From Repeated Splitting to a Finite-Stage Bound
Caratheodory measurability gives an exact rule for splitting any test set by one measurable set. When several measurable sets are pairwise disjoint, the rule can be applied successively: at each stage, one intersection is separated from what remains. The resulting identity records all the pieces separated so far and the remainder still present. Replacing that finite remainder by the part outside the entire family gives the finite-stage lower bound.
Throughout, let \(\mu^*\) be an outer measure on a set \(X\). Thus \(\mu^*: \mathcal{P}(X)\to[0,\infty]\), \(\mu^*(\varnothing)=0\), \(\mu^*\) is monotone, and it is countably subadditive. The values may be infinite. A set \(E\subseteq X\) is Caratheodory measurable if, for every \(A\subseteq X\),
We use this criterion as the definition of measurability here. In particular, it applies to every subset \(A\), not only to measurable test sets. An outer measure is not assumed to be a measure on all of \(\mathcal{P}(X)\); we do not assume that an arbitrary union of measurable sets is measurable. The argument below only uses finitely many measurable sets at a time.
Proof. We prove the statement by induction on \(n\). For \(n=1\), it is precisely the Caratheodory criterion applied to \(A\) and \(E_1\). Suppose the identity holds for \(n\) pairwise disjoint measurable sets. Set \(B=A\setminus\bigcup_{i=1}^{n}E_i\). Apply the criterion for \(E_{n+1}\) to the set \(B\): \(\mu^*(B)=\mu^*(B\cap E_{n+1})+\mu^*(B\setminus E_{n+1})\). Pairwise disjointness implies \(B\cap E_{n+1}=A\cap E_{n+1}\): a point of \(E_{n+1}\) cannot belong to any of \(E_1,\ldots,E_n\). Also \(B\setminus E_{n+1}=A\setminus\bigcup_{i=1}^{n+1}E_i\). Substitute this expression for \(\mu^*(B)\) in the induction hypothesis to obtain the claimed identity for \(n+1\). This substitution is valid for extended nonnegative values, since addition is associative in \([0,\infty]\). The induction proves the identity for every finite \(n\). \(\square\)
The theorem gives an equality, which is stronger than the lower bound often needed. Its final term is the remainder after removing just the first \(n\) sets. If the family continues, the part of \(A\) outside every set is contained in that finite-stage remainder. Monotonicity of the outer measure therefore gives the following estimate.
Proof. Write \(R_n=A\setminus\bigcup_{i=1}^{n}E_i\) and \(R=A\setminus\bigcup_{i=1}^{\infty}E_i\). If \(x\in R\), then \(x\in A\) and \(x\notin E_i\) for every \(i\), so \(x\in R_n\). Thus \(R\subseteq R_n\), and monotonicity gives \(\mu^*(R)\leq\mu^*(R_n)\). The finite-stage identity yields \(\mu^*(A)=\sum_{i=1}^{n}\mu^*(A\cap E_i)+\mu^*(R_n)\). Replacing the last term by the no-larger value \(\mu^*(R)\) proves the estimate. This argument uses only the inclusion \(R\subseteq R_n\); it does not apply a measure property to the infinite union. \(\square\)
Worked Calculations
Worked Example: Weighted Counting Measure on a Finite Set
Let \(X=\{a,b,c,d,e\}\), and define \(\mu^*(S)=\sum_{x\in S}w(x)\) for every \(S\subseteq X\), with weights \(w(a)=2\), \(w(b)=3\), \(w(c)=5\), \(w(d)=7\), and \(w(e)=11\). This is a measure, hence also an outer measure; every subset is Caratheodory measurable. Take \(A=\{a,b,c,d\}\), \(E_1=\{a,c\}\), \(E_2=\{b\}\), and \(E_3=\{d,e\}\). These three sets are pairwise disjoint.
For the first stage, \(A\cap E_1=\{a,c\}\), with outer measure \(2+5=7\). The remainder is \(A\setminus E_1=\{b,d\}\), with outer measure \(3+7=10\). The identity reads \(17=7+10\), since \(\mu^*(A)=2+3+5+7=17\).
At stage two, \(A\cap E_2=\{b\}\) has outer measure \(3\), and \(A\setminus(E_1\cup E_2)=\{d\}\) has outer measure \(7\). The sum is \(7+3+7=17\). At stage three, \(A\cap E_3=\{d\}\) has outer measure \(7\), and the remainder is empty, of outer measure \(0\). The sum is \(7+3+7+0=17\). Check: each point of \(A\) occurs in exactly one of the listed intersections or the final remainder, so the total weight is \(2+5+3+7=17\), the original value \(\mu^*(A)\).
Worked Example: Lebesgue Outer Measure on the Real Line
Take Lebesgue outer measure \(m^*\) on \(\mathbb{R}\), \(A=[0,4]\), \(E_1=[0,1]\), \(E_2=(1,2]\), and \(E_3=(3,4]\). These are pairwise disjoint Borel sets, so they are Caratheodory measurable for \(m^*\). Their endpoints do not affect their Lebesgue measures.
The test-set measure is \(m^*(A)=4\). At stage one, \(A\cap E_1=[0,1]\) has measure \(1\), and the remainder \(A\setminus E_1=(1,4]\) has measure \(3\); the identity gives \(4=1+3\). At stage two, the two intersections have measures \(1\) and \(1\), while \(A\setminus(E_1\cup E_2)=(2,4]\) has measure \(2\). Thus \(4=1+1+2\). At stage three, \(A\cap E_3=(3,4]\) has measure \(1\), and the remainder is \((2,3]\), also of measure \(1\). The identity gives \(4=1+1+1+1\).
For the full-remainder estimate at stage two, \(A\setminus(E_1\cup E_2\cup E_3)=(2,3]\), so the right-hand side is \(1+1+1=3\), and \(4\geq3\). Check: the finite-stage remainder \((2,4]\) has measure \(2\), while the full remainder \((2,3]\) has measure \(1\); replacing \(2\) by \(1\) lowers the right-hand side from \(4\) to \(3\), as monotonicity requires.
Worked Example: Why Pairwise Disjointness Matters
On \(X=\{a,b,c\}\), use ordinary counting measure, so \(\mu^*(S)\) is the number of points of \(S\). Let \(A=\{a,b,c\}\), \(E_1=\{a,b\}\), and \(E_2=\{b,c\}\). Both sets are measurable, but they are not disjoint: their intersection is \(\{b\}\).
A tempting but incorrect calculation is \(\mu^*(A)\geq\mu^*(A\cap E_1)+\mu^*(A\cap E_2)\). The left side is \(3\), whereas the right side is \(2+2=4\), so the claimed inequality fails. The error is that the point \(b\) has been counted twice. The finite-stage theorem cannot be applied to this overlapping pair.
For a valid splitting, replace \(E_2\) by \(F_2=E_2\setminus E_1=\{c\}\). Now \(E_1\) and \(F_2\) are disjoint measurable sets. The intersections with \(A\) have measures \(2\) and \(1\), and the remainder \(A\setminus(E_1\cup F_2)\) is empty, of measure \(0\). The identity is \(3=2+1+0\). Check: the corrected pieces contain \(a,b\) in the first piece and \(c\) in the second, so every point of \(A\) is counted once and the total is \(3\).
What the Estimate Does—and Does Not—Say
The finite-stage bound is useful because it keeps two kinds of information at once: the outer measures of the first \(n\) intersections and the portion of \(A\) that has not been captured by any set in the family. The exact identity uses the remainder after the first \(n\) sets; the estimate retains only the smaller remainder outside the whole family. For every fixed finite \(n\), this comparison is justified by set inclusion and monotonicity alone.
There are several limits to keep in view. First, disjointness is essential for identifying the new piece at each induction step with \(A\cap E_{n+1}\). Second, the estimate is a statement about an outer measure, not a claim that the infinite union is measurable or that the outer measure is countably additive on it. Third, no finiteness assumption is imposed: if one of the terms is \(+\infty\), the identities and inequalities are interpreted in the extended nonnegative reals. The proof never subtracts infinite quantities. The empty finite family, if considered, has empty sum \(0\) and remainder \(A\), giving the identity \(\mu^*(A)=0+\mu^*(A)\).
A reliable way to use the result is to keep the finite identity intact until the last step. Establish the intersections using pairwise disjointness, identify the exact finite remainder, and only then compare that remainder with the desired set by inclusion. This prevents an unjustified appeal to properties of an infinite union and makes the direction of the inequality transparent.
Check Your Understanding
In each question, distinguish the exact finite-stage identity from the estimate that uses the remainder outside the whole family.
- State the finite-stage splitting identity for three pairwise disjoint Caratheodory-measurable sets and an arbitrary test set \(A\subseteq X\).
- In the proof of the induction step, why is \((A\setminus(E_1\cup E_2))\cap E_3=A\cap E_3\) when the three sets are pairwise disjoint?
- Explain why \(A\setminus\bigcup_{i\geq1}E_i\subseteq A\setminus\bigcup_{i=1}^{n}E_i\), and state which outer-measure property turns this inclusion into the desired comparison.
- For counting measure on \(X=\{a,b,c\}\), let \(A=X\), \(E_1=\{a\}\), and \(E_2=\{b,c\}\). Compute both intersection measures and the finite-stage remainder, then verify the identity.
- Give an example of overlapping measurable sets for which the sum of the outer measures of their intersections with \(A\) exceeds \(\mu^*(A)\). Identify the point counted more than once.