Verifying the Induced Outer Measure

From a Premeasure to a Covering Infimum

Let \(\mathcal{R}\) be a ring of subsets of a set \(X\): it contains \(\varnothing\), is closed under finite unions, and is closed under differences. Let \(\mu_0:\mathcal{R}\to[0,\infty]\) be a premeasure. This means that \(\mu_0(\varnothing)=0\), and whenever pairwise disjoint \(A_1,A_2,\ldots\in\mathcal{R}\) have union \(A\in\mathcal{R}\), then \(\mu_0(A)=\sum_{n=1}^{\infty}\mu_0(A_n)\). The sum is allowed to be \(+\infty\).

For any \(E\subseteq X\), a countable \(\mathcal{R}\)-cover of \(E\) is a sequence \((A_n)_{n\ge1}\) in \(\mathcal{R}\) such that \(E\subseteq\bigcup_{n\ge1}A_n\). Finite covers are included by padding them with empty sets. Define the cost of a cover to be \(\sum_{n=1}^{\infty}\mu_0(A_n)\), and take the infimum over all such covers. If there is no cover, the infimum over the empty collection is \(+\infty\).

\mu^*(E)=\inf\left\{\sum_{n=1}^{\infty}\mu_0(A_n): A_n\in\mathcal{R},\quad E\subseteq\bigcup_{n=1}^{\infty}A_n\right\}, \qquad E\subseteq X.

The function \(\mu^*\) is defined on the entire power set of \(X\), whether or not its subsets belong to \(\mathcal{R}\). At this stage it is an outer measure, not a measure on a sigma-algebra: no measurability assertion is being made.

Definition: A set function \(\eta:\mathcal{P}(X)\to[0,\infty]\) is an outer measure if \(\eta(\varnothing)=0\), \(\eta(E)\le\eta(F)\) whenever \(E\subseteq F\), and \(\eta(\bigcup_{j=1}^{\infty}E_j)\le\sum_{j=1}^{\infty}\eta(E_j)\) for every sequence of subsets of \(X\).

Approximate Covers When the Infimum Is Not Attained

The infimum in the definition need not be the cost of any one cover. The proof of countable subadditivity therefore cannot assume that each set has a cheapest cover. It uses covers whose costs are within a prescribed positive error of the infimum.

Lemma: If \(E\subseteq X\) and \(\mu^*(E)<\infty\), then for every \(\varepsilon>0\) there is a countable \(\mathcal{R}\)-cover of \(E\) with cost less than \(\mu^*(E)+\varepsilon\).

Proof. The set of costs of covers of \(E\) is nonempty, since its infimum is finite, and it is bounded below by \(0\). By the defining property of an infimum, some cost is strictly less than \(\mu^*(E)+\varepsilon\). The corresponding cover gives the claim. This argument does not assert that a cover of cost exactly \(\mu^*(E)\) exists. \(\square\)

When \(\mu^*(E)=\infty\), this lemma makes no claim about a finite-cost approximate cover. That distinction is important: a finite error above \(+\infty\) gives no useful bound.

The Induced Set Function Is an Outer Measure

Theorem: For a ring \(\mathcal{R}\) on \(X\) and a premeasure \(\mu_0\) on \(\mathcal{R}\), the covering infimum \(\mu^*\) defined above is an outer measure on \(X\).

Proof. First, the sequence \(A_n=\varnothing\) covers \(\varnothing\) and has cost \(\sum_{n=1}^{\infty}\mu_0(\varnothing)=0\). All cover costs are nonnegative, so \(0\le\mu^*(\varnothing)\le0\), and hence \(\mu^*(\varnothing)=0\).

Next, let \(E\subseteq F\subseteq X\). Every cover of \(F\) also covers \(E\). Thus the collection of costs available for covers of \(F\) is contained in the collection available for covers of \(E\). Taking infima gives \(\mu^*(E)\le\mu^*(F)\). This also covers the case where \(F\) has no cover: then \(\mu^*(F)=+\infty\), and the inequality holds in the extended order.

Finally, let \(E_1,E_2,\ldots\subseteq X\). If \(\sum_{j=1}^{\infty}\mu^*(E_j)=+\infty\), the desired inequality holds because every value of \(\mu^*\) is at most \(+\infty\). Otherwise the sum is finite, so each \(\mu^*(E_j)\) is finite. Fix \(\varepsilon>0\). For each \(j\), apply the lemma with error \(\varepsilon 2^{-j}\) to choose a cover \((A_{j,k})_{k\ge1}\) of \(E_j\) such that

\sum_{k=1}^{\infty}\mu_0(A_{j,k}) < \mu^*(E_j)+\varepsilon 2^{-j}.

The family indexed by pairs \((j,k)\) is countable, and its sets cover \(\bigcup_{j=1}^{\infty}E_j\). A nonnegative double series can be summed in either order: both orders are the supremum of sums over finite subsets of pairs. Therefore the cost of this combined cover is

\sum_{j=1}^{\infty}\sum_{k=1}^{\infty}\mu_0(A_{j,k}) \le \sum_{j=1}^{\infty}\mu^*(E_j) +\varepsilon\sum_{j=1}^{\infty}2^{-j} = \sum_{j=1}^{\infty}\mu^*(E_j)+\varepsilon.

Since \(\mu^*(\bigcup_j E_j)\) is the infimum over cover costs, it is no larger than this cost. The resulting bound holds for every \(\varepsilon>0\), so letting \(\varepsilon\) decrease to \(0\) gives \(\mu^*(\bigcup_j E_j)\le\sum_j\mu^*(E_j)\). All three outer-measure axioms follow. \(\square\)

Worked Examples

Worked Example: A Finite Ring with Atomic Weights

Let \(X=\{a,b,c\}\), let \(\mathcal{R}=\mathcal{P}(X)\), and define \(\mu_0(A)=2\chi_A(a)+0\chi_A(b)+3\chi_A(c)\). This is a premeasure: for disjoint sets, each point occurs in at most one set, so the weight of their union is the sum of their weights. Compute \(\mu^*(E)\) for \(E=\{a,c\}\).

The set \(E\) itself is a cover, with cost \(\mu_0(\{a,c\})=2+3=5\), so \(\mu^*(E)\le5\). For any countable cover \((A_n)\) of \(E\), at least one \(A_n\) contains \(a\) and at least one contains \(c\). In the sum of costs, the weight \(2\) for \(a\) is counted at least once and the weight \(3\) for \(c\) is counted at least once. All other contributions are nonnegative. Thus \(\sum_n\mu_0(A_n)\ge2+3=5\). Taking the infimum gives \(\mu^*(E)\ge5\), so \(\mu^*(\{a,c\})=5\). Check: the cover consisting of \(\{a,c\}\) has cost \(2+3=5\), matching the lower bound.

Worked Example: A Covering Infimum Equal to Zero but Never Attained

Let \(X=\{p\}\cup\mathbb{N}\), where \(\mathbb{N}=\{1,2,\ldots\}\), and set \(A_n=\{p\}\cup\{k\in\mathbb{N}:k\ge n\}\). Let \(\mathcal{R}\) consist of finite subsets of \(\mathbb{N}\) and sets \(A_n\cup F\), where \(F\) is a finite subset of \(\{1,\ldots,n-1\}\). This is a ring: finite unions and differences of tails and finite sets again have one of these forms.

Define \(\mu_0(B)=\sum_{k\in B\cap\mathbb{N}}1/(k(k+1))\) for \(B\in\mathcal{R}\). The weights are nonnegative, so partitioning the integer points among disjoint sets proves countable additivity whenever their union is in \(\mathcal{R}\); the point \(p\) has weight zero. In particular, by \(1/(k(k+1))=1/k-1/(k+1)\),

\mu_0(A_n)=\sum_{k=n}^{\infty}\left(\frac1k-\frac1{k+1}\right)=\frac1n.

Each \(A_n\) covers \(\{p\}\), so \(\mu^*(\{p\})\le1/n\) for every \(n\). Since \(\mu^*(\{p\})\ge0\), these bounds give \(\mu^*(\{p\})=0\). But any ring set containing \(p\) has the form \(A_n\cup F\), and its cost is at least \(1/n>0\). Every cover of \(\{p\}\) contains at least one such set, so every cover has strictly positive cost. The infimum \(0\) is not attained. Check: the covers \(A_1,A_2,A_4\) have costs \(1,1/2,1/4\), respectively, and the formula \(1/n\) tends to \(0\); none of these costs is \(0\).

Worked Example: Countable Subadditivity with Equality

Use the same \(X,\mathcal{R},\mu_0\) as in the preceding example. For each \(j\ge1\), set \(E_j=\{j\}\). First, \(\mu^*(E_j)=1/(j(j+1))\): the singleton itself gives the upper bound, and every cover of it has total cost at least its weight, since \(j\) must occur in at least one covering set.

The union of the \(E_j\) is \(\mathbb{N}\). Covering \(\mathbb{N}\) by all its singletons costs

\sum_{j=1}^{\infty}\mu_0(\{j\}) = \sum_{j=1}^{\infty}\left(\frac1j-\frac1{j+1}\right) =1,

so \(\mu^*(\mathbb{N})\le1\). Conversely, for any cover of \(\mathbb{N}\), its total cost is at least the sum of the weights of the first \(m\) integers, for every \(m\), because each of those integers is covered at least once. That finite sum is \(1-1/(m+1)\). Letting \(m\) tend to infinity shows every cover has cost at least \(1\). Hence \(\mu^*(\mathbb{N})=1\). Check: \(\sum_j\mu^*(E_j)=\sum_j1/(j(j+1))=1=\mu^*(\bigcup_jE_j)\), so countable subadditivity is sharp here.

A Common Pitfall and the Role of the Axioms

A tempting but invalid step in a subadditivity proof is to choose, for each \(E_j\), a cover whose cost is exactly \(\mu^*(E_j)\). An infimum need not be attained. In the example above, \(\mu^*(\{p\})=0\), but every cover of \(\{p\}\) has positive cost. There is no cover of cost exactly \(0\) to choose. The correction is to fix \(\varepsilon>0\), use covers with costs less than \(\mu^*(E_j)+\varepsilon 2^{-j}\), and then let \(\varepsilon\) tend to zero after combining them.

The outer-measure axioms are the basic structural properties needed before applying the Carathéodory criterion. They alone do not say which subsets are measurable, or that the induced outer measure agrees with the original premeasure on \(\mathcal{R}\). Those are separate questions; neither follows merely from the cover-combination argument.

Check Your Understanding

Use the definition of the covering infimum and identify where approximation, rather than an exact minimizing cover, is needed.

Why does the sequence consisting entirely of \(\varnothing\) establish the value of \(\mu^*(\varnothing)\)? Which premeasure property is used?
If \(E\subseteq F\), explain the inclusion between the collections of covers that gives monotonicity.
In the tail example, why does \(\mu^*(\{p\})=0\) not imply that some cover of \(\{p\}\) has cost zero?
When proving countable subadditivity and \(\sum_j\mu^*(E_j)<\infty\), why is it valid to choose a finite-cost approximate cover for every \(E_j\)?
What summable sequence of errors would you use if the error budget is \(\varepsilon\), and what is the sum of that sequence?