From Finite Splittings to a Countable Union
Caratheodory measurability is tested by asking whether every set splits exactly into its part inside a candidate set and its part outside. For a finite family of measurable sets, repeated splitting gives a finite-stage estimate. The countable case requires an additional step: compare the part remaining after finitely many stages with the part outside the full union, and use monotonicity of the outer measure. This argument works even when some of the outer measures are infinite.
Throughout, 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. No finiteness or sigma-finiteness assumption is imposed. All sets considered below are subsets of \(X\); there is not yet an ambient measure space of measurable sets.
The criterion must hold for every test set \(A\), not just for \(A=X\). For measurable \(E\), it can therefore be applied to a remainder such as \(A\setminus F\), where \(F\) is a union of sets already split off. That repeated use is the finite-stage estimate.
The Finite-Stage Estimate
Proof. We use induction on \(n\). For \(n=1\), the displayed inequality is equality by the definition of measurability of \(E_1\). Suppose the inequality holds for \(n-1\), and set \(F_{n-1}=\bigcup_{i=1}^{n-1}E_i\). The induction hypothesis gives $$ \mu^*(A)\geq\sum_{i=1}^{n-1}\mu^*(A\cap E_i)+\mu^*(A\setminus F_{n-1}). $$ Apply the measurability criterion for \(E_n\) to the test set \(A\setminus F_{n-1}\). Since \(E_n\) is disjoint from \(F_{n-1}\), its intersection with this test set is \(A\cap E_n\). Hence $$ \mu^*(A\setminus F_{n-1}) =\mu^*(A\cap E_n)+\mu^*\!\left(A\setminus\bigcup_{i=1}^{n}E_i\right). $$ Substitution proves the inequality for \(n\). All terms are nonnegative extended real numbers, so adding the finite sums introduces no subtraction or indeterminate expression. \(\square\)
In fact, this induction gives equality in the finite-stage estimate. We will only need the inequality. Its useful feature is that it keeps a remainder term, so that the same test set \(A\) can be used at every stage.
Countable Disjoint Unions Are Measurable
Proof. Fix an arbitrary test set \(A\subseteq X\). Define \(F_n=\bigcup_{i=1}^{n}E_i\) and \(S_n=\sum_{i=1}^{n}\mu^*(A\cap E_i)\). The finite-stage estimate gives $$ \mu^*(A)\geq S_n+\mu^*(A\setminus F_n). $$ Since \(E\) contains \(F_n\), taking complements within \(A\) gives \(A\setminus E\subseteq A\setminus F_n\). By monotonicity, $$ \mu^*(A\setminus E)\leq\mu^*(A\setminus F_n). $$ Consequently, for every \(n\), $$ \mu^*(A)\geq S_n+\mu^*(A\setminus E). $$ The sequence \(S_n\) is nondecreasing, and its limit is the extended nonnegative series \(S=\sum_{i=1}^{\infty}\mu^*(A\cap E_i)\). Passing to the limit in these numerical inequalities yields $$ \mu^*(A)\geq S+\mu^*(A\setminus E). $$ This passage is valid also when a term or the limit is infinite: if the left side is finite, the inequalities bound every partial sum and the fixed outside term by that finite value; if it is infinite, the displayed lower bound holds automatically.
Now \(A\cap E=\bigcup_{i=1}^{\infty}(A\cap E_i)\). Countable subadditivity gives \(\mu^*(A\cap E)\leq S\). Therefore $$ \mu^*(A)\geq\mu^*(A\cap E)+\mu^*(A\setminus E). $$ For the reverse inequality, \(A=(A\cap E)\cup(A\setminus E)\), so countable subadditivity (in particular, its two-set case) gives $$ \mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\setminus E). $$ The two inequalities establish the Caratheodory criterion for this \(A\). Since \(A\subseteq X\) was arbitrary, \(E\) is measurable. \(\square\)
The proof does not assert that the outer measure of the union is the sum of the outer measures of the sets; that is a separate countable-additivity statement for a measure. Here the conclusion is precisely the splitting criterion. Countable subadditivity supplies the upper bound for the part inside the union, while the finite-stage estimates and monotonicity supply the lower bound.
Worked Example: Counting Measure and a Finite Test Set
Let \(X=\mathbb{N}=\{1,2,\ldots\}\), and let \(\mu^*(B)\) be the counting outer measure: it is the number of elements of \(B\) if \(B\) is finite and \(+\infty\) if \(B\) is infinite. Set \(E_i=\{2i\}\) for \(i\geq1\), and test the union \(E\) with \(A=\{1,2,3,4,5\}\). The sets \(E_i\) are pairwise disjoint because distinct indices give distinct even integers.
Here \(A\cap E=\{2,4\}\), so \(\mu^*(A\cap E)=2\); also \(A\setminus E=\{1,3,5\}\), so \(\mu^*(A\setminus E)=3\). Thus the full splitting has right side \(2+3=5=\mu^*(A)\). At stage \(n=1\), the intersection contribution is \(1\), and the remainder \(A\setminus E_1=\{1,3,4,5\}\) has measure \(4\), giving \(1+4=5\). At stage \(n=2\), the two intersection contributions total \(2\), and the remainder \(\{1,3,5\}\) has measure \(3\), again giving \(2+3=5\). For \(n\geq3\), the additional intersections with \(A\) are empty, and the remainder still has measure \(3\).
Check: the count of the five elements of \(A\) is accounted for exactly: two lie in the union of the even singletons and three lie outside it. Every membership test here uses the sets as written; in particular, \(6\notin A\), so it contributes nothing.
Worked Example: Weighted Atoms and the Limit of the Remainder
On \(X=\mathbb{N}\), define \(\mu^*(B)=\sum_{k\in B}2^{-k}\), with a nonnegative series understood in \([0,\infty]\). This is an outer measure: the empty sum is zero, inclusion can only increase the sum, and a sum over a countable union is at most the sum of the separate sums. In fact, every subset \(B\) is Caratheodory measurable, since the series over \(B\) splits into its terms in \(B\cap C\) and \(B\setminus C\) for any \(C\subseteq\mathbb{N}\).
Consider the disjoint sets \(E_i=\{2i\}\), their union \(E\) of even positive integers, and \(A=\mathbb{N}\). The measure of the full set is $$ \mu^*(A)=\sum_{k=1}^{\infty}2^{-k}=1. $$ The sum of the weights on the even integers is $$ \mu^*(A\cap E)=\sum_{i=1}^{\infty}2^{-2i} =\sum_{i=1}^{\infty}4^{-i}=\frac{1/4}{1-1/4}=\frac13. $$ The outside set consists of the odd integers, and its weight is $$ \mu^*(A\setminus E)=\sum_{i=1}^{\infty}2^{-(2i-1)} =2\sum_{i=1}^{\infty}4^{-i}=2\cdot\frac13=\frac23. $$ Thus the two parts sum to \(1/3+2/3=1\).
At a finite stage \(n\), the contribution from the first \(n\) even singletons is \(\sum_{i=1}^n4^{-i}\). The remainder includes all odd integers and all even integers \(2i\) with \(i>n\), so its weight is $$ \frac23+\sum_{i=n+1}^{\infty}4^{-i} =\frac23+\frac{4^{-n}}{3}. $$ The finite-stage total is therefore $$ \sum_{i=1}^n4^{-i}+\frac23+\frac{4^{-n}}3 =\frac{1-4^{-n}}3+\frac23+\frac{4^{-n}}3=1. $$ For example, at \(n=2\), the first two weights total \(1/4+1/16=5/16\), and the remainder is \(2/3+1/48=11/16\); their sum is \(16/16=1\). In the limit, the tail \(4^{-n}/3\) tends to zero, leaving the outside weight \(2/3\), as required.
Worked Example: Why Overlapping Sets Cannot Be Counted as Separate Pieces
Take \(X=\{a\}\) with counting outer measure, and let \(E_i=\{a\}\) for every \(i\geq1\). Each \(E_i\) is measurable: for any \(A\subseteq X\), the two pieces \(A\cap E_i\) and \(A\setminus E_i\) are \(A\) and the empty set in some order, so their measures sum to \(\mu^*(A)\). But the sequence is not pairwise disjoint, since \(E_1\cap E_2=\{a\}\).
For \(A=\{a\}\), the union has measure \(1\), while \(\sum_{i=1}^{\infty}\mu^*(A\cap E_i)=\sum_{i=1}^{\infty}1=+\infty\). The union is measurable, but this sum is not its measure. The finite-stage splitting lemma cannot be applied to this sequence: the same point is being counted at every stage. This illustrates why pairwise disjointness is a hypothesis of the theorem, not a cosmetic condition.
A Monotonicity Pitfall
The crucial inclusion in the proof is \(A\setminus E\subseteq A\setminus F_n\), not the reverse. Thus monotonicity gives \(\mu^*(A\setminus E)\leq\mu^*(A\setminus F_n)\). Reversing this inequality would invalidate the step that replaces the finite-stage remainder by a fixed term before taking the limit.
The weighted example gives a concrete check. At \(n=1\), \(A\setminus F_1\) contains all odd integers and the even integers \(4,6,8,\ldots\), whereas \(A\setminus E\) contains only the odd integers. Their weights are \(3/4\) and \(2/3\), respectively: the first is larger by the remaining weight \(\sum_{i=2}^{\infty}4^{-i}=1/12\). The incorrect reversed inequality would say \(3/4\leq2/3\), which is false. The correct direction lets the finite-stage lower bound imply a lower bound involving the outside part of the full union.
The argument covers degenerate cases as well. If every \(E_i\) is empty, then \(E=\varnothing\) and the criterion reduces to \(\mu^*(A)=\mu^*(A)+0\). If some \(E_i=X\), pairwise disjointness forces every other \(E_j\) to be empty, and the union is \(X\), which satisfies the criterion. Infinite outer-measure values are permitted throughout; no step subtracts one infinite quantity from another.
Check Your Understanding
For each question, identify where the finite-stage estimate, monotonicity, or countable subadditivity is used.
- In the theorem’s proof, why does \(A\setminus E\subseteq A\setminus F_n\) hold for every \(n\)?
- Which direction of monotonicity converts the finite-stage remainder into a bound involving \(\mu^*(A\setminus E)\)?
- For counting outer measure on \(\mathbb{N}\), take \(E_i=\{i\}\) and \(A=\{1,2,3,4\}\). Compute the two parts of the Caratheodory split.
- In the weighted example, compute the first two even-atom weights and the weight of the odd atoms, then verify the full splitting.
- Why does the example with \(E_i=\{a\}\) for every \(i\) not contradict the theorem?