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.
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!)\(\quad \square\)
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.
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.\(\quad \square\)
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 \(A=E^{n}\)).
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).
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.\(\quad \square\)
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 \(\bigcup_{k} B_{k}=B\).
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.\(\quad \square\)
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.
(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).