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# 7.5: Nonnegative Set Functions. Premeasures. Outer Measures

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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.

## Definition 1

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.

## Corollary $$\PageIndex{1}$$

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

## Corollary $$\PageIndex{2}$$

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

## Definition 2

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.

## Examples

(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.)

## Definition 3

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, $$m^{*} \emptyset=\mu \emptyset=0$$.

## Theorem $$\PageIndex{1}$$

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.

## Theorem $$\PageIndex{2}$$

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.

## Definition 4

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.

## Definition 5

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.

## Theorem $$\PageIndex{3}$$

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).$$\quad \square$$