7.2: $\mathcal{C}_{\sigma}$-Sets. Countable Additivity. Permutable Series

We now want to further extend the definition of volume by considering countable unions of intervals, called $\mathcal{C}_{\sigma}$-sets ($\mathcal{C}$ being the semiring of all intervals in $E^{n}$).

We also ask, if $A$ is split into countably many such sets, does additivity still hold? This is called countable additivity or $\sigma$-additivity (the $\sigma$ is used whenever countable unions are involved).

We need two lemmas in addition to that of §1.

lemma 1

If $B$ is a nonempty interval in $E^{n},$ then given $\varepsilon>0,$ there is an open interval $C$ and a closed one $A$ such that

$$A \subseteq B \subseteq C$$

and

$$v C-\varepsilon<v B<v A+\varepsilon.$$

Proof

Let the endpoints of $B$ be

$$\overline{a}=\left(a_{1}, \ldots, a_{n}\right) \text { and } \overline{b}=\left(b_{1}, \ldots, b_{n}\right).$$

For each natural number $i$, consider the open interval $C_{i},$ with endpoints

$$\left(a_{1}-\frac{1}{i}, a_{2}-\frac{1}{i}, \ldots, a_{n}-\frac{1}{i}\right) \text { and }\left(b_{1}+\frac{1}{i}, b_{2}+\frac{1}{i}, \ldots, b_{n}+\frac{1}{i}\right).$$

Then $B \subseteq C_{i}$ and

$$v C_{i}=\prod_{k=1}^{n}\left[b_{k}+\frac{1}{i}-\left(a_{k}-\frac{1}{i}\right)\right]=\prod_{k=1}^{n}\left(b_{k}-a_{k}+\frac{2}{i}\right).$$

Making $i \rightarrow \infty,$ we get

$$\lim _{i \rightarrow \infty} v C_{i}=\prod_{k=1}^{n}\left(b_{k}-a_{k}\right)=v B.$$

(Why?) Hence by the sequential limit definition, given $\varepsilon>0,$ there is a natural $i$ such that

$$v C_{i}-v B<\varepsilon,$$

or

$$v C_{i}-\varepsilon<v B.$$

As $C_{i}$ is open and $\supseteq B,$ it is the desired interval $C.$

Similarly, one finds the closed interval $A \subseteq B.$ (Verify!)$\quad \square$

lemma 2

Any open set $G \subseteq E^{n}$ is a countable union of open cubes $A_{k}$ and also a disjoint countable union of half-open intervals.

(See also Problem 2 below.)

Proof

If $G=\emptyset,$ take all $A_{k}=\emptyset$.

If $G \neq \emptyset,$ every point $p \in G$ has a cubic neighborhood

$$C_{p} \subseteq G,$$

centered at $p$ (Problem 3 in Chapter 3, §12). By slightly shrinking this $C_{p},$ one can make its endpoints rational, with $p$ still in it (but not necessarily its center), and make $C_{p}$ open, half-open, or closed, as desired. (Explain!)

Choose such a cube $C_{p}$ for every $p \in G;$ so

$$G \subseteq \bigcup_{p \in G} C_{p}.$$

But by construction, $G$ contains all $C_{p},$ so that

$$G=\bigcup_{p \in G} C_{p}.$$

Moreover, because the coordinates of the endpoints of all $C_{p}$ are rational, the set of ordered pairs of endpoints of the $C_{p}$ is countable, and thus, while the set of all $p \in G$ is uncountable, the set of distinct $C_{p}$ is countable. Thus one can put the family of all $C_{p}$ in a sequence and rename it $\left\{A_{k}\right\}$:

$$G=\bigcup_{k=1}^{\infty} A_{k}.$$

If, further, the $A_{k}$ are half-open, we can use Corollary 1 and Note 3, both from §1, to make the union disjoint (half-open intervals form a semiring!).$\quad \square$

Now let $\mathcal{C}_{\sigma}$ be the family of all possible countable unions of intervals in $E^{n},$ such as $G$ in Lemma 2 (we use $\mathcal{C}_{s}$ for all finite unions). Thus $A \in \mathcal{C}_{\sigma}$ means that $A$ is a $\mathcal{C}_{\sigma}$-set, i.e.,

$$A=\bigcup_{i=1}^{\infty} A_{i}$$

for some sequence of intervals $\left\{A_{i}\right\}.$ Such are all open sets in $E^{n},$ but there also are many other $\mathcal{C}_{\sigma}$-sets.

We can always make the sequence $\left\{A_{i}\right\}$ infinite (add null sets or repeat a term!).

By Corollary 1 and Note 3 of §1, we can decompose any $\mathcal{C}_{\sigma}$-set $A$ into countably many disjoint intervals. This can be done in many ways. However, we have the following result.

Theorem $\PageIndex{1}$

If

$$A=\bigcup_{i=1}^{\infty} A_{i} \text { (disjoint)} =\bigcup_{k=1}^{\infty} B_{k} \text { (disjoint)}$$

for some intervals $A_{i}, B_{k}$ in $E^{n},$ then

$$\sum_{i=1}^{\infty} v A_{i}=\sum_{k=1}^{\infty} v B_{k}.$$

Thus we can (and do) unambiguously define either of these sums to be the volume $v A$ of the $\mathcal{C}_{\sigma}$-set $A.$

Proof

We shall use the Heine-Borel theorem (Problem 10 in Chapter 4, §6; review it!).

Seeking a contradiction, let (say)

$$\sum_{i=1}^{\infty} v A_{i}>\sum_{k=1}^{\infty} v B_{k},$$

so, in particular,

$$\sum_{k=1}^{\infty} v B_{k}<+\infty.$$

As

$$\sum_{i=1}^{\infty} v A_{i}=\lim _{m \rightarrow \infty} \sum_{i=1}^{m} v A_{i},$$

there is an integer $m$ for which

$$\sum_{i=1}^{m} v A_{i}>\sum_{k=1}^{\infty} v B_{k}.$$

We fix that $m$ and set

$$2 \varepsilon=\sum_{i=1}^{m} v A_{i}-\sum_{k=1}^{\infty} v B_{k}>0.$$

Dropping "empties" (if any), we assume $A_{i} \neq \emptyset$ and $B_{k} \neq \emptyset$.

Then Lemma 1 yields open intervals $Y_{k} \supseteq B_{k},$ with

$$v B_{k}>v Y_{k}-\frac{\varepsilon}{2^{k}}, \quad k=1,2, \ldots,$$

and closed ones $X_{i} \subseteq A_{i},$ with

$$v X_{i}+\frac{\varepsilon}{m}>v A_{i};$$

so

$$\begin{aligned} 2 \varepsilon=\sum_{i=1}^{m} v A_{i}-\sum_{k=1}^{\infty} v B_{k} &<\sum_{i=1}^{m}\left(v X_{i}+\frac{\varepsilon}{m}\right)-\sum_{k=1}^{\infty}\left(v Y_{k}-\frac{\varepsilon}{2^{k}}\right) \\ &=\sum_{i=1}^{m} v X_{i}-\sum_{k=1}^{\infty} v Y_{k}+2 \varepsilon. \end{aligned}$$

Thus

$$\sum_{i=1}^{m} v X_{i}>\sum_{k=1}^{\infty} v Y_{k}.$$

(Explain in detail!)

Now, as

$$X_{i} \subseteq A_{i} \subseteq A=\bigcup_{k=1}^{\infty} B_{k} \subseteq \bigcup_{k=1}^{\infty} Y_{k},$$

each of the closed intervals $X_{i}$ is covered by the open sets $Y_{k}$.

By the Heine-Borel theorem, $\bigcup_{i=1}^{m} X_{i}$ is already covered by a finite number of the $Y_{k},$ say,

$$\bigcup_{i=1}^{m} X_{i} \subseteq \bigcup_{k=1}^{p} Y_{k}.$$

The $X_{i}$ are disjoint, for even the larger sets $A_{i}$ are. Thus by Lemma 1(ii) in §1,

$$\sum_{i=1}^{m} v X_{i} \leq \sum_{k=1}^{p} v Y_{k} \leq \sum_{k=1}^{\infty} v Y_{k},$$

contrary to (1). This contradiction completes the proof.$\quad \square$

Corollary $\PageIndex{1}$

If

$$A=\bigcup_{k=1}^{\infty} B_{k} \text { (disjoint)}$$

for some intervals $B_{k},$ then

$$v A=\sum_{k=1}^{\infty} v B_{k}.$$

Indeed, this is simply the definition of $v A$ contained in Theorem 1.

Note 1. In particular, Corollary 1 holds if $A$ is an interval itself. We express this by saying that the volume of intervals is $\sigma$-additive or countably additive. This also shows that our previous definition of volume (for intervals) agrees with the definition contained in Theorem 1 (for $\mathcal{C}_{\sigma}$-sets).

Note 2. As all open sets are $\mathcal{C}_{\sigma}$-sets (Lemma 2), volume is now defined for any open set $A \subseteq E^{n}$ (in particular, for $A=E^{n}$).

Corollary $\PageIndex{2}$

If $A_{i}, B_{k}$ are intervals in $E^{n},$ with

$$\bigcup_{i=1}^{\infty} A_{i} \subseteq \bigcup_{k=1}^{\infty} B_{k},$$

then provided the $A_{i}$ are mutually disjoint,

$$\sum_{i=1}^{\infty} v A_{i} \leq \sum_{k=1}^{\infty} v B_{k}.$$

Proof

The proof is as in Theorem 1 (but the $B_{k}$ need not be disjoint here).

Corollary $\PageIndex{3}$ ("$\sigma$-subadditivity" of the volume)

If

$$A \subseteq \bigcup_{k=1}^{\infty} B_{k},$$

where $A \in \mathcal{C}_{\sigma}$ and the $B_{k}$ are intervals in $E^{n},$ then

$$v A \leq \sum_{k=1}^{\infty} v B_{k}.$$

Proof

Set

$$A=\bigcup_{i=1}^{\infty} A_{i}(\text { disjoint }), A_{i} \in \mathcal{C},$$

and use Corollary 2.$\quad \square$

Corollary $\PageIndex{4}$ ("monotonicity")

If $A, B \in \mathcal{C}_{\sigma},$ with

$$A \subseteq B,$$

then

$$v A \leq v B.$$

("Larger sets have larger volumes.")

This is simply Corollary 3, with $\bigcup_{k} B_{k}=B$.

Corollary $\PageIndex{5}$

The volume of all of $E^{n}$ is $\infty$ (we write $\infty$ for $+\infty$).

Proof

We have $A \subseteq E^{n}$ for any interval $A$.

Thus, by Corollary 4, $v A \leq v E^{n}$.

As $v A$ can be chosen arbitrarily large, $v E^{n}$ must be infinite.$\quad \square$

Corollary $\PageIndex{6}$

For any countable set $A \subset E^{n}, v A=0.$ In particular, $v \emptyset=0$.

Proof

First let $A=\{\overline{a}\}$ be a singleton. Then we may treat $A$ as a degenerate interval $[\overline{a}, \overline{a}].$ As all its edge lengths are $0,$ we have $v A=0$.

Next, if $A=\left\{\overline{a}_{1}, \overline{a}_{2}, \ldots\right\}$ is a countable set, then

$$A=\bigcup_{k}\left\{\overline{a}_{k}\right\};$$

so

$$v A=\sum_{k} v\left\{\overline{a}_{k}\right\}=0$$

by Corollary 1.

Finally, $\emptyset$ is the degenerate open interval $(\overline{a}, \overline{a});$ so $v \emptyset=0. \quad \square$

Note 3. Actually, all these propositions hold also if all sets involved are $\mathcal{C}_{\sigma}$-sets, not just intervals (split each $\mathcal{C}_{\sigma}$-set into disjoint intervals!).

Permutable Series. Since $\sigma$-additivity involves countable sums, it appears useful to generalize the notion of a series.

We say that a series of constants,

$$\sum a_{n},$$

is permutable iff it has a definite (possibly infinite) sum obeying the general commutative law:

Given any one-one map

$$u : N \stackrel{\mathrm{onto}}{\longleftrightarrow} N$$

($N=$ the naturals), we have

$$\sum_{n} a_{n}=\sum_{n} a_{u_{n}},$$

where $u_{n}=u(n)$.

(Such are all positive and all absolutely convergent series in a complete space $E;$ see Chapter 4, §13.) If the series is permutable, the sum does not depend on the choice of the map $u.$

Thus, given any $u : N \stackrel{\mathrm{onto}}{\longleftrightarrow} J$ (where $J$ is a countable index set) and a set

$$\left\{a_{i} | i \in J\right\} \subseteq E$$

(where $E$ is $E^{*}$ or a normed space), we can define

$$\sum_{i \in J} a_{i}=\sum_{n=1}^{\infty} a_{u_{n}}$$

if $\sum_{n} a_{u_{n}}$ is permutable.

In particular, if

$$J=N \times N$$

(a countable set, by Theorem 1 in Chapter 1, §9), we call

$$\sum_{i \in J} a_{i}$$

a double series, denoted by symbols like

$$\sum_{n, k} a_{k n} \quad(k, n \in N).$$

Note that

$$\sum_{i \in J}\left|a_{i}\right|$$

is always defined (being a positive series).

If

$$\sum_{i \in J}\left|a_{i}\right|<\infty,$$

we say that $\sum_{i \in J} a_{i}$ converges absolutely.

For a positive series, we obtain the following result.

Theorem $\PageIndex{2}$

(i) All positive series in $E^{*}$ are permutable.

(ii) For positive double series in $E^{*},$ we have

$$\sum_{n, k=1}^{\infty} a_{n k}=\sum_{n=1}^{\infty}\left(\sum_{k=1}^{\infty} a_{n k}\right)=\sum_{k=1}^{\infty}\left(\sum_{n=1}^{\infty} a_{n k}\right).$$

Proof

(i) Let

$$s=\sum_{n=1}^{\infty} a_{n} \text { and } s_{m}=\sum_{n=1}^{m} a_{n} \quad\left(a_{n} \geq 0\right).$$

Then clearly

$$s_{m+1}=s_{m}+a_{m+1} \geq s_{m};$$

i.e., $\left\{s_{m}\right\} \uparrow$, and so

$$s=\lim _{m \rightarrow \infty} s_{m}=\sup _{m} s_{m}$$

by Theorem 3 in Chapter 3, §15.

Hence $s$ certainly does not exceed the lub of all possible sums of the form

$$\sum_{i \in I} a_{i},$$

where $I$ is a finite subset of $N$ (the partial sums $s_{m}$ are among them). Thus

$$s \leq \sup \sum_{i \in I} a_{i},$$

over all finite sets $I \subset N$.

On the other hand, every such $\sum_{i \in I} a_{i}$ is exceeded by, or equals, some $s_{m}$. Hence in (4), the reverse inequality holds, too, and so

$$s=\sup \sum_{i \in I} a_{i}.$$

But sup $\sum_{i \in I} a_{i}$ clearly does not depend on any arrangement of the a_{i}. Therefore, the series $\sum a_{n}$ is permutable, and assertion (i) is proved.

Assertion (ii) follows similarly by considering sums of the form $\sum_{i \in I} a_{i}$ where $I$ is a finite subset of $N \times N,$ and showing that the lub of such sums equals each of the three expressions in (3). We leave it to the reader.$\quad \square$

A similar formula holds for absolutely convergent series (see Problems).