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

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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.

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).

This page titled 7.2: $$\mathcal{C}_{\sigma}$$-Sets. Countable Additivity. Permutable Series is shared under a CC BY license and was authored, remixed, and/or curated by Elias Zakon (The Trilla Group (support by Saylor Foundation)) .