From Splitting to a Complete Measure
Caratheodory’s construction begins with an outer measure, which is defined on every subset of a set \(X\), but an outer measure need not be countably additive on all subsets. The splitting criterion identifies the subsets on which it behaves additively: a set is measurable when cutting any test set along it splits the outer measure exactly. The theorem proved here shows that these sets form a sigma-algebra, and that the outer measure restricted to this sigma-algebra is a complete measure. No finiteness or sigma-finiteness assumption is needed.
The values may be infinite, and all sets in this definition are arbitrary subsets, not necessarily members of a previously specified sigma-algebra. The splitting equality is meaningful even when one or both terms on the right are infinite. Countable subadditivity always gives \(\mu^*(T)\le\mu^*(T\cap E)+\mu^*(T\cap E^c)\); measurability requires the reverse inequality as well.
The Caratheodory Theorem
\(\varnothing\) satisfies the criterion because \(T\cap\varnothing=\varnothing\) and \(T\cap\varnothing^c=T\), so the required equality is \( \mu^*(T)=0+\mu^*(T)\). For any \(E\subseteq X\), the criterion for \(E^c\) is the same equality as for \(E\), with its two terms reversed. Hence \(E\in\mathcal{C}_{\mu^*}\) if and only if \(E^c\in\mathcal{C}_{\mu^*}\).
Suppose \(A,B\in\mathcal{C}_{\mu^*}\), and fix any \(T\subseteq X\). Split \(T\) first along \(A\), and then split each resulting piece along \(B\). Applying the criterion twice gives $$ \mu^*(T)=\mu^*(T\cap A\cap B)+\mu^*(T\cap A\cap B^c)+\mu^*(T\cap A^c\cap B)+\mu^*(T\cap A^c\cap B^c). $$ The last term is \(\mu^*(T\cap(A\cup B)^c)\). The first three terms sum to at least \(\mu^*(T\cap(A\cup B))\), by countable subadditivity applied to their union. Therefore \(\mu^*(T)\ge\mu^*(T\cap(A\cup B))+\mu^*(T\cap(A\cup B)^c)\). The reverse inequality follows from countable subadditivity, since those two sets partition \(T\). This proves \(A\cup B\in\mathcal{C}_{\mu^*}\). With complements, it also proves closure under finite intersections and set differences.
Let \(B_1,B_2,\ldots\) be pairwise disjoint members of \(\mathcal{C}_{\mu^*}\), let \(B=\bigcup_{i=1}^{\infty}B_i\), and fix \(T\subseteq X\). Repeatedly split along \(B_1,\ldots,B_N\). For every positive integer \(N\), this gives $$ \mu^*(T)=\sum_{i=1}^{N}\mu^*(T\cap B_i)+\mu^*\left(T\cap\left(\bigcup_{i=1}^{N}B_i\right)^c\right) \ge \sum_{i=1}^{N}\mu^*(T\cap B_i)+\mu^*(T\cap B^c). $$ The inequality uses monotonicity, since \(T\cap B^c\subseteq T\cap(\bigcup_{i=1}^{N}B_i)^c\). Taking the supremum over \(N\) yields \(\mu^*(T)\ge\sum_{i=1}^{\infty}\mu^*(T\cap B_i)+\mu^*(T\cap B^c)\), with extended values allowed. Countable subadditivity gives \(\mu^*(T\cap B)\le\sum_{i=1}^{\infty}\mu^*(T\cap B_i)\), so \(\mu^*(T)\ge\mu^*(T\cap B)+\mu^*(T\cap B^c)\). In the other direction, subadditivity applied to the partition \(T=(T\cap B)\cup(T\cap B^c)\) gives \(\mu^*(T)\le\mu^*(T\cap B)+\mu^*(T\cap B^c)\). Thus \(B\) is measurable.
For any sequence \(A_1,A_2,\ldots\) in \(\mathcal{C}_{\mu^*}\), define \(B_1=A_1\) and \(B_n=A_n\cap(\bigcup_{i=1}^{n-1}A_i)^c\) for \(n\ge2\). Finite-union and complement closure show that each \(B_n\) is measurable. These sets are pairwise disjoint and have the same union as the \(A_n\). Step 3 therefore proves that \(\bigcup_n A_n=\bigcup_n B_n\) is measurable. To prove additivity, take any pairwise disjoint measurable \(B_n\) and put \(B=\bigcup_nB_n\). The finite splitting identity with \(T=B\) gives \(\mu^*(B)\ge\sum_{i=1}^{N}\mu^*(B_i)\) for every \(N\), because the remaining term is nonnegative. Taking the supremum over \(N\) gives \(\mu^*(B)\ge\sum_{i=1}^{\infty}\mu^*(B_i)\). Countable subadditivity gives the reverse inequality. Hence the restriction is countably additive.
Suppose \(N\in\mathcal{C}_{\mu^*}\) and \(\mu^*(N)=0\). Let \(S\subseteq N\) be arbitrary and fix \(T\subseteq X\). Monotonicity gives \(\mu^*(T\cap S)\le\mu^*(N)=0\), so \(\mu^*(T\cap S)=0\). Also \(\mu^*(T\cap S^c)\le\mu^*(T)\). These inequalities give \(\mu^*(T\cap S)+\mu^*(T\cap S^c)\le\mu^*(T)\). Countable subadditivity applied to the partition of \(T\) gives the reverse inequality. Therefore \(S\) satisfies the splitting criterion and is measurable. Its measure is zero because \(\mu^*(S)\le\mu^*(N)=0\). This is precisely completeness of the restricted measure.
The argument used only the three outer-measure axioms. In particular, the partial sums in the disjoint-union argument were not assumed finite: their supremum is the extended nonnegative series, and if the left side is finite the inequalities bound every partial sum; if it is infinite the lower bound is automatic. The empty set and zero-measure cases are included throughout.
Three Finite Examples
Worked Example: Counting Outer Measure
Let \(X=\{a,b,c\}\) and define \(\mu^*(A)=|A|\) for every \(A\subseteq X\). This is an outer measure: the empty set has value zero, inclusion cannot decrease cardinality, and the cardinality of a union is at most the sum of the cardinalities. For any \(E,T\subseteq X\), the sets \(T\cap E\) and \(T\cap E^c\) are disjoint and their union is \(T\). Therefore $$ \mu^*(T)=|T|=|T\cap E|+|T\cap E^c| =\mu^*(T\cap E)+\mu^*(T\cap E^c). $$ Every subset is measurable, so \(\mathcal{C}_{\mu^*}=\mathcal{P}(X)\). For example, with \(T=\{a,b,c\}\) and \(E=\{a,c\}\), the split gives \(3=2+1\). The restricted measure is counting measure and is complete: its only zero-measure set is \(\varnothing\), whose only subset is itself. The theorem establishes the same conclusion for arbitrary \(X\), including infinite counting measure.
Check: the displayed split has left side \(|\{a,b,c\}|=3\), and right side \(|\{a,c\}|+|\{b\}|=2+1=3\).
Worked Example: A Zero-One Outer Measure
On \(X=\{a,b,c\}\), define \(\mu^*(A)=0\) if \(A=\varnothing\), and \(\mu^*(A)=1\) otherwise. It is monotone. For a sequence of sets, if their union is empty then every set in the sequence is empty and both sides of the subadditivity inequality are zero. If their union is nonempty, at least one set is nonempty, so the sum on the right is at least \(1\), while the union has outer measure \(1\). Thus this is an outer measure.
The sets \(\varnothing\) and \(X\) satisfy the splitting criterion: either split has one empty part and one part equal to \(T\). No nonempty proper subset \(E\) is measurable. Indeed, take \(T=X\). Then \(\mu^*(T)=1\), while both \(T\cap E\) and \(T\cap E^c\) are nonempty, giving $$ \mu^*(T\cap E)+\mu^*(T\cap E^c)=1+1=2\ne1. $$ Hence \(\mathcal{C}_{\mu^*}=\{\varnothing,X\}\). The restriction is a measure with \(\mu(X)=1\); it is complete because its only measure-zero set is \(\varnothing\).
Check: for \(E=\{a\}\), \(T=X\), the proposed split is \(1=1+1\), which is false. This concrete failure shows why an outer measure need not make every subset measurable.
Worked Example: A Nonempty Null Set and Its Subsets
Let \(X=\{u,v,z\}\) and define \(\mu^*(A)=|A\cap\{u,v\}|\). The empty set has value zero and the function is monotone. For every sequence \((A_n)\), the active points of \(\bigcup_n A_n\) are the union of the active points of the \(A_n\), so their number is at most \(\sum_n|A_n\cap\{u,v\}|\). This proves countable subadditivity, including when the sum is infinite.
For any \(E,T\subseteq X\), each of \(u,v\) that belongs to \(T\) belongs to exactly one of \(T\cap E\) and \(T\cap E^c\). The number of active points therefore splits exactly, so every \(E\subseteq X\) is measurable. The set \(N=\{z\}\) has measure \(|\{z\}\cap\{u,v\}|=0\), and its subsets are \(\varnothing\) and \(\{z\}\). Both are measurable and have measure zero. For instance, with \(T=\{u,z\}\) and \(E=\{z\}\), the split reads \(1=0+1\).
Check: \(\mu^*(T)=|\{u,z\}\cap\{u,v\}|=1\); \(\mu^*(T\cap E)=\mu^*(\{z\})=0\); and \(\mu^*(T\cap E^c)=\mu^*(\{u\})=1\). Thus the equality is \(1=0+1\).
What the Theorem Does—and a Common Mistake
The theorem separates two roles of the same set function. The outer measure \(\mu^*\) is defined on all subsets of \(X\), but the measure \(\mu\) is its restriction to \(\mathcal{C}_{\mu^*}\). Countable additivity is asserted for measurable sets, not for arbitrary subsets. Completeness is also a statement about this restricted measure: if \(N\) is measurable and \(\mu(N)=0\), then every subset of \(N\), without any prior measurability assumption, belongs to \(\mathcal{C}_{\mu^*}\).
Common mistake: assuming that a small outer measure makes every set measurable. In the zero-one example, a nonempty proper \(E\) has \(\mu^*(E)=1\), and its complement also has outer measure \(1\). Testing only the values of \(E\) and \(E^c\), or assuming their union has measure \(1\), misses the required condition: the split must work for every \(T\). With \(T=X\), the incorrect calculation would claim \(1=1\) by looking only at the whole space; the actual split is \(1=1+1=2\), so \(E\) is not measurable. Always test the defining equality on the test set \(T\), not just on \(X\).
Check Your Understanding
In each question, use the splitting criterion and keep track of which function is defined on all subsets and which is restricted to the measurable sigma-algebra.
- For an arbitrary outer measure \(\mu^*\), explain why the measurability criterion for \(E\) is equivalent to the criterion for \(E^c\).
- In the zero-one example, test \(E=\{a,c\}\) using \(T=X\). Is \(E\) measurable? Show the two sides of the criterion.
- Why does the proof for a disjoint countable union use the supremum of finite partial sums, rather than assuming the series of outer measures is finite?
- Suppose \(N\in\mathcal{C}_{\mu^*}\) and \(\mu^*(N)=0\). Identify the inequalities that prove an arbitrary subset \(S\subseteq N\) is measurable.
- In the active-point example on \(\{u,v,z\}\), compute the splitting equality for \(T=\{v,z\}\) and \(E=\{z\}\).