Verifying Outer Measure Axioms in Explicit Examples

What Must Be Checked

An outer measure assigns a size to every subset of a set, including subsets that might not belong to a separately specified sigma-algebra. In this tutorial, the two set functions have domain \(\mathcal{P}(X)\), the power set of a set \(X\). No measure space, sigma-finiteness assumption, or completeness assumption is needed to check their outer measure axioms.

All three conditions matter. The first fixes the size of the empty set, the second says that enlarging a set cannot reduce its size, and the third bounds the size of a countable union by the sum of the sizes of its pieces. The sets \(A_n\) need not be disjoint. In particular, countable subadditivity is not the same condition as countable additivity.

Counting Measure on a Power Set

For any set \(X\), define the counting set function \(\#\) on \(\mathcal{P}(X)\) by \(\#A=|A|\) when \(A\) is finite, and \(\#A=\infty\) when \(A\) is infinite. Thus every infinite set receives the same extended value, whether it is countable or uncountable. On \((X,\mathcal{P}(X))\), this set function is also a measure; the outer measure verification can be made directly.

Proof. First, \(\#\varnothing=0\), since the empty set has zero elements. If \(A\subseteq B\), then when \(B\) is finite, \(A\) is finite and has no more elements than \(B\), so \(\#A\le\#B\). When \(B\) is infinite, \(\#B=\infty\), and \(\#A\le\infty\) for every \(A\). This proves monotonicity.

For countable subadditivity, take any sequence \((A_n)\) of subsets of \(X\). If \(\sum_n\#A_n=\infty\), then \(\#(\bigcup_n A_n)\le\infty=\sum_n\#A_n\), as required. If \(\sum_n\#A_n<\infty\), no \(A_n\) can be infinite, and only finitely many of them can be nonempty: infinitely many nonempty sets would contribute at least \(1\) each, making the sum infinite. The union is therefore finite. Each element of the union belongs to at least one \(A_n\), so the number of distinct elements in the union is at most the total number of element-occurrences counted across the \(A_n\). Consequently, \(\#(\bigcup_n A_n)\le\sum_n\#A_n\). This proves all three axioms. \(\square\)

The Zero-One Set Function

Define \(\eta:\mathcal{P}(X)\to[0,\infty]\) by \(\eta(\varnothing)=0\) and \(\eta(A)=1\) for every nonempty \(A\subseteq X\). This function records only whether a set is empty; it does not measure how many points it contains.

Proof. The definition gives \(\eta(\varnothing)=0\). To check monotonicity, suppose \(A\subseteq B\). If \(A=\varnothing\), then \(\eta(A)=0\le\eta(B)\). If \(A\ne\varnothing\), then \(B\ne\varnothing\) as well, so \(\eta(A)=1=\eta(B)\). Thus \(\eta(A)\le\eta(B)\) in either case.

Now take any sequence \((A_n)\). If \(\bigcup_n A_n=\varnothing\), every \(A_n\) is empty, so \(\eta(\bigcup_n A_n)=0=\sum_n\eta(A_n)\). If \(\bigcup_n A_n\ne\varnothing\), at least one \(A_k\) is nonempty. Thus \(\eta(\bigcup_n A_n)=1\), while \(\sum_n\eta(A_n)\ge\eta(A_k)=1\). In both cases, \(\eta(\bigcup_n A_n)\le\sum_n\eta(A_n)\). All three outer measure axioms hold. \(\square\)

Outer Measure Does Not Mean Countable Additivity

A measure on a sigma-algebra satisfies countable additivity for pairwise disjoint measurable sets: the measure of their union equals the sum of their measures. An outer measure is only required to satisfy the inequality in the definition. Counting measure on \((X,\mathcal{P}(X))\) is a measure, but the zero-one set function generally is not.

Proof. If \(X\) has at least two elements, choose distinct \(a,b\in X\). The singleton sets \(\{a\}\) and \(\{b\}\) are disjoint, and their union \(\{a,b\}\) is nonempty. Thus \(\eta(\{a,b\})=1\), whereas \(\eta(\{a\})+\eta(\{b\})=1+1=2\). Countable additivity fails even for this finite disjoint pair.

If \(X=\varnothing\), its only subset is empty and \(\eta\) is identically zero, which is countably additive. If \(X\) has one element, its subsets are the empty set and \(X\); on these sets \(\eta\) equals counting measure. Counting measure is countably additive, so \(\eta\) is a measure in this case as well. \(\square\)

For comparison, counting measure on \((X,\mathcal{P}(X))\) is finite exactly when \(X\) is finite. It is sigma-finite exactly when \(X\) is countable: if \(X\) is countable, it is a countable union of singleton sets of measure \(1\); conversely, a countable union of finite-measure sets is a countable union of finite sets and hence is countable. It is complete because every subset of a counting-measure null set is measurable: the only null set is the empty set. These measure-space properties are additional facts about counting measure, not extra outer measure axioms.

A Common Miscalculation

A frequent error is to apply countable additivity to sets that overlap, or to assume that every outer measure must be additive on disjoint sets. For example, for the zero-one set function and \(A=\{a\}\), \(B=\{a,b\}\), a wrong calculation is \(\eta(A\cup B)=\eta(A)+\eta(B)=1+1=2\). The left side is actually \(1\), since \(A\cup B=\{a,b\}\) is nonempty. The correct outer-measure statement is \(\eta(A\cup B)\le\eta(A)+\eta(B)\), here \(1\le2\).

Even disjointness does not repair additivity for \(\eta\): for distinct \(a,b\), the disjoint singletons give \(\eta(\{a\}\cup\{b\})=1\), but \(\eta(\{a\})+\eta(\{b\})=2\). By contrast, counting measure gives \(2=1+1\) for those sets. The point of checking each axiom separately is to identify exactly what has been established: both functions are outer measures, but only counting measure is always countably additive.