A Finite Outer Measure with Few Measurable Sets

A Coarse Outer Measure

Outer measure need not distinguish between different nonempty sets. On a finite set, consider the rule that charges every nonempty subset the same amount, namely one. This example is useful because the Caratheodory criterion can be evaluated exactly: a set is measurable only if it never splits any test set into two nonempty pieces. The criterion is a condition on every subset of the space, so a single test set that splits across both sides is enough to prove nonmeasurability.

Here \(\mu^*\) is an outer measure, not initially a measure on a specified sigma-algebra. In particular, its definition on \(\mathcal P(X)\) does not mean that every subset satisfies the Caratheodory criterion. The outer measure is finite: if \(X\ne\varnothing\), then \(\mu^*(X)=1\); if \(X=\varnothing\), then \(\mu^*(X)=0\).

Proof. We check the outer-measure axioms. First, \(\mu^*(\varnothing)=0\) by definition. For monotonicity, suppose \(A\subseteq B\subseteq X\). If \(A=\varnothing\), then \(\mu^*(A)=0\le\mu^*(B)\). If \(A\ne\varnothing\), then \(B\ne\varnothing\), so \(\mu^*(A)=\mu^*(B)=1\). Thus \(\mu^*(A)\le\mu^*(B)\) in every case.

Now let \(A_1,A_2,\ldots\) be any sequence of subsets of \(X\). If \(\bigcup_{n=1}^\infty A_n=\varnothing\), then every \(A_n=\varnothing\), so \(\mu^*(\bigcup_n A_n)=0=\sum_n\mu^*(A_n)\). If the union is nonempty, at least one \(A_j\) is nonempty. Therefore \(\sum_n\mu^*(A_n)\ge\mu^*(A_j)=1=\mu^*(\bigcup_n A_n)\). This proves countable subadditivity, including when the sum is infinite. Hence \(\mu^*\) is an outer measure. \(\square\)

How a Test Set Detects a Failed Split

For a fixed \(E\subseteq X\), each test set \(A\subseteq X\) is divided into \(A\cap E\) and \(A\setminus E\). Under this outer measure, each piece contributes either zero or one. Thus the criterion can fail only when both pieces are nonempty: then the split pieces cost two in total, while the original nonempty set costs one.

To verify the calculation, first take \(A=\varnothing\). Both intersections are empty, so the difference is \(0+0-0=0\). Next take nonempty \(A\) with \(A\subseteq E\). Then \(A\cap E=A\) is nonempty and \(A\setminus E=\varnothing\), giving \(1+0-1=0\). If instead \(A\subseteq E^c\), the two terms are \(0\) and \(1\), again giving \(0+1-1=0\). Finally, if \(A\) meets both sides, then each part is nonempty, and the difference is \(1+1-1=1\). These cases exhaust all possibilities.

Proof. First, let \(E=\varnothing\). For every \(A\subseteq X\), \(A\cap E=\varnothing\) and \(A\setminus E=A\), so \(\mu^*(A\cap E)+\mu^*(A\setminus E)=0+\mu^*(A)=\mu^*(A)\). Thus \(\varnothing\) is measurable. If \(E=X\), then \(A\cap E=A\) and \(A\setminus E=\varnothing\), giving the same equality. These arguments also cover \(X=\varnothing\).

For the converse, suppose \(X\ne\varnothing\) and \(E\) is neither empty nor all of \(X\). Choose \(x\in E\) and \(y\in E^c\); these choices are possible because both sets are nonempty. Set \(A=\{x,y\}\). Then \(A\) is nonempty, \(A\cap E=\{x\}\) is nonempty, and \(A\setminus E=\{y\}\) is nonempty. Consequently, \(\mu^*(A)=1\), whereas \(\mu^*(A\cap E)+\mu^*(A\setminus E)=1+1=2\). The criterion fails for this \(A\), so \(E\) is not measurable. If \(X=\varnothing\), there is no subset other than \(\varnothing=X\), already shown measurable. This proves the assertion in all cases. \(\square\)

What “Few Measurable Sets” Means Here

For every nonempty \(X\), the measurable collection is \(\{\varnothing,X\}\), even though the outer measure is defined on the entire power set. This is a genuine restriction: if \(X\) has at least two elements, there are proper nonempty subsets, and the theorem shows that none of them is measurable. The reason is not that their outer measure is undefined. Rather, the uniform cost of one for each nonempty set prevents a nontrivial split from preserving the cost.

On the measurable collection \(\{\varnothing,X\}\), the restriction of \(\mu^*\) is a measure: it assigns zero to \(\varnothing\) and, for nonempty \(X\), one to \(X\). To check countable additivity on this collection, a disjoint sequence can contain at most one term equal to \(X\); all its other terms must be empty. If no term equals \(X\), the union and the sum both have measure zero. If one term equals \(X\), the union has measure one and the sum is one. When \(X=\varnothing\), every term and the union are empty, so both are zero. This observation describes the restriction without asserting that \(\mu^*\) is a measure on all of \(\mathcal P(X)\).

Common pitfall. For \(X=\{a,b,c\}\), \(E=\{a\}\), and \(A=\{a,b\}\), a mistaken calculation might assign \(\mu^*(A\setminus E)=0\) because the piece is “outside” \(E\). That uses membership in the wrong set: \(A\setminus E=\{b\}\) is nonempty, so its outer measure is \(1\), not \(0\). The correct computation is \(\mu^*(A)=1\), \(\mu^*(A\cap E)=\mu^*(\{a\})=1\), and \(\mu^*(A\setminus E)=\mu^*(\{b\})=1\). Thus the split has total \(2\), and \(E\) fails the criterion.

The decisive test is always whether some \(A\) has a nonempty part on each side. If such an \(A\) exists, the two pieces cost \(1+1\), while \(A\) itself costs \(1\). If no such test set exists, every nonempty test set lies on one side, and its cost is unchanged by splitting. For a nonempty space, this second situation occurs for \(E\) exactly when \(E\) is empty or all of \(X\).