measure of the whole space is finite is needed.

Definition: An outer measure on a set \(X\) is a function \(\mu^*:\mathcal{P}(X)\to[0,\infty]\) such that \(\mu^*(\varnothing)=0\), \(\mu^*(E)\le\mu^*(F)\) whenever \(E\subseteq F\), and \(\mu^*(\bigcup_{n=1}^\infty E_n)\le\sum_{n=1}^\infty\mu^*(E_n)\) for every sequence of subsets of \(X\). A set \(A\subseteq X\) is Caratheodory measurable if, for every \(T\subseteq X\), \(\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\cap A^c)\). All these quantities take values in \([0,\infty]\).

The outer-measure axioms give the two elementary inequalities used below. Monotonicity is part of the definition. Subadditivity for two sets follows by applying countable subadditivity to the sequence \(E,F,\varnothing,\varnothing,\ldots\), giving \(\mu^*(E\cup F)\le\mu^*(E)+\mu^*(F)\). These statements hold whether the values are finite or infinite.

The Two-Set Splitting Argument

Theorem: Let \(\mu^*\) be an outer measure on \(X\), and let \(A,B\subseteq X\) be Caratheodory measurable. Then \(A\cup B\) is Caratheodory measurable. No finiteness assumption on \(\mu^*(X)\), \(\mu^*(T)\), or the measures of the pieces is required.

Proof. Fix an arbitrary set \(T\subseteq X\). We must prove the splitting equality for \(A\cup B\). Since \(A\) is measurable, splitting \(T\) by \(A\) gives

$$ \mu^*(T)=\mu^*(T\cap A)+\mu^*(T\setminus A). $$

Now apply the measurability of \(B\) to the test set \(T\setminus A\). This gives

$$ \mu^*(T\setminus A) =\mu^*((T\setminus A)\cap B) +\mu^*((T\setminus A)\setminus B). $$

Define \(C=T\cap A\), \(D=(T\setminus A)\cap B\), and \(F=T\setminus(A\cup B)\). The first two sets form a disjoint partition of \(T\cap(A\cup B)\), while \(F=T\cap(A\cup B)^c\). Substituting the second splitting equality into the first therefore yields

$$ \mu^*(T)=\mu^*(C)+\mu^*(D)+\mu^*(F). $$

By subadditivity, \(\mu^*(T\cap(A\cup B))=\mu^*(C\cup D)\le\mu^*(C)+\mu^*(D)\). Addition is monotone on \([0,\infty]\), so the displayed equality implies \(\mu^*(T)\ge\mu^*(T\cap(A\cup B))+\mu^*(F)\). For the reverse inequality, \(T\) is the disjoint union of \(T\cap(A\cup B)\) and \(F\), and subadditivity gives \(\mu^*(T)\le\mu^*(T\cap(A\cup B))+\mu^*(F)\). The two inequalities give the required splitting equality. This argument uses no subtraction, including when one or more terms equal \(+\infty\). Since \(T\) was arbitrary, \(A\cup B\) is measurable. \(\square\)

The splitting order is important: after splitting by \(A\), only the remainder \(T\setminus A\) is split by \(B\). Thus the three pieces are disjoint, and their outer measures can be added without counting the overlap of \(A\) and \(B\) twice.

Worked Example: Overlapping Intervals

Take \(X=\mathbb{R}\) with Lebesgue outer measure \(m^*\), and let \(T=[0,4]\), \(A=[0,2]\), and \(B=[1,3]\). Intervals are Caratheodory measurable for Lebesgue outer measure. Apply the repeated splitting to \(T\).

First, \(T\cap A=[0,2]\) has outer measure \(2\), and \(T\setminus A=(2,4]\) has outer measure \(2\). Splitting the remainder by \(B\), we obtain \((2,4]\cap[1,3]=(2,3]\), of outer measure \(1\), and \((2,4]\setminus[1,3]=(3,4]\), also of outer measure \(1\). Thus the successive pieces have measures \(2,1,1\), and their sum is \(2+1+1=4=m^*(T)\).

The union is \(A\cup B=[0,3]\). Its intersection with \(T\) has measure \(3\), and \(T\setminus(A\cup B)=(3,4]\) has measure \(1\). Hence the union split gives \(3+1=4=m^*(T)\), as required. The overlap \([1,2]\) is included in the first piece \(T\cap A\); it is not counted again when the remainder is split by \(B\).

Why Infinite Outer Measure Causes No Obstruction

The equality in the definition is an equality of extended nonnegative numbers. It remains meaningful when a term is infinite. The proof above compares sums by monotonicity and subadditivity; it never tries to recover a finite quantity by subtracting \(+\infty\). This distinction matters on an infinite-measure space, where splitting an infinite set can produce an equality such as \(+\infty=2+(+\infty)\).

Worked Example: A Split of the Whole Real Line

Again take Lebesgue outer measure \(m^*\) on \(\mathbb{R}\), now with \(T=\mathbb{R}\), \(A=(0,1)\), and \(B=(1,2)\). These intervals are Caratheodory measurable. The first split gives \(T\cap A=(0,1)\), of measure \(1\), and \(T\setminus A=(-\infty,0]\cup[1,\infty)\), of infinite measure. Therefore \(m^*(T)=1+\infty=\infty\).

Within \(T\setminus A\), the part lying in \(B\) is \((1,2)\), of measure \(1\). The remainder is \((-\infty,0]\cup\{1\}\cup[2,\infty)\), of infinite measure. The successive split therefore gives \(m^*(T)=1+1+\infty=\infty\). The union \(A\cup B\) has measure \(2\), while its complement has infinite measure, so the final splitting equality is \(m^*(T)=2+\infty=\infty\).

Check: The complement is unbounded, so its Lebesgue measure is infinite; adding the two interval lengths gives \(1+1=2\). Both sides of the union splitting equality are therefore \(\infty\). No finite value is inferred by subtracting infinities.

Worked Example: Repeated Splitting with Counting Measure

Let \(X=\mathbb{N}\), and let \(\mu^*(E)\) be the counting measure, namely the number of elements of \(E\) if \(E\) is finite and \(+\infty\) otherwise. This is an outer measure, and every subset is Caratheodory measurable: for any \(T,A\subseteq\mathbb{N}\), the disjoint sets \(T\cap A\) and \(T\setminus A\) partition \(T\), so their counts add to the count of \(T\), with an infinite count allowed.

Use \(T=\{1,2,\ldots,12\}\), let \(A=\{2,4,6,8,10,12\}\), and let \(B=\{3,6,9,12\}\). Splitting first by \(A\) gives \(T\cap A=A\), with \(6\) elements, and \(T\setminus A=\{1,3,5,7,9,11\}\), also with \(6\) elements. The part of this remainder in \(B\) is \(\{3,9\}\), with \(2\) elements. The part outside \(B\) is \(\{1,5,7,11\}\), with \(4\) elements. The successive counts add to \(6+2+4=12\).

The union \(A\cup B=\{2,3,4,6,8,9,10,12\}\) has \(8\) elements, and its complement in \(T\) is \(\{1,5,7,11\}\), with \(4\) elements. The union splitting check is \(8+4=12\), agreeing with the original count of \(T\).

Finite Unions and Common Errors

Corollary: For an outer measure \(\mu^*\) on \(X\), every finite union of Caratheodory-measurable subsets of \(X\) is Caratheodory measurable.

Proof. The empty union is \(\varnothing\), which is measurable because \(\mu^*(T)=\mu^*(T)+0\) for every \(T\). A union of one measurable set is measurable. Suppose the union of \(n\) measurable sets is measurable. The two-set theorem applied to that union and one more measurable set shows that the union of \(n+1\) sets is measurable. Induction proves the assertion for every finite number of sets. \(\square\)

Common mistake: splitting the original set twice. In the counting example, \(A\) has \(6\) elements and \(B\) has \(4\) elements. The incorrect calculation \(6+4=10\) counts the shared elements \(6\) and \(12\) twice; the union actually has \(8\) elements. The correction is to split \(T\setminus A\) by \(B\): only \(B\setminus A=\{3,9\}\) contributes new elements, so \(6+2=8\).

Common mistake: subtracting infinite outer measures. From an equality such as \(\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\setminus A)\), one cannot subtract an infinite term to isolate the other term. For instance, the equality \(\infty=1+\infty\) does not permit the conclusion \(\infty-\infty=1\). The proof instead compares the desired pieces using subadditivity and monotonicity, both of which apply to extended values.

The method is useful because it proves a closure property directly from the defining criterion, without requiring a finite total outer measure. It also gives a reliable bookkeeping rule: split the part not yet assigned to the union, so the pieces are disjoint. Countable unions require additional arguments; the finite induction here makes no claim about them.

Check Your Understanding

Use the splitting criterion and keep track of which pieces are disjoint.

  1. For measurable \(A,B\subseteq X\), identify the three successive pieces obtained by splitting \(T\) first by \(A\) and then by \(B\).
  2. In the proof of the two-set theorem, where is subadditivity used to bound \(\mu^*(T\cap(A\cup B))\)?
  3. Why does the proof remain valid if \(\mu^*(T)=+\infty\)? Name the operations it uses instead of subtraction.
  4. For the counting-measure example, compute the number of elements in \(T\cap(A\cup B)\) and in its complement in \(T\).
  5. Explain why knowing \(\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\setminus A)\) does not justify subtracting \(\mu^*(T\setminus A)\) when it is infinite.

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