Restricting an Outer Measure to Measurable Sets

From an Outer Measure to a Measure

An outer measure assigns a size to every subset of a space, but its values need not be additive on disjoint sets. Caratheodory measurability identifies the sets that split every test set exactly: for those sets, the outer measure has the additivity needed to define a measure. The restriction construction is simple to state, but its key point is proving countable additivity, including when some of the values are infinite.

Definition: An outer measure on a set \(X\) is a function \(\mu^*:\mathcal{P}(X)\to[0,\infty]\) such that \(\mu^*(\varnothing)=0\), \(\mu^*(A)\le\mu^*(B)\) whenever \(A\subseteq B\subseteq X\), and \(\mu^*(\bigcup_{n=1}^{\infty}A_n)\le\sum_{n=1}^{\infty}\mu^*(A_n)\) for every sequence of subsets \(A_n\subseteq X\). A set \(E\subseteq X\) is Caratheodory measurable if \(\mu^*(T)=\mu^*(T\cap E)+\mu^*(T\setminus E)\) for every \(T\subseteq X\). Write \(\mathcal{M}^*\) for the family of Caratheodory-measurable sets.

The defining equality for \(E\) is an exact splitting condition, required for every test set \(T\), not just for \(T=X\). The earlier result that \(\mathcal{M}^*\) is a sigma-algebra lets us use it as the domain of a measure. The outer measure axioms also imply the splitting inequality \(\mu^*(T)\le\mu^*(T\cap E)+\mu^*(T\setminus E)\); measurability requires equality.

The Restricted Set Function

Definition: Given an outer measure \(\mu^*\) on \(X\), its restriction to the Caratheodory sigma-algebra is the set function \(\mu:\mathcal{M}^*\to[0,\infty]\) defined by \(\mu(E)=\mu^*(E)\) for every \(E\in\mathcal{M}^*\). A measure on \((X,\mathcal{M}^*)\) is a nonnegative set function with value \(0\) at \(\varnothing\) that is countably additive on pairwise disjoint sequences in \(\mathcal{M}^*\).

Theorem: For any set \(X\) and outer measure \(\mu^*\) on \(X\), the restriction \(\mu=\mu^*|_{\mathcal{M}^*}\) is a measure on \((X,\mathcal{M}^*)\). No finiteness assumption on \(\mu^*(X)\) is required; the measure may take the value \(+\infty\).

Proof. Since \(\mu^*(\varnothing)=0\), the restriction satisfies \(\mu(\varnothing)=0\), and its values lie in \([0,\infty]\). It remains to prove countable additivity. Let \((E_n)_{n\ge1}\) be pairwise disjoint sets in \(\mathcal{M}^*\), and put \(E=\bigcup_{n\ge1}E_n\). The sigma-algebra property gives \(E\in\mathcal{M}^*\). Outer-measure countable subadditivity gives \(\mu^*(E)\le\sum_{n\ge1}\mu^*(E_n)\).

For the reverse inequality, fix a positive integer \(N\). Define \(R_0=E\) and \(R_k=E\setminus\bigcup_{i=1}^{k}E_i\) for \(1\le k\le N\). Each \(R_{k-1}\) is a subset of \(X\), so the Caratheodory equality for \(E_k\) applies to this test set. Disjointness gives \(R_{k-1}\cap E_k=E_k\), hence \(\mu^*(R_{k-1})=\mu^*(E_k)+\mu^*(R_k)\). Iterating these finite equalities yields \(\mu^*(E)=\sum_{k=1}^{N}\mu^*(E_k)+\mu^*(R_N)\), so \(\mu^*(E)\ge\sum_{k=1}^{N}\mu^*(E_k)\). This inequality holds for every \(N\). Taking the increasing limit of the partial sums, whose limit defines the extended nonnegative series, gives \(\mu^*(E)\ge\sum_{n\ge1}\mu^*(E_n)\). Together with the subadditivity inequality, this proves \(\mu(E)=\sum_{n\ge1}\mu(E_n)\), including the case where the sum is infinite. Thus \(\mu\) is a measure. \(\square\)

The proof uses two different properties for the two inequalities. Outer-measure subadditivity gives the upper bound; exact splitting by measurable sets gives the lower bound. The finite splitting argument is applied to each initial segment, and the infinite sum appears only after taking the limit of those partial sums.

Worked Examples

Worked Example: Counting Measure on the Natural Numbers

Let \(X=\mathbb{N}=\{1,2,\ldots\}\) and define \(\mu^*(A)\) to be the number of elements of \(A\) if \(A\) is finite, and \(+\infty\) if \(A\) is infinite. Determine \(\mathcal{M}^*\) and the restricted measure, then check additivity for the singleton sets.

This is an outer measure: the empty set has value \(0\), inclusion cannot decrease the number of elements, and a union covered by sets has at most the sum of their numbers of elements when that sum is finite. If the sum is infinite, the subadditivity inequality holds because its right side is \(+\infty\). For every \(E,T\subseteq\mathbb{N}\), the two sets \(T\cap E\) and \(T\setminus E\) are disjoint and their union is \(T\). If \(T\) is finite, their cardinalities add to the cardinality of \(T\); if \(T\) is infinite, at least one of them is infinite or both are finite with an impossible finite union. Thus the splitting equality holds, with the sum on the right equal to \(+\infty\) when \(T\) is infinite. Every set is measurable, so \(\mathcal{M}^*=\mathcal{P}(\mathbb{N})\), and the restriction is counting measure.

For \(E_n=\{n\}\), the sets are pairwise disjoint, \(\mu(E_n)=1\) for every \(n\), and \(\bigcup_{n\ge1}E_n=\mathbb{N}\). Therefore \(\sum_{n\ge1}\mu(E_n)=\sum_{n\ge1}1=+\infty=\mu(\mathbb{N})\). Check: the first \(N\) terms sum to \(N\), and their limit as \(N\to\infty\) is \(+\infty\), matching \(\mu(\mathbb{N})\).

Worked Example: A Unit-Cost Outer Measure on Two Points

Let \(X=\{a,b\}\), and set \(\mu^*(A)=0\) if \(A=\varnothing\) and \(\mu^*(A)=1\) otherwise. Find the Caratheodory sigma-algebra and its restricted measure.

This is an outer measure. It is monotone, and for any sequence \((A_n)\), if the union is empty then every \(A_n\) is empty and both sides of subadditivity are \(0\). If the union is nonempty, at least one \(A_n\) is nonempty, so \(\sum_n\mu^*(A_n)\ge1=\mu^*(\bigcup_nA_n)\).

For \(E=\varnothing\) or \(E=X\), every test set \(T\) is split into \(T\) and \(\varnothing\), so the Caratheodory equality holds. For \(E=\{a\}\), take \(T=X\): then \(\mu^*(T)=1\), while \(\mu^*(T\cap E)+\mu^*(T\setminus E)=\mu^*(\{a\})+\mu^*(\{b\})=1+1=2\). Thus \(\{a\}\) is not measurable. Taking \(E=\{b\}\) and \(T=X\) gives the same calculation, so \(\{b\}\) is not measurable. Consequently \(\mathcal{M}^*=\{\varnothing,X\}\), and the restricted measure has \(\mu(\varnothing)=0\) and \(\mu(X)=1\). Check: the only disjoint sequence in this two-set sigma-algebra whose union is \(X\) must include \(X\) exactly once; all other terms are empty, so the sum of the measures is \(1=\mu(X)\).

Worked Example: An Outer Measure from a Partition

Let \(X=\{a,b,c\}\), partitioned into \(B_1=\{a,b\}\) and \(B_2=\{c\}\). Define \(\mu^*(A)\) as the number of blocks \(B_i\) that intersect \(A\). Determine the measurable sets and compute the restricted measure.

The empty set meets zero blocks, and inclusion can only increase the number of blocks met. If a union meets a block, at least one of the sets in the union meets that block; counting the two blocks therefore gives \(\mu^*(\bigcup_n A_n)\le\sum_n\mu^*(A_n)\). So this is an outer measure. If \(E\) splits a block into a nonempty part in \(E\) and a nonempty part outside \(E\), use that whole block as the test set \(T\). Then \(\mu^*(T)=1\), but both \(T\cap E\) and \(T\setminus E\) meet the block, so the sum of their outer measures is \(2\). Such an \(E\) is not measurable.

Conversely, suppose \(E\) is a union of whole blocks. For any test set \(T\), each block met by \(T\) is met on exactly one side of the split: its whole block lies in \(E\), or its whole block lies outside \(E\). Thus the blocks counted by \(\mu^*(T)\) are counted exactly once in \(\mu^*(T\cap E)+\mu^*(T\setminus E)\), proving the splitting equality. The measurable sets are therefore \(\varnothing,\{a,b\},\{c\},X\). Their restricted measures are \(0,1,1,2\), respectively. For the disjoint measurable sets \(\{a,b\}\) and \(\{c\}\), the union is \(X\) and \(1+1=2=\mu(X)\). Check: \(\mu^*(X)\) counts both blocks, so its value is \(2\), as computed by the sum of the two restricted measures.

A Common Mistake: Using Additivity Outside the Domain

In the partition example, it is tempting to treat the restriction as though it were defined on every subset and to write \(\mu^*(\{a\}\cup\{b\})=\mu^*(\{a\})+\mu^*(\{b\})\). The left side is \(1\), because \(\{a,b\}\) meets one block; the right side is \(1+1=2\), because each singleton meets that same block. This is not a contradiction to the theorem: \(\{a\}\) and \(\{b\}\) are not Caratheodory measurable. Indeed, using \(T=\{a,b\}\), splitting by \(E=\{a\}\) gives \(1\ne1+1\). Countable additivity is asserted for disjoint sets in \(\mathcal{M}^*\), not for arbitrary subsets on which the outer measure happens to be defined.

The restriction construction is useful because it turns the splitting criterion into an ordinary measure on a sigma-algebra. The measure retains the outer measure's values on measurable sets, while its domain records exactly where the additivity proof applies. The theorem does not assert that every subset is measurable, that the resulting measure is finite, or that it is complete.

Check Your Understanding

For each question, distinguish carefully between the outer measure on all subsets and the restricted measure on \(\mathcal{M}^*\).

In the two-point unit-cost example, which test set shows that \(\{a\}\) is not Caratheodory measurable?
Explain why outer-measure subadditivity gives one direction of countable additivity for disjoint measurable sets.
For the partition example, compute \(\mu^*(\{a,c\})\) and determine whether \(\{a,c\}\) is Caratheodory measurable.
In the proof of countable additivity, why is it valid to take the limit of the finite partial sums even when the measure of the union is infinite?
Does the theorem require \(\mu^*(X)<\infty\)? State which part of the proof permits infinite values.

Restricting an Outer Measure to Measurable Sets

What You'll Learn