From the Definition to a Finite Checklist
On an infinite space, the Caratheodory condition quantifies over every test set, which can make it difficult to decide measurability directly. On a finite space, the quantifier becomes an explicit finite computation: list the subsets, evaluate both sides of the splitting identity, and keep a proposed set precisely when every identity holds. This procedure uses the outer measure itself; it does not assume that the outer measure is already a measure on all subsets.
Throughout this tutorial, \(X\) is a finite set, \(\mathcal{P}(X)\) is its power set, and \(\mu^*:\mathcal{P}(X)\to[0,\infty]\) is an outer measure. Thus \(\mu^*(\varnothing)=0\), \(\mu^*\) is monotone, and it is countably subadditive. The measurable sets are defined by the Caratheodory splitting condition below. All equalities in the finite tables and tests are exact, not almost-everywhere statements.
The last assertion is the standard conclusion of Caratheodory's theorem, proved earlier in the course. Here the task is to apply its defining condition on a finite space. For any fixed \(A\), each test \(E\) gives one equality. The outer-measure axioms already imply \(\mu^*(E)\leq\mu^*(E\cap A)+\mu^*(E\setminus A)\), because \(E\) is the union of those two disjoint pieces. Measurability requires equality, not merely this inequality.
Proof. If \(E\cap A=\varnothing\), then \(E\setminus A=E\), so the right side is \(0+\mu^*(E)=\mu^*(E)\). If \(E\setminus A=\varnothing\), then \(E\cap A=E\), and the right side is \(\mu^*(E)+0=\mu^*(E)\). Therefore every such test passes. For each remaining \(E\), both pieces are nonempty and its equality must be checked. Since \(X\) has finitely many subsets, this is a finite list of exact tests, and passing every test is exactly the defining condition. \(\square\)
There is also a useful symmetry: testing \(A\) or \(A^c\) gives the same right-hand side, since the two pieces are exchanged. Therefore \(A\) is measurable if and only if \(A^c\) is measurable. This can cut the number of proposed sets in half, but it does not remove the requirement to check every relevant \(E\).
Worked Examples of the Test
Worked Example: The Unit Outer Measure on a Two-Point Space
Let \(X=\{p,q\}\) and set \(\mu^*(\varnothing)=0\), while \(\mu^*(E)=1\) for every nonempty \(E\subseteq X\). This is an outer measure: it is monotone, and a union is empty only when every set in the union is empty; otherwise its value is \(1\), which is at most the sum of the nonnegative values of any cover of that union.
The complete table of values is \(\mu^*(\varnothing)=0\), \(\mu^*(\{p\})=1\), \(\mu^*(\{q\})=1\), and \(\mu^*(X)=1\). The candidates \(\varnothing\) and \(X\) are measurable: splitting by either one leaves one piece equal to \(E\) and the other empty, for every \(E\subseteq X\).
For \(A=\{p\}\), the four tests are as follows. For \(E=\varnothing\), the identity is \(0=0+0\). For \(E=\{p\}\), it is \(1=1+0\). For \(E=\{q\}\), it is \(1=0+1\). For \(E=X\), it is \(1=1+1=2\), which is false. Hence \(\{p\}\) is not measurable. By the complement symmetry, \(\{q\}\) is not measurable either. The measurable sets are exactly \(\varnothing\) and \(X\).
Check: the only failed split has two nonempty pieces, as the finite test predicts. The table also confirms that the outer measure is not additive on all subsets: \(\mu^*(X)=1\), whereas \(\mu^*(\{p\})+\mu^*(\{q\})=2\).
Worked Example: A Partition That Determines the Measurable Sets
Let \(X=\{a,b,c\}\), and define \(\mu^*(E)=\mathbf{1}_{\{E\cap\{a,b\}\ne\varnothing\}}+2\mathbf{1}_{\{c\in E\}}\), where \(\mathbf{1}\) is \(1\) when its stated condition holds and \(0\) otherwise. The first term charges one unit for meeting the block \(\{a,b\}\); the second charges two units for containing \(c\). This is an outer measure: it vanishes on the empty set, is monotone, and each block hit by a union is hit by at least one set in that union, so the charge of the union is at most the sum of the charges.
Listing all eight subsets gives \(\mu^*(\varnothing)=0\), \(\mu^*(\{a\})=1\), \(\mu^*(\{b\})=1\), \(\mu^*(\{c\})=2\), \(\mu^*(\{a,b\})=1\), \(\mu^*(\{a,c\})=3\), \(\mu^*(\{b,c\})=3\), and \(\mu^*(X)=3\).
For \(A=\{a,b\}\), every \(E\subseteq X\) splits into its portion in the block \(\{a,b\}\) and its portion in \(\{c\}\). The first portion contributes \(1\) if it is nonempty and \(0\) otherwise; the second contributes \(2\) if it contains \(c\) and \(0\) otherwise. These contributions add to exactly the defining value of \(\mu^*(E)\). Thus \(\{a,b\}\) passes every test. Its complement \(\{c\}\) also passes every test by complement symmetry.
To rule out the other nontrivial candidates, test \(A=\{a\}\) with \(E=\{a,b\}\): the identity would require \(1=1+1=2\), so it fails. The same test works for \(A=\{b\}\). For \(A=\{a,c\}\), use \(E=\{a,b\}\); its two pieces are \(\{a\}\) and \(\{b\}\), so the same false equality \(1=1+1\) results. For \(A=\{b,c\}\), this test again gives \(1=1+1\). Along with the always-measurable \(\varnothing\) and \(X\), this classifies the measurable sets as \(\varnothing,\{a,b\},\{c\},X\).
Check: for \(E=X\) and \(A=\{a,b\}\), the split reads \(3=1+2\). For \(E=\{a,c\}\), it reads \(3=1+2\) as well. The block formula explains why the identity holds for every \(E\), not just these two checks.
Worked Example: A Whole-Space Check Can Miss Failure
Let \(X=\{a,b,c\}\) and define \(\mu^*\) by the following values: \(\mu^*(\varnothing)=0\); each singleton has value \(1\); \(\mu^*(\{a,b\})=1\), \(\mu^*(\{b,c\})=1\), and \(\mu^*(\{a,c\})=2\); finally \(\mu^*(X)=2\). These values are monotone. To check subadditivity, it suffices first to check disjoint pieces: two singletons have union value at most \(2\); a singleton disjoint from a pair has union \(X\), of value \(2\), while the sum of their values is either \(1+1=2\) or \(1+2=3\); and the union of three singletons has value \(2\leq3\). Monotonicity then gives subadditivity for overlapping pieces as well. For a countable union in a finite \(X\), one can select finitely many sets from the union that already cover it (choose one set containing each point); finite subadditivity therefore gives countable subadditivity.
Consider \(A=\{a\}\). The test \(E=X\) passes: \(\mu^*(X)=2\), \(\mu^*(A)=1\), and \(\mu^*(X\setminus A)=\mu^*(\{b,c\})=1\), so \(2=1+1\). But \(E=\{a,b\}\) fails: \(\mu^*(E)=1\), while \(\mu^*(E\cap A)+\mu^*(E\setminus A)=\mu^*(\{a\})+\mu^*(\{b\})=1+1=2\). Thus \(\{a\}\) is not measurable even though its split of the whole space succeeds.
For \(A=\{b\}\), the test \(E=\{a,b\}\) also gives \(1\ne1+1\). For \(A=\{c\}\), use \(E=\{b,c\}\) to obtain \(1\ne1+1\). Every two-point candidate has a singleton complement, and measurability is invariant under complementation, so none of those candidates is measurable either. The empty set and \(X\) are measurable; they are the only measurable subsets.
Check: the listed values satisfy the failed equality calculations directly. The example shows why a single test, even \(E=X\), cannot replace the full finite checklist.
A Reliable Step-by-Step Procedure
For a specified outer measure on a finite set, organize the computation so that no test set is silently omitted.
Record \(\mu^*(E)\) for every subset \(E\subseteq X\), including \(\varnothing\). If the values are supplied by a formula, evaluate that formula on each subset.
Record its complement. The two candidates \(A\) and \(A^c\) have identical pass-or-fail status, so one may test only one representative from each complementary pair.
If \(E\subseteq A\) or \(E\subseteq A^c\), one split piece is empty and the identity holds. Keep every \(E\) that meets both sides.
For each remaining \(E\), calculate \(\mu^*(E)\), \(\mu^*(E\cap A)\), and \(\mu^*(E\setminus A)\). A single unequal result rules out \(A\); \(A\) is measurable only if all tests agree.
Collect the candidates that passed every test. In particular, include \(\varnothing\) and \(X\), which always pass.
The main sources of error are treating subadditivity as if it were equality, testing only a few convenient sets \(E\), and failing to distinguish the proposed measurable set \(A\) from the test set \(E\). The required equality concerns every test set \(E\), with the two pieces formed by intersecting with \(A\) and removing \(A\). The examples also show that an outer measure may be nonadditive on some subsets while having a nontrivial Caratheodory sigma-algebra.
Check Your Understanding
For each question, make the splitting pieces explicit before deciding whether the identity holds.
- On \(X=\{r,s\}\), set \(\mu^*(\varnothing)=0\), \(\mu^*(\{r\})=2\), \(\mu^*(\{s\})=3\), and \(\mu^*(X)=5\). Test all subsets \(A\subseteq X\) for Caratheodory measurability.
- Explain why the splitting identity always holds when \(E\subseteq A\), regardless of the outer-measure value of \(E\).
- In the second worked example, compute the splitting identity for \(A=\{a,b\}\) and \(E=\{b,c\}\).
- In the third worked example, which test set demonstrates that \(A=\{c\}\) is not measurable? Write both sides of the failed identity.
- Why does checking \(E=X\) alone not establish that a proposed set is Caratheodory measurable?