Measuring the Failure of Exact Splitting
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. For a fixed set \(E\subseteq X\), the Caratheodory criterion asks whether every test set \(A\subseteq X\) splits exactly across \(E\): the outer measure of \(A\) must equal the sum of the outer measures of \(A\cap E\) and \(A\setminus E\). The splitting defect records the amount by which that sum exceeds \(\mu^*(A)\).
There is one detail to settle before subtracting: outer measures may be infinite. If \(\mu^*(A)=\infty\), subadditivity forces the sum of the two pieces to be \(\infty\) as well. Exact splitting then holds as an equality of extended nonnegative real numbers, but the expression \(\infty-\infty\) is undefined. We define the defect to be zero in this case, so it measures only a genuine failure of the criterion.
The criterion is an equality, while the defect is a nonnegative number in \([0,\infty]\). In the finite-total case it measures the excess of the sum of the pieces over the original outer measure. In the infinite-total case the criterion cannot fail: the sum of the pieces must be infinite, so we set the defect to zero rather than form an undefined difference.
Why the Defect Is Nonnegative
Proof. If \(\mu^*(A)=\infty\), the definition gives \(\Delta_E(A)=0\), which is nonnegative. Suppose instead that \(\mu^*(A)<\infty\). Since \(A=(A\cap E)\cup(A\setminus E)\), countable subadditivity, applied to these two sets (and, if needed, the empty set as a third set), gives $$ \mu^*(A)\leq \mu^*(A\cap E)+\mu^*(A\setminus E). $$ Subtracting the finite number \(\mu^*(A)\) from the right-hand side preserves the inequality and gives \(\Delta_E(A)\geq0\). If the sum on the right is infinite, the defect is \(+\infty\), which is also in \([0,\infty]\). \(\square\)
This lemma explains the word “defect”: subadditivity already guarantees that splitting cannot make the sum smaller than the original outer measure. The defect is the nonnegative gap, with the convention above ensuring that it is defined in every case.
The Vanishing-Defect Characterization
Proof. First suppose \(E\) is Caratheodory measurable. Fix any \(A\subseteq X\). If \(\mu^*(A)<\infty\), the defining equality gives $$ \Delta_E(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)-\mu^*(A)=0. $$ If \(\mu^*(A)=\infty\), the definition of the defect gives \(\Delta_E(A)=0\). Thus the defect vanishes for every \(A\).
Conversely, suppose \(\Delta_E(A)=0\) for every \(A\subseteq X\). Fix any such \(A\). If \(\mu^*(A)<\infty\), the definition of \(\Delta_E(A)\) and the assumption that it is zero yield $$ \mu^*(A\cap E)+\mu^*(A\setminus E)-\mu^*(A)=0, $$ so \(\mu^*(A)=\mu^*(A\cap E)+\mu^*(A\setminus E)\). If \(\mu^*(A)=\infty\), subadditivity gives \(\infty=\mu^*(A)\leq\mu^*(A\cap E)+\mu^*(A\setminus E)\). The sum is therefore \(\infty\), and the required equality holds in the extended nonnegative real numbers. We have verified the splitting equality for every \(A\subseteq X\), so \(E\) is Caratheodory measurable. \(\square\)
The quantifier “for every \(A\)” is essential. A zero defect for one particular test set says only that this one split is exact; measurability requires exactness against every subset of \(X\). Conversely, a single positive defect supplies a specific witness that \(E\) is not measurable.
Worked Example: Exact Splitting for a Weighted Outer Measure
Let \(X=\{a,b,c\}\) and define \(\mu^*(S)=\sum_{x\in S}w(x)\) for \(S\subseteq X\), where \(w(a)=1\), \(w(b)=2\), and \(w(c)=4\). This is an outer measure: the empty set has weight zero, inclusion cannot decrease a sum of nonnegative weights, and the weight of a union is at most the sum of the weights of any countable family covering it. Take \(E=\{a,c\}\).
For any \(A\subseteq X\), the sets \(A\cap E\) and \(A\setminus E\) are disjoint and their union is \(A\). Each point of \(A\) occurs in exactly one of these two pieces, so its weight is counted exactly once in their sum. Hence $$ \mu^*(A\cap E)+\mu^*(A\setminus E)=\mu^*(A), $$ and the defect is zero for every \(A\). For example, if \(A=\{a,b\}\), then \(A\cap E=\{a\}\) has outer measure \(1\), and \(A\setminus E=\{b\}\) has outer measure \(2\). Also \(\mu^*(A)=1+2=3\), so \(\Delta_E(A)=1+2-3=0\). If \(A=X\), the two pieces have outer measures \(1+4=5\) and \(2\); the total is \(1+2+4=7\), and the defect is \(5+2-7=0\). Check: adding the weights in the two pieces for \(A=X\) gives \(5+2=7\), the same total as adding all three point weights, \(1+2+4=7\).
Worked Example: A Positive Defect Witnesses Failure
Let \(X=\{a,b\}\) and define \(\mu^*(S)=0\) if \(S=\varnothing\), and \(\mu^*(S)=1\) if \(S\ne\varnothing\). This is an outer measure. It is monotone. For countable subadditivity, if a union is empty, every set in the union is empty and both sides are zero; if a union is nonempty, at least one set in the covering family is nonempty, so the sum of the outer measures is at least \(1\), the outer measure of the union.
Take \(E=\{a\}\). With test set \(A=X\), the intersection \(A\cap E=\{a\}\) and the difference \(A\setminus E=\{b\}\) are both nonempty. Their outer measures are \(1\) and \(1\), while \(\mu^*(A)=1\). Therefore $$ \Delta_E(X)=1+1-1=1. $$ This positive defect shows that \(E\) is not Caratheodory measurable. To see why other test sets alone would not expose the failure, compute them too: for \(A=\varnothing\), all three outer measures are zero and the defect is \(0+0-0=0\); for \(A=\{a\}\), the pieces have outer measures \(1\) and \(0\), giving defect \(1+0-1=0\); for \(A=\{b\}\), they have outer measures \(0\) and \(1\), giving defect \(0+1-1=0\). Thus it is specifically the test set \(X\) that witnesses failure. Check: for \(A=X\), the two pieces have total outer measure \(2\), exactly one more than \(\mu^*(X)=1\), matching the computed defect.
Worked Example: Infinite Outer Measure and the Defect Convention
Let \(X=\{a,b\}\) and define \(\mu^*(S)=0\) for \(S=\varnothing\) and \(\mu^*(S)=\infty\) for every nonempty \(S\). This is an outer measure: it is monotone, and any countable family whose union is nonempty contains a nonempty member, so the sum of its outer measures is \(\infty\). Let \(E=\{a\}\).
For the test set \(A=X\), both pieces are nonempty, so \(\mu^*(A\cap E)=\infty\) and \(\mu^*(A\setminus E)=\infty\). Their sum is \(\infty\), as is \(\mu^*(A)\). The splitting equality therefore holds, and the defined defect is \(\Delta_E(X)=0\). For the other test sets, \(A=\varnothing,\{a\},\{b\}\), the pieces consist of one empty set and one set equal to \(A\); the sum of the two outer measures equals \(\mu^*(A)\) in each case. Thus the defect is zero for every test set. Check: when \(A=X\), the equality is \(\infty=\infty+\infty=\infty\) in the extended nonnegative real numbers; the defect is zero by definition, not by subtracting infinities.
A Common Mistake and a Practical Test
A tempting but invalid calculation in the infinite example is $$ \Delta_E(X)=\infty+\infty-\infty=0. $$ The conclusion happens to agree with the definition, but the displayed subtraction is not a valid calculation: \(\infty-\infty\) is undefined. The correction is to use the definition by cases. Since \(\mu^*(X)=\infty\), it assigns \(\Delta_E(X)=0\); separately, the splitting equality is checked as \(\infty=\infty\). Keeping these two steps distinct prevents an undefined expression from being used as evidence.
For a fixed \(E\), a useful test is to select a set \(A\) whose two pieces can be computed exactly. If \(\mu^*(A)\) is finite, compute both piece outer measures and subtract the finite total from their sum. A positive result immediately disproves measurability. If the defect is zero for a handful of test sets, that is not enough to prove measurability: the theorem requires vanishing for every \(A\subseteq X\). In the infinite-total case, do not try to measure a numerical gap; subadditivity already makes the splitting equality automatic.
Check Your Understanding
Use the definition by cases and the all-test-sets quantifier when answering.
- For an outer measure \(\mu^*\), explain why \(\mu^*(A)=\infty\) implies \(\mu^*(A\cap E)+\mu^*(A\setminus E)=\infty\).
- Suppose \(\mu^*(A)=5\), \(\mu^*(A\cap E)=2\), and \(\mu^*(A\setminus E)=4\). Compute \(\Delta_E(A)\). What does this value say about \(E\)?
- For the outer measure on \(\{a,b\}\) that assigns \(1\) to each nonempty set, compute the defect for \(E=\{a\}\) using \(A=\{a\}\), and compare it with the defect using \(A=\{a,b\}\).
- Why is \(\infty+\infty-\infty\) not a valid way to compute a splitting defect? State the definition that applies instead.
- State both directions of the vanishing-defect characterization, including the quantifier over test sets.