When an Outer Measure Is Not a Measure

Finite Subadditivity Is Not Enough

An outer measure is designed to control countable covers, not just finite ones. That distinction matters: the defining countable subadditivity condition cannot be replaced by finite subadditivity, even when the set function is monotone and assigns zero to the empty set. We will see this on the natural numbers, where a countable cover by singletons makes the failure exact.

There is a second distinction to keep in view. An outer measure is defined on every subset of its underlying set, but it need not be countably additive on all those subsets. Caratheodory's theorem identifies a sigma-algebra of sets on which its restriction is a measure. Thus a set function can fail to be an outer measure because it lacks countable subadditivity; a genuine outer measure, in turn, need not itself be a measure on its entire domain.

Here \(\mathcal{P}(X)\) denotes the collection of all subsets of \(X\). In contrast, a measure is defined on a sigma-algebra and is countably additive on pairwise disjoint members of that sigma-algebra. The domain and the required inequality are both part of the distinction.

A Set Function That Passes Every Finite Test

Let \(X=\mathbb{N}=\{1,2,3,\ldots\}\), with domain \(\mathcal{P}(\mathbb{N})\). Define \(c:\mathcal{P}(\mathbb{N})\to[0,\infty)\) by assigning zero to finite sets and one to infinite sets. In particular, \(c(\varnothing)=0\). We first verify carefully that \(c\) is monotone and finitely subadditive.

Proof. Suppose \(A\subseteq B\). If \(A\) is finite, then \(c(A)=0\le c(B)\), since \(c(B)\) is either zero or one. If \(A\) is infinite, then \(B\) is infinite, so \(c(A)=c(B)=1\). This proves monotonicity.

Now let \(A_1,\ldots,A_k\subseteq\mathbb{N}\), with \(k\) finite. If their union is finite, then \(c(\bigcup_{j=1}^k A_j)=0\le\sum_{j=1}^k c(A_j)\). If their union is infinite, at least one of the \(A_j\) must be infinite: a finite union of finite sets is finite. Thus at least one summand on the right is one, and \(c(\bigcup_{j=1}^k A_j)=1\le\sum_{j=1}^k c(A_j)\). For the empty finite union, both sides are zero. This proves finite subadditivity, including the degenerate case.

For the countable family \(A_n=\{n\}\), every \(A_n\) is finite, so \(c(A_n)=0\) for every \(n\). But \(\bigcup_{n=1}^{\infty}A_n=\mathbb{N}\), which is infinite. Therefore \(c(\bigcup_{n=1}^{\infty}A_n)=1\), whereas \(\sum_{n=1}^{\infty}c(A_n)=0\). The countable-subadditivity inequality would say \(1\le0\), which is false. Hence \(c\) is not an outer measure. \(\square\)

Why the Finite Argument Does Not Pass to the Limit

Finite subadditivity controls each finite partial union. It does not, by itself, control the value assigned to the union of all sets. In the singleton example, every finite partial union has value zero, but the full union has value one. The missing step would be a justified relation between the values on the partial unions and the value on their increasing union. Finite subadditivity provides no such relation.

For comparison, every countably subadditive set function with value zero at the empty set is finitely subadditive. Given \(A_1,\ldots,A_k\), apply countable subadditivity to the sequence \(A_1,\ldots,A_k,\varnothing,\varnothing,\ldots\). The union remains \(\bigcup_{j=1}^k A_j\), while the extra terms contribute zero. This establishes the finite inequality. The converse fails, as \(c\) shows.

A Genuine Outer Measure Need Not Be a Measure Everywhere

The preceding function \(c\) is not an outer measure. To see the other side of the distinction, we now construct a genuine outer measure on \(\mathbb{N}\) that is not a measure on all of \(\mathcal{P}(\mathbb{N})\). Give point \(n\) weight \(w_n=2^{-n}\), and for \(A\subseteq\mathbb{N}\) define \(\mu^*(A)\) to be the supremum of the weights of points in \(A\), with supremum zero when \(A=\varnothing\).

Proof. The empty-set axiom holds by definition. If \(A\subseteq B\), every weight represented in the supremum for \(A\) is also represented for \(B\), so \(\mu^*(A)\le\mu^*(B)\). For countable subadditivity, let \(A\subseteq\bigcup_{j=1}^{\infty}A_j\). If the sum \(\sum_j\mu^*(A_j)\) is infinite, the desired inequality holds because \(\mu^*(A)\le1/2\). If it is finite, take any \(n\in A\). There is an index \(j\) with \(n\in A_j\), and hence \(2^{-n}\le\mu^*(A_j)\le\sum_{i=1}^{\infty}\mu^*(A_i)\). Taking the supremum over \(n\in A\) gives \(\mu^*(A)\le\sum_i\mu^*(A_i)\). For \(A=\varnothing\), the same inequality holds because its left side is zero. Thus all outer-measure axioms hold.

To show that \(\mu^*\) is not a measure on the entire power set, use the pairwise disjoint sets \(\{n\}\), whose union is \(\mathbb{N}\). We have \(\mu^*(\{n\})=2^{-n}\), so countable additivity would require \(\mu^*(\mathbb{N})=\sum_{n=1}^{\infty}2^{-n}\). But \(\mu^*(\mathbb{N})=\sup_{n\ge1}2^{-n}=1/2\), while \(\sum_{n=1}^{\infty}2^{-n}=1\). These values differ, so countable additivity fails on \(\mathcal{P}(\mathbb{N})\). \(\square\)

What the Distinctions Tell Us

For the genuine outer measure \(\mu^*\), the failure of additivity on all subsets does not contradict Caratheodory's theorem. That theorem says that the Caratheodory measurable subsets form a sigma-algebra and that the restriction of the outer measure to that sigma-algebra is a measure. The conclusion concerns the restriction, not countable additivity on every subset in the original domain.

Keep the two failures separate:

The common mistake is to treat “finite subadditive” as a shortened version of “countably subadditive.” The singleton cover disproves that inference concretely. The reverse confusion is also important: even after countable subadditivity has been checked and an outer measure has been established, countable additivity on its whole domain does not follow. The relevant measure is the restriction to the Caratheodory measurable sets.