7.4: Set Functions. Additivity. Continuity

I. The letter "$v$" in $v A$ may be treated as a certain function symbol that assigns a numerical value (called "volume") to the set $A.$ So far we have defined such "volumes" for all intervals, then for $\mathcal{C}$-simple sets, and even for $\mathcal{C}_{\sigma}$-sets in $E^{n}$.

Mathematically this means that the volume function $v$ has been defined first on $\mathcal{C}$ (the intervals), then on $\mathcal{C}_{s}^{\prime}$ ($\mathcal{C}$-simple sets), and finally on $\mathcal{C}_{\sigma}.$

Thus we have a function $v$ which assigns values ("volumes") not just to single points, as ordinary "point functions" do, but to whole sets, each set being treated as one thing.

In other words, the domain of the function $v$ is not just a set of points, but a set family ($\mathcal{C}, \mathcal{C}_{s}^{\prime}$, or $\mathcal{C}_{\sigma}$).

The "volumes" assigned to such sets are the function values (for $\mathcal{C}$ and $\mathcal{C}_{s}^{\prime}$-sets they are real numbers; for $\mathcal{C}_{\sigma}$-sets they may reach $+\infty$).This is symbolized by

$$v : \mathcal{C} \rightarrow E^{1}$$

or

$$v : \mathcal{C}_{\sigma} \rightarrow E^{*};$$

more precisely,

$$v : \mathcal{C}_{\sigma} \rightarrow[0, \infty],$$

since volume is nonnegative.

It is natural to call $v$ a set function (as opposed to ordinary point functions). As we shall see, there are many other set functions. The function values need not be real; they may be complex numbers or vectors. This agrees with our general definition of a function as a certain set of ordered pairs (Definition 3 in Chapter 1, §§4-7); e.g.,

$$v=\left(\begin{array}{cccc}{A} & {B} & {C} & {\cdots} \\ {v A} & {v B} & {v C} & {\cdots}\end{array}\right).$$

Here the domain consists of certain sets $A, B, C, \ldots$. This leads us to the following definition.

Definition 1

A set function is a mapping

$$s : \mathcal{M} \rightarrow E$$

whose domain is a set family $\mathcal{M}$.

The range space $E$ is assumed to be $E^{1}, E^{*}, C$ (the complex field), $E^{n}$, or another normed space. Thus $s$ may be real, extended real, complex, or vector valued.

To each set $X \in \mathcal{M},$ the function $s$ assigns a unique function value denoted $s(X)$ or $s X$ (which is an element of the range space $E).$

We say that $s$ is finite on a set family $\mathcal{N} \subseteq \mathcal{M}$ iff

$$(\forall X \in \mathcal{N}) \quad|s X|<\infty;$$

briefly, $|s|<\infty$ on $\mathcal{N}$. (This is automatic if $s$ is complex or vector valued.)

We call s semifinite if at least one of $\pm \infty$ is excluded as function value, e.g., if $s \geq 0$ on $\mathcal{M};$ i.e.,

$$s : \mathcal{M} \rightarrow[0, \infty].$$

(The symbol $\infty$ stands for $+\infty$ throughout).

Definition 2

A set function

$$s : \mathcal{M} \rightarrow E$$

is called additive (or finitely additive) on $\mathcal{N} \subseteq \mathcal{M}$ iff for any finite disjoint union $\bigcup_{k} A_{k},$ we have

$$\sum_{k} s A_{k}=s\left(\bigcup_{k} A_{k}\right),$$

provided $\bigcup_{k} A_{k}$ and all the $A_{k}$ are $\mathcal{N}$-sets.

If this also holds for countable disjoint unions, $s$ is called $\sigma$-additive (or countably additive or completely additive) on $\mathcal{N}$.

If $\mathcal{N}=\mathcal{M}$ here, we simply say that $s$ is additive ($\sigma$-additive, respectively).

Note 1. As $\bigcup A_{k}$ is independent of the order of the $A_{k}, \sigma$-additivity pre-supposes and implies that the series

$$\sum s A_{k}$$

is permutable (§2) for any disjoint sequence

$$\left\{A_{k}\right\} \subseteq \mathcal{N}.$$

(The partial sums do exist, by our conventions (2*) in Chapter 4, §4.)

The set functions in the examples below are additive; $v$ is even $\sigma$-additive (Corollary 1 in §2).

Examples (b)-(d) show that set functions may arise from ordinary "point functions."

Examples

(a) The volume function $v : \mathcal{C} \rightarrow E^{1}$ on $\mathcal{C}$ (= intervals in $E^{n}$), discussed above, is called the Lebesgue premeasure (in $E^{n}$).

(b) Let $\mathcal{M}=\{$all finite intervals $I \subset E^{1}\}$.

Given $f : E^{1} \rightarrow E,$ set

$$(\forall I \in \mathcal{M}) \quad s I=V_{f}[\overline{I}],$$

the total variation of $f$ on the closure of $I$ (Chapter 5, §7).

Then $s : \mathcal{M} \rightarrow[0, \infty]$ is additive by Theorem 1 of Chapter 5, §7.

(c) Let $\mathcal{M}$ and $f$ be as in Example (b).

Suppose $f$ has an antiderivative (Chapter 5, §5) on $E^{1}.$ For each interval $X$ with endpoints $a, b \in E^{1}(a \leq b),$ set

$$s X=\int_{a}^{b} f.$$

This yields a set function $s : \mathcal{M} \rightarrow E$ (real, complex, or vector valued), additive by Corollary 6 in Chapter 5, §5.

(d) Let $\mathcal{C}=\{$all finite intervals in $E^{1}\}$.

Suppose

$$\alpha : E^{1} \rightarrow E^{1}$$

has finite one-sided limits

$$\alpha(p+) \text { and } \alpha(p-)$$

at each $p \in E^{1}.$ The Lebesgue-Stieltjes $(L S)$ function

$$s_{\alpha} : \mathcal{C} \rightarrow E^{1}$$

(important for Lebesgue-Stieltjes integration) is defined as follows.

Set $s_{\alpha} \emptyset=0.$ For nonvoid intervals, including $[a, a]=\{a\},$ set

$$\begin{aligned} s_{\alpha}[a, b] &=\alpha(b+)-\alpha(a-), \\ s_{\alpha}(a, b] &=\alpha(b+)-\alpha(a+), \\ s_{\alpha}[a, b) &=\alpha(b-)-\alpha(a-), \text { and } \\ s_{\alpha}(a, b) &=\alpha(b-)-\alpha(a+). \end{aligned}$$

For the properties of $s_{\alpha}$ see Problem 7ff. , below.

(e) Let $m X$ be the mass concentrated in the part $X$ of the physical space $S$. Then $m$ is a nonnegative set function defined on

$$2^{S}=\{\text { all subsets } X \subseteq S\} \text{ (§3).}$$

If instead $m X$ were the electric load of $X,$ then $m$ would be sign changing.

II. The rest of this section is redundant for a "limited approach."

lemmas

Let $s : \mathcal{M} \rightarrow E$ be additive on $\mathcal{N} \subseteq \mathcal{M}.$ Let

$$A, B \in \mathcal{N}, A \subseteq B.$$

Then we have the following.

(1) If $|s A|<\infty$ and $B-A \in \mathcal{N},$ then

$$s(B-A)=s B-s A \text { ("subtractivity").}$$

(2) If $\emptyset \in \mathcal{N},$ then $s \emptyset=0$ provided $|s X|<\infty$ for at least one $X \in \mathcal{N}$.

(3) If $\mathcal{N}$ is a semiring, then $s A=\pm \infty$ implies $|s B|=\infty.$ Hence

$$|s B|<\infty \Rightarrow|s A|<\infty.$$

If further $s$ is semifinite then

$$s A=\pm \infty \Rightarrow s B=\pm \infty$$

(same sign).

Proof

(1) As $B \supseteq A,$ we have

$$B=(B-A) \cup A \text { (disjoint);}$$

so by additivity,

$$s B=s(B-A)+s A.$$

If $|s A|<\infty,$ we may transpose to get

$$s B-s A=s(B-A),$$

as claimed.

(2) Hence

$$s \emptyset=s(X-X)=s X-s X=0$$

if $X, \emptyset \in \mathcal{N},$ and $|s X|<\infty$.

(3) If $\mathcal{N}$ is a semiring, then

$$B-A=\bigcup_{k=1}^{n} A_{k} \text { (disjoint)}$$

for some $\mathcal{N}$-sets $A_{k};$ so

$$B=A \cup \bigcup_{k=1}^{n} A_{k} \text { (disjoint).}$$

By additivity,

$$s B=s A+\sum_{k=1}^{n} s A_{k};$$

so by our conventions,

$$|s A|=\infty \Rightarrow|s B|=\infty.$$

If, further, $s$ is semifinite, one of $\pm \infty$ is excluded. Thus $s A$ and $s B,$ if infinite, must have the same sign. This completes the proof.$\quad\square$

In §§1 and 2, we showed how to extend the notion of volume from intervals to a larger set family, preserving additivity. We now generalize this idea.

Theorem $\PageIndex{1}$

If

$$s : \mathcal{C} \rightarrow E$$

is additive on $\mathcal{C},$ an arbitrary semiring, there is a unique set function

$$\overline{s} : \mathcal{C}_{s} \rightarrow E,$$

additive on $\mathcal{C}_{s},$ with $\overline{s}=s$ on $\mathcal{C},$ i.e.,

$$\overline{s} X=s X \text { for } X \in \mathcal{C}.$$

We call $\overline{s}$ the additive extension of $s$ to $\mathcal{C}_{s}=\mathcal{C}_{s}^{\prime}$ (Corollary 2 in §3).

Proof

If $s \geq 0(s : \mathcal{C} \rightarrow[0, \infty]),$ proceed as in Lemma 1 and Corollary 2, all of §1.

The general proof (which may be omitted or deferred) is as follows.

Each $X \in \mathcal{C}_{s}^{\prime}$ has the form

$$X=\bigcup_{i=1}^{m} X_{i}(\text {disjoint}), \quad X_{i} \in \mathcal{C}.$$

Thus if $\overline{s}$ is to be additive, the only way to define it is to set

$$\overline{s} X=\sum_{i=1}^{m} s X_{i}.$$

This already makes $\overline{s}$ unique, provided we show that

$$\sum_{i=1}^{m} s X_{i}$$

does not depend on the particular decomposition

$$X=\bigcup_{i=1}^{m} X_{i}$$

(otherwise, all is ambiguous).

Then take any other decomposition

$$X=\bigcup_{k=1}^{n} Y_{k} \text { (disjoint)}, \quad Y_{k} \in \mathcal{C}.$$

Additivity implies

$$(\forall i, k) \quad s X_{i}=\sum_{k=1}^{n} s\left(X_{i} \cap Y_{k}\right) \text { and } s Y_{k}=\sum_{i=1}^{m} s\left(X_{i} \cap Y_{k}\right).$$

(Verify!) Hence

$$\sum_{i=1}^{m} s X_{i}=\sum_{i, k} s\left(X_{i} \cap Y_{k}\right)=\sum_{k=1}^{n} s Y_{k}.$$

Thus, indeed, it does not matter which particular decomposition we choose, and our definition of $\overline{s}$ is unambiguous.

If $X \in \mathcal{C},$ we may choose (say)

$$X=\bigcup_{i=1}^{1} X_{i}, X_{1}=X;$$

so

$$\overline{s} X=s X_{1}=s X;$$

i.e., $\overline{s}=s$ on $\mathcal{C},$ as required.

Finally, for the additivity of $\overline{s},$ let

$$A=\bigcup_{k=1}^{m} B_{k} \text { (disjoint)}, \quad A, B_{k} \in \mathcal{C}_{s}^{\prime}.$$

Here we may set

$$B_{k}=\bigcup_{i=1}^{n_{k}} C_{k i} \text { (disjoint)}, \quad C_{k i} \in \mathcal{C}.$$

Then

$$A=\bigcup_{k, i} C_{k i} \text { (disjoint);}$$

so by our definition of $\overline{s}$,

$$\overline{s} A=\sum_{k, i} s C_{k i}=\sum_{k=1}^{m}\left(\sum_{i=1}^{n_{k}} s C_{k i}\right)=\sum_{k=1}^{m} \overline{s} B_{k},$$

as required.$\quad \square$

Continuity. We write $X_{n} \nearrow X$ to mean that

$$X=\bigcup_{n=1}^{\infty} X_{n}$$

and $\left\{X_{n}\right\} \uparrow,$ i.e.,

$$X_{n} \subseteq X_{n+1}, \quad n=1,2, \ldots.$$

Similarly, $X_{n} \searrow X$ iff

$$X=\bigcap_{n=1}^{\infty} X_{n}$$

and $\left\{X_{n}\right\} \downarrow,$ i.e.,

$$X_{n} \supseteq X_{n+1}, \quad n=1,2, \ldots.$$

In both cases, we set

$$X=\lim _{n \rightarrow \infty} X_{n}.$$

This suggests the following definition.

Definition 3

A set function $s : \mathcal{M} \rightarrow E$ is said to be

(i) left continuous (on $\mathcal{M})$ iff

$$s X=\lim _{n \rightarrow \infty} s X_{n}$$

whenever $X_{n} \nearrow X$ and $X, X_{n} \in \mathcal{M}$;

(ii) right continuous iff

$$s X=\lim _{n \rightarrow \infty} s X_{n}$$

whenever $X_{n} \searrow X,$ with $X, X_{n} \in \mathcal{M}$ and $\left|s X_{j}\right|<\infty$.

Thus in case (i),

$$\lim _{n \rightarrow \infty} s X_{n}=s \bigcup_{n=1}^{\infty} X_{n}$$

if all $X_{n}$ and $\bigcup_{n=1}^{\infty} X_{n}$ are $\mathcal{M}$-sets.

In case (ii),

$$\lim _{n \rightarrow \infty} s X_{n}=s \bigcap_{n=1}^{\infty} X_{n}$$

if all $X_{n}$ and $\bigcap_{n=1}^{\infty} X_{n}$ are in $\mathcal{M},$ and $\left|s X_{1}\right|<\infty$.

Note 2. The last restriction applies to right continuity only. (We choose simply to exclude from consideration sequences $\left\{X_{n}\right\} \downarrow,$ with $\left|s X_{1}\right|=\infty;$ see Problem 4.)

Theorem $\PageIndex{2}$

If $s : \mathcal{C} \rightarrow E$ is $\sigma$-additive and semifinite on $\mathcal{C}, a$ semiring, then $s$ is both left and right continuous (briefly, continuous).

Proof

We sketch the proof for rings; for semirings, see Problem 1.

Left continuity. Let $X_{n} \nearrow X$ with $X_{n}, X \in \mathcal{C}$ and

$$X=\bigcup_{n=1}^{\infty} X_{n}.$$

If $s X_{n}=\pm \infty$ for some $n,$ then (Lemma 3)

$$s X=s X_{m}=\pm \infty \text { for } m \geq n,$$

since $X \supseteq X_{m} \supseteq X_{n};$ so

$$\operatorname{lims} X_{m}=\pm \infty=s X,$$

as claimed.

Thus assume all $s X_{n}$ finite; so $s \emptyset=0,$ by Lemma 2.

Set $X_{0}=\emptyset.$ As is easily seen,

$$X=\bigcup_{n=1}^{\infty} X_{n}=\bigcup_{n=1}^{\infty}\left(X_{n}-X_{n-1}\right) \text { (disjoint),}$$

and

$$(\forall n) \quad X_{n}-X_{n-1} \in \mathcal{C} \text { (a ring).}$$

Also,

$$(\forall m \geq n) \quad X_{m}=\bigcup_{n=1}^{m}\left(X_{n}-X_{n-1}\right) \text { (disjoint).}$$

(Verify!) Thus by additivity,

$$s X_{m}=\sum_{n=1}^{m} s\left(X_{n}-X_{n-1}\right),$$

and by the assumed $\sigma$-additivity,

$$\begin{aligned} s X=s \bigcup_{n=1}^{\infty}\left(X_{n}-X_{n-1}\right) &=\sum_{n=1}^{\infty} s\left(X_{n}-X_{n-1}\right) \\ &=\lim _{m \rightarrow \infty} \sum_{n=1}^{m} s\left(X_{n}-X_{n-1}\right)=\lim _{m \rightarrow \infty} s X_{m}, \end{aligned}$$

as claimed.

Right continuity. Let $X_{n} \searrow X$ with $X, X_{n} \in \mathcal{C}$,

$$X=\bigcap_{n=1}^{\infty} X_{n},$$

and

$$\left|s X_{1}\right|<\infty.$$

As $X \subseteq X_{n} \subseteq X_{1},$ Lemma 3 yields that

$$(\forall n) \quad\left|s X_{n}\right|<\infty$$

and $|s X|<\infty$.

As

$$X=\bigcap_{k=1}^{\infty} X_{k},$$

we have

$$(\forall n) \quad X_{n}=X \cup \bigcup_{k=n+1}^{\infty}\left(X_{k-1}-X_{k}\right) \text { (disjoint).}$$

(Verify!) Thus by $\sigma$-additivity,

$$(\forall n) \quad s X_{n}=s X+\sum_{k=n+1}^{\infty} s\left(X_{k-1}-X_{k}\right),$$

with $|s X|<\infty,\left|s X_{n}\right|<\infty$ (see above).

Hence the sum

$$\sum_{k=n+1}^{\infty} s\left(X_{k-1}-X_{k}\right)=s X_{n}-s X$$

is finite. Therefore, it tends to $0$ as $n \rightarrow \infty$ (being the "remainder term" of a convergent series). Thus $n \rightarrow \infty$ yields

$$\lim _{n \rightarrow \infty} s X_{n}=s X+\lim \sum_{k=n+1}^{\infty} s\left(X_{k-1}-X_{k}\right)=s X,$$

as claimed.$\quad \square$