Exact Splitting and the Sigma-Algebra Goal
Caratheodory measurability is defined by how a set splits the outer measure of every test set. The key consequence is structural: although an outer measure is defined on every subset of a space, the subsets that split it exactly form a sigma-algebra. The countable-union step needs care. An arbitrary outer measure need not be continuous from below, so the proof uses finite-stage splitting and countable subadditivity instead.
Throughout, \(X\) is a set and \(\mu^*:\mathcal{P}(X)\to[0,\infty]\) is an outer measure. This means \(\mu^*(\varnothing)=0\), \(\mu^*\) is monotone, and it is countably subadditive. No finiteness or sigma-finiteness assumption is made. All sets considered are subsets of \(X\). The two properties used repeatedly are, for \(B,C\subseteq X\), \(\mu^*(B)\le\mu^*(C)\) if \(B\subseteq C\), and \(\mu^*(B\cup C)\le\mu^*(B)+\mu^*(C)\).
The criterion is required for every test set \(A\), not merely for measurable test sets. The values may be infinite; sums and equalities below are interpreted in \([0,\infty]\). In particular, no subtraction of infinite quantities will be used.
The Empty Set, Complements, and Finite Unions
The empty set and its complement satisfy the criterion directly. Complement closure follows from the symmetry of the two terms. For the countable-union argument, we also need finite unions, so we establish that closure explicitly.
Proof. For every \(A\subseteq X\), the splitting expression for \(\varnothing\) is \(\mu^*(A\cap\varnothing)+\mu^*(A\cap X)=0+\mu^*(A)=\mu^*(A)\). For \(X\), it is \(\mu^*(A)+\mu^*(\varnothing)=\mu^*(A)\). Thus both sets are measurable. If \(E\in\mathcal{C}\), then \(E^c\cap A=A\cap E^c\) and \((E^c)^c=E\), so the defining equality for \(E^c\) is exactly the defining equality for \(E\), with its two summands reversed. Hence \(E^c\in\mathcal{C}\). \(\square\)
For finite unions, let \(E,F\in\mathcal{C}\) and put \(U=E\cup F\). Splitting first by \(E\) and then splitting the part outside \(E\) by \(F\) gives, for every \(A\subseteq X\),
Splitting \(A\cap U\) by \(E\) gives \(\mu^*(A\cap U)=\mu^*(A\cap E)+\mu^*(A\cap E^c\cap F)\), while \(A\cap U^c=A\cap E^c\cap F^c\). Substitution proves the Caratheodory criterion for \(U\). Therefore \(\mathcal{C}\) is closed under unions of two sets, and induction gives closure under finite unions. Complements and finite unions also give finite intersections and differences: for example, \(E\cap F=(E^c\cup F^c)^c\), and \(E\setminus F=E\cap F^c\).
Worked Example: An Outer Measure with Only Trivial Measurable Sets
Let \(X=\{a,b,c\}\), and define \(\mu^*(S)=0\) if \(S=\varnothing\) and \(\mu^*(S)=1\) otherwise. This is an outer measure: it is monotone, vanishes at the empty set, and for any countable family, if its union is nonempty then at least one member is nonempty, so the sum of the outer measures is at least \(1\).
The sets \(\varnothing\) and \(X\) are measurable by the proposition. To check any other candidate \(E\), it is nonempty and proper. Choose \(a_0\in E\) and \(b_0\in E^c\), and take the test set \(A=\{a_0,b_0\}\). Then \(\mu^*(A)=1\), but \(A\cap E\) and \(A\cap E^c\) are both nonempty, so their outer measures sum to \(1+1=2\). The criterion fails. Thus \(\mathcal{C}=\{\varnothing,X\}\), which is a sigma-algebra.
Check: for \(E=\{a\}\), the particular test \(A=X\) gives \(\mu^*(X)=1\), whereas \(\mu^*(X\cap E)+\mu^*(X\cap E^c)=1+1=2\), confirming failure directly.
Worked Example: Weighted Counting Outer Measure
Let \(X=\{a,b,c\}\), assign weights \(w(a)=2\), \(w(b)=3\), and \(w(c)=5\), and set \(\mu^*(S)=\sum_{x\in S}w(x)\). This is an outer measure: the empty sum is zero, inclusion can only add nonnegative weights, and for any countable family \((S_j)\), each point in the union is counted at least once in \(\sum_j\mu^*(S_j)\). Hence \(\mu^*(\bigcup_jS_j)\le\sum_j\mu^*(S_j)\).
Take \(E=\{b,c\}\) and test it with \(A=\{a,b\}\). The left side is \(\mu^*(A)=2+3=5\). The two pieces are \(A\cap E=\{b\}\), of outer measure \(3\), and \(A\cap E^c=\{a\}\), of outer measure \(2\). Their sum is \(3+2=5\). More generally, for every \(A\subseteq X\), \(A\cap E\) and \(A\cap E^c\) partition \(A\), and each point’s weight occurs in exactly one piece. Therefore every subset \(E\subseteq X\) is measurable in this example.
Check: for \(A=\{b,c\}\), the left side is \(3+5=8\); the split gives \(8+0=8\). This also checks the case where the test set lies entirely inside \(E\).
Countable Unions: The Finite-Stage Estimate
Let \((E_n)_{n\ge1}\) be a sequence in \(\mathcal{C}\). Its members need not be disjoint. Define \(D_1=E_1\) and \(D_n=E_n\setminus\bigcup_{i=1}^{n-1}E_i\) for \(n\ge2\). The finite union and difference properties proved above show that each \(D_n\) is measurable. The \(D_n\) are pairwise disjoint, and their union is \(U=\bigcup_{n=1}^{\infty}E_n\): a point in \(U\) belongs to the first \(E_n\) that contains it, and therefore belongs to that \(D_n\).
Proof. For \(N=1\), this is the defining splitting equality for \(D_1\). Suppose the formula holds for \(N\). Apply the defining equality for \(D_{N+1}\) to the last set \(B=A\setminus\bigcup_{i=1}^{N}D_i\). Since the \(D_i\) are disjoint, \(B\cap D_{N+1}=A\cap D_{N+1}\), and \(B\cap D_{N+1}^c=A\setminus\bigcup_{i=1}^{N+1}D_i\). Replacing \(\mu^*(B)\) by the sum of these two outer measures proves the formula for \(N+1\). Induction proves it for every \(N\). \(\square\)
Proof. The empty set belongs to \(\mathcal{C}\), and \(\mathcal{C}\) is closed under complements, as already proved. It remains to prove closure under countable unions. Take \(E_n\in\mathcal{C}\), form the disjoint measurable sets \(D_n\) above, and write \(U=\bigcup_nD_n=\bigcup_nE_n\). Fix an arbitrary test set \(A\subseteq X\), and put \(R=A\setminus U\).
For each \(N\), the finite-stage estimate and the inclusion \(R\subseteq A\setminus\bigcup_{i=1}^{N}D_i\) give $$ \mu^*(A)\ge \sum_{i=1}^{N}\mu^*(A\cap D_i)+\mu^*(R). $$ The partial sums increase to \(S=\sum_{i=1}^{\infty}\mu^*(A\cap D_i)\), where the series is allowed to be \(+\infty\). Taking their supremum yields \(\mu^*(A)\ge S+\mu^*(R)\). Countable subadditivity applied to \(A\cap U=\bigcup_i(A\cap D_i)\) gives \(\mu^*(A\cap U)\le S\). Consequently, $$ \mu^*(A)\ge\mu^*(A\cap U)+\mu^*(R). $$ On the other hand, \(A=(A\cap U)\cup R\), so countable subadditivity (in particular, its two-set case) gives the reverse inequality \(\mu^*(A)\le\mu^*(A\cap U)+\mu^*(R)\). The two inequalities prove the exact splitting equality for \(U\). Since \(A\) was arbitrary, \(U\in\mathcal{C}\). Thus \(\mathcal{C}\) contains the empty set and is closed under complements and countable unions; it is a sigma-algebra. \(\square\)
Why the Countable-Union Argument Is Careful
A tempting but invalid shortcut is to assert that \(\mu^*(A\cap U)\) is the limit of the outer measures of the finite-stage pieces. Continuity from below is a property of measures, not a general axiom for outer measures. The proof above instead bounds \(\mu^*(A\cap U)\) by the sum of the disjoint pieces using countable subadditivity, while the finite-stage splitting estimate supplies the lower bound needed for exact splitting.
Worked Example: An Outer Measure That Is Not Continuous from Below
On \(X=\mathbb{N}\), define \(\mu^*(S)=0\) for \(S=\varnothing\), \(\mu^*(S)=1\) for nonempty finite \(S\), and \(\mu^*(S)=2\) for infinite \(S\). This is an outer measure. It is monotone. For countable subadditivity, if the union is finite and nonempty, at least one set is nonempty and the sum is at least \(1\); if the union is infinite and at least one set is infinite, the sum is at least \(2\); if the union is infinite and every set is finite, infinitely many sets must be nonempty, so the sum is \(+\infty\). These cases verify the required inequality, and the empty-union case has both sides zero.
Let \(A_n=\{1,\ldots,n\}\). Then \(A_n\subseteq A_{n+1}\), every \(A_n\) is nonempty and finite, and \(\mu^*(A_n)=1\) for every \(n\). But \(\bigcup_nA_n=\mathbb{N}\), so \(\mu^*(\bigcup_nA_n)=2\), not \(\lim_n\mu^*(A_n)=1\). This example shows the specific mistake in replacing the finite-stage outer measures by the outer measure of their union. The sigma-algebra proof avoids that replacement.
Check: the calculation uses the strict distinction in the definition: each \(A_n\) is finite, while its union is infinite. The values are therefore \(1\) at every finite stage and \(2\) at the union.
The sigma-algebra result identifies the sets on which an outer measure has exact splitting behavior. It does not yet assert that restricting the outer measure to this collection gives a countably additive measure; that is a separate property. For this structural step, the essential ingredients are the exact splitting criterion, finite-stage splitting, and countable subadditivity.
Check Your Understanding
Use the exact-splitting definition and keep arbitrary test sets in view.
- Verify directly that \(X\) is Caratheodory measurable for any outer measure \(\mu^*\).
- In the finite-union proof, identify the three disjoint pieces of \(A\) and state which one is \(A\cap(E\cup F)^c\).
- Why does the finite-stage splitting estimate remain valid if one of its terms has outer measure \(+\infty\)?
- In the countable-union proof, which inequality uses countable subadditivity to bound \(\mu^*(A\cap U)\)?
- For the outer measure in the final example, compute \(\mu^*(\{1,2\})\) and \(\mu^*(\mathbb{N})\), and explain why these values disprove continuity from below for \((A_n)\).