# 7.5: Nonnegative Set Functions. Premeasures. Outer Measures

- Page ID
- 19208

We now concentrate on nonnegative set functions

\[m : \mathcal{M} \rightarrow[0, \infty]\]

(we mostly denote them by \(m\) or \(\mu\)). Such functions have the advantage that

\[\sum_{n=1}^{\infty} m X_{n}\]

exists and is permutable (Theorem 2 in §2) for any sets \(X_{n} \in \mathcal{M},\) since \(m X_{n} \geq\) \(0.\) Several important notions apply to such functions (only). They "mimic" §§1 and 2.

A set function

\[m : \mathcal{M} \rightarrow[0, \infty]\]

is said to be

(i) monotone (on \(\mathcal{M}\)) iff

\[m X \leq m Y\]

whenever

\[X \subseteq Y \text { and } X, Y \in \mathcal{M};\]

(ii) (finitely) subadditive (on \(\mathcal{M}\)) iff for any finite union

\[\bigcup_{k=1}^{n} Y_{k},\]

we have

\[m X \leq \sum_{k=1}^{m} m Y_{k}\]

whenever \(X, Y_{k} \in \mathcal{M}\) and

\[X \subseteq \bigcup_{k=1}^{n} Y_{k} \text { (disjoint or not);}\]

(iii) \(\sigma\)-subadditive (on \(\mathcal{M}\)) iff (1) holds for countable unions, too.

Recall that \(\left\{Y_{k}\right\}\) is called a covering of \(X\) iff

\[X \subseteq \bigcup_{k} Y_{k}.\]

We call it an \(\mathcal{M}\)-covering of \(X\) if all \(Y_{k}\) are \(\mathcal{M}\)-sets. We now obtain the following corollary.

Subadditivity implies monotonicity.

Take \(n=1\) in formula (1).

If \(m : \mathcal{C} \rightarrow[0, \infty]\) is additive (\(\sigma\)-additive) on \(\mathcal{C}\), a semiring, then \(m\) is also subadditive (\(\sigma\)-subadditive, respectively), hence monotone, on \(\mathcal{C}\).

**Proof**-
The proof is a mere repetition of the argument used in Lemma 1 in §1.

Taking \(n=1\) in formula (ii) there, we obtain finite subadditivity.

For \(\sigma\)-subadditivity, one only has to use countable unions instead of finite ones.

**Note 1.** The converse fails: subadditivity does not imply additivity.

**Note 2.** Of course, Corollary 2 applies to rings, too (see Corollary 1 in §3).

A premeasures is a set function

\[\mu : \mathcal{C} \rightarrow[0, \infty]\]

such that

\[\emptyset \in \mathcal{C} \text { and } \mu \emptyset=0.\]

(\(\mathcal{C}\) may, but need not, be a semiring.)

A premeasure space is a triple

\[(S, \mathcal{C}, \mu),\]

where \(\mathcal{C}\) is a family of subsets of \(S\) (briefly, \(\mathcal{C} \subseteq 2^{S})\) and

\[\mu : \mathcal{C} \rightarrow[0, \infty]\]

is a premeasure. In this case, \(\mathcal{C}\)-sets are also called basic sets.

If

\[A \subseteq \bigcup_{n} B_{n},\]

with \(B_{n} \in \mathcal{C},\) the sequence \(\left\{B_{n}\right\}\) is called a basic covering of \(A,\) and

\[\sum_{n} \mu B_{n}\]

is a basic covering value of \(A;\left\{B_{n}\right\}\) may be finite or infinite.

(a) The volume function \(v\) on \(\mathcal{C}\) (= intervals in \(E^{n}\)) is a premeasure, as \(v \geq 0\) and \(v \emptyset=0.\) (\(E^{n}, \mathcal{C}, v\)) is the Lebesgue premeasure space.

(b) The LS set function \(s_{\alpha}\) is a premeasure if \(\alpha \uparrow\) (see Problem 7 in §4). We call it the \(\alpha\)-induced Lebesgue-Stieltjes \((L S)\) premeasure in \(E^{1}\).

We now develop a method for constructing \(\sigma\)-subadditive premeasures. (This is a first step toward achieving \(\sigma\)-additivity; see §4.)

For any premeasure space \((S, \mathcal{C}, \mu),\) we define the \(\mu\)-induced outer measure \(m^{*}\) on \(2^{S}\) (= all subsets of \(S\)) by setting, for each \(A \subseteq S\),

\[m^{*} A=\inf \left\{\sum_{n} \mu B_{n} | A \subseteq \bigcup_{n} B_{n}, B_{n} \in \mathcal{C}\right\},\]

i.e., \(m^{*} A\) (called the outer measure of \(A\)) is the glb of all basic covering values of \(A.\)

If \(\mu=v, m^{*}\) is called the Lebesgue outer measure in \(E^{n}\).

**Note 3.** If \(A\) has no basic coverings, we set \(m^{*} A=\infty.\) More generally, we make the convention that inf \(\emptyset=+\infty\).

**Note 4.** By the properties of the glb, we have

\[(\forall A \subseteq S) \quad 0 \leq m^{*} A.\]

If \(A \in \mathcal{C},\) then \(\{A\}\) is a basic covering; so

\[m^{*} A \leq \mu A.\]

In particular, \(m^{*} \emptyset=\mu \emptyset=0\).

The set function \(m^{*}\) so defined is \(\sigma\)-subadditive on \(2^{S}\).

**Proof**-
Given

\[A \subseteq \bigcup_{n} A_{n} \subset S,\]

we must show that

\[m^{*} A \leq \sum_{n} m^{*} A_{n}.\]

This is trivial if \(m^{*} A_{n}=\infty\) for some \(n.\) Thus assume

\[(\forall n) \quad m^{*} A_{n}<\infty\]

and fix \(\varepsilon>0\).

By Note 3, each \(A_{n}\) has a basic covering

\[\left\{B_{n k}\right\}, \quad k=1,2, \ldots\]

(otherwise, \(m^{*} A_{n}=\infty.\)) By properties of the glb, we can choose the \(B_{n k}\) so that

\[(\forall n) \quad \sum_{k} \mu B_{n k}<m^{*} A_{n}+\frac{\varepsilon}{2^{n}}.\]

(Explain from (2)). The sets \(B_{n k}\) (for all \(n\) and all \(k )\) form a countable basic covering of all \(A_{n},\) hence of \(A.\) Thus by Definition 3,

\[m^{*} A \leq \sum_{n}\left(\sum_{k} \mu B_{n k}\right) \leq \sum_{n}\left(m^{*} A_{n}+\frac{\varepsilon}{2^{n}}\right) \leq \sum^{n} m^{*} A_{n}+\varepsilon.\]

As \(\varepsilon\) is arbitrary, we can let \(\varepsilon \rightarrow 0\) to obtain the desired result.\(\quad \square\)

**Note 5.** In view of Theorem 1, we now generalize the notion of an outer measure in \(S\) to mean any \(\sigma\)-subadditive premeasure defined on all of \(2^{S}\).

By Note 4, \(m^{*} \leq \mu\) on \(\mathcal{C},\) not \(m^{*}=\mu\) in general. However, we obtain the following result.

With \(m^{*}\) as in Definition 3, we have \(m^{*}=\mu\) on \(\mathcal{C}\) iff \(\mu\) is \(\sigma\)-subadditive on \(\mathcal{C}.\) Hence, in this case, \(m^{*}\) is an extension of \(\mu.\)

**Proof**-
Suppose \(\mu\) is \(\sigma\)-subadditive and fix any \(A \in \mathcal{C}.\) By Note 4,

\[m^{*} A \leq \mu A.\]

We shall show that

\[\mu A \leq m^{*} A,\]

too, and hence \(\mu A=m^{*} A\).

Now, as \(A \in \mathcal{C}, A\) surely has basic coverings, e.g., \(\{A\}.\) Take any basic covering:

\[A \subseteq \bigcup_{n} B_{n}, \quad B_{n} \in \mathcal{C}.\]

As \(\mu\) is \(\sigma\)-subadditive,

\[\mu A \leq \sum_{n} \mu B_{n}.\]

Thus \(\mu A\) does not exceed any basic covering values of \(A;\) so it cannot exceed their glb, \(m^{*} A.\) Hence \(\mu=m^{*},\) indeed.

Conversely, if \(\mu=m^{*}\) on \(\mathcal{C},\) then the \(\sigma\)-subadditivity of \(m^{*}\) (Theorem 1) implies that of \(\mu\) (on \(\mathcal{C}\)). Thus all is proved.\(\quad \square\)

**Note 6.** If, in (2), we allow only finite basic coverings, then the \(\mu\)-induced set function is called the \(\mu\)-induced outer content, \(c^{*}.\) It is only finitely subadditive, in general.

In particular, if \(\mu=v\) (Lebesgue premeasure), we speak of the Jordan outer content in \(E^{n}.\) (It is superseded by Lebesgue theory but still occurs in courses on Riemann integration.)

We add two more definitions related to the notion of coverings.

A set function \(s : \mathcal{M} \rightarrow E\left(\mathcal{M} \subseteq 2^{S}\right)\) is called \(\sigma\)-finite iff every \(X \in \mathcal{M}\) can be covered by a sequence of \(\mathcal{M}\)-sets \(X_{n},\) with

\[\left|s X_{n}\right|<\infty \quad(\forall n).\]

Any set \(A \subseteq S\) which can be so covered is said to be \(\sigma\)-finite with respect to \(s\) (briefly, (\(s\)) \(\sigma\)-finite).

If the whole space \(S\) can be so covered, we say that \(s\) is totally \(\sigma\)-finite.

For example, the Lebesgue premeasure \(v\) on \(E^{n}\) is totally \(\sigma\)-finite.

A set function \(s : \mathcal{M} \rightarrow E^{*}\) is said to be regular with respect to a set family \(\mathcal{A}\) (briefly, \(\mathcal{A}\)-regular) iff for each \(A \in \mathcal{M}\),

\[s A=\inf \{s X | A \subseteq X, X \in \mathcal{A}\};\]

that is, \(s A\) is the glb of all \(s X,\) with \(A \subseteq X\) and \(X \in \mathcal{A}\).

These notions are important for our later work. At present, we prove only one theorem involving Definitions 3 and 5.

For any premeasure space \((S, \mathcal{C}, \mu),\) the \(\mu\)-induced outer measure \(m^{*}\) is \(\mathcal{A}\)-regular whenever

\[\mathcal{C}_{\sigma} \subseteq \mathcal{A} \subseteq 2^{S}.\]

Thus in this case,

\[(\forall A \subseteq S) \quad m^{*} A=\inf \left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]

**Proof**-
As \(m^{*}\) is monotone, \(m^{*} A\) is surely a lower bound of

\[\left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]

We must show that there is no greater lower bound.

This is trivial if \(m^{*} A=\infty\).

Thus let \(m^{*} A<\infty;\) so \(A\) has basic coverings (Note 3). Now fix any \(\varepsilon>0\).

By formula (2), there is a basic covering \(\left\{B_{n}\right\} \subseteq \mathcal{C}\) such that

\[A \subseteq \bigcup_{n} B_{n}\]

and

\[m^{*} A+\varepsilon>\sum_{n} \mu B_{n} \geq \sum_{n} m^{*} B_{n} \geq m^{*} \bigcup_{n} B_{n}.\]

(\(m^{*}\) is \(\sigma\)-subadditive!)

Let

\[X=\bigcup_{n} B_{n}.\]

Then \(X\) is in \(\mathcal{C}_{\sigma},\) hence in \(\mathcal{A},\) and \(A \subseteq X.\) Also,

\[m^{*} A+\varepsilon>m^{*} X.\]

Thus \(m^{*} A+\varepsilon\) is not a lower bound of

\[\left\{m^{*} X | A \subseteq X, X \in \mathcal{A}\right\}.\]

This proves (4).\(\quad \square\)