7.5: Nonnegative Set Functions. Premeasures. Outer Measures

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 $\PageIndex{1}$

Subadditivity implies monotonicity.

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

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). $\quad \square$