Extending the Reals

Having discussed several results in discrete probability, we now turn our eyes to continuous probability. It’s a nontrivial task, and we don’t need to stray too far to see why. Recall that counting measure: given any finite set $\Omega$ , the map $| \cdot | : \mathcal P(\Omega) \to [0, \infty)$ counts the number of elements in a subset (e.g. $|K|$ counts the number of elements in $K$ ).

What is $|\mathbb N|$ ? Intuitively, since $\{1,\dots,n\} \subseteq \mathbb N$ , if we accept the properties of a measure, we must conclude that $|\mathbb N| \geq |\{1,\dots,n\}| = n$ . However, since $n \in \mathbb N$ is arbitrary, we must conclude that $|\mathbb N|$ is not finite. Can we say that it is infinite? What do we even mean by infinite?

The answer is yes. We will extend the reals in a meaningful way so as to allow for infinite measures. We need to control this extension, but extend we can, so that we can say that $|\mathbb N| = \infty$ in a meaningful way. Later on, we want to say that the length $\lambda(\mathbb R)$ of $\mathbb R$ also equals $\infty$ .

Before discussing measures, it helps to discuss $\sigma$ -algebras in their unadulterated countably infinite setting.

Definition 1. For any set $\Omega$ , a collection of subsets $\mathcal F \subseteq \mathcal P(\Omega)$ is a $\sigma$ -algebra on $\Omega$ if it satisfies the following properties:

$\emptyset \in \Omega$ ,
for any $K \in \mathcal F$ , $\Omega \backslash K \in \mathcal F$ ,
for any $K_1,K_2,\dots \in \mathcal F$ , $\bigcup_{i=1}^\infty K_i \in \mathcal F$ .

For example, $\mathcal P(\Omega)$ is a $\sigma$ -algebra on $\Omega$ . The pair $(\Omega, \mathcal F)$ is called a measurable space. A function $f : (\Omega, \mathcal F) \to (\Psi, \mathcal G)$ is called $\mathcal F/\mathcal G$ -measurable if for any $K \in \mathcal G$ , $f^{-1}(K) \in \mathcal F$ .

Lemma 1. For any $K_1,K_2,\dots \in \mathcal F$ , $\bigcap_{i=1}^\infty K_i \in \mathcal F$ .

Lemma 2. For any $\mathcal K \subseteq \mathcal P(\Omega)$ , there exists a unique $\sigma$ -algebra $\sigma(\mathcal K)$ that contains $\mathcal K$ . Furthermore, $\sigma(\mathcal K)$ is the “smallest” in the following sense: for any $\sigma$ -algebra $\mathcal F$ containing $\sigma(\mathcal K)$ , $\sigma(\mathcal K) \subseteq \mathcal F$ .

Proof. Let $\Sigma$ denote all $\sigma$ -algebras that contain $K$ , which is nonempty since $\mathcal P(\Omega) \in \Sigma$ . Verify that

$\displaystyle \sigma(\mathcal K) := \bigcap_{\mathcal F \in \Sigma} \mathcal F$

is a $\sigma$ -algebra, and by construction, it must be the smallest.

We call $\sigma(\mathcal K)$ the $\sigma$ -algebra generated by $\mathcal K$ .

Example 1. For any $n$ , let $\mathcal B$ denote the usual topological basis on $\mathbb R^n$ (resp. $\mathbb R^\omega)$ generated by the Euclidean metric (resp. product topology). For example $\mathcal B = \{(a,b) : a, b \in \mathbb Q\}$ . Define the Borel $\sigma$ -algebra of $\mathbb R^n$ by $\frak{B}(\mathbb R^n) := \sigma(\mathcal B)$ . Define $\frak{B}(\mathbb R^\omega)$ are similarly. Henceforth, we regard $(\mathbb R^n, \frak{B}(\mathbb R^n))$ as a measurable space.

Henceforth, let $(\Omega, \mathcal F)$ be a measurable space.

We now want to set up a meaningful arithmetic for $[0, \infty]$ that agrees with past intuitions. If we can do so, then we can define the measure on $(\Omega, \mathcal F)$ .

Definition 2. Assume that we have defined $[0, \infty]$ . A measure on $(\Omega, \mathcal F)$ is a map $\mu : \mathcal F \to [0, \infty]$ that satisfies the following properties:

$\mu(\emptyset) = 0$ ,
$\mu\left(\bigsqcup_{i=1}^\infty K_i \right) = \sum_{i=1}^\infty \mu(K_i)$ .

We call $(\Omega, \mathcal F, \mu) \equiv \Omega$ a measure space and abbreviate to the right-hand side when there is no ambiguity. We call $\mu = \mathbb P$ a probability measure if $\mu(\Omega) = 1$ .

To construct $[0, \infty]$ meaningfully, we will take inspiration from the notation

$\displaystyle \sum_{i=1}^\infty a_i := \lim_{n \to \infty} \sum_{i=1}^n a_i = \infty,\quad a_i \geq 0,$

whenever the left-hand side diverges. In particular, for any $c \geq 0$ ,

$\displaystyle \sum_{i=1}^\infty (c \cdot a_i) = c \cdot \sum_{i=1}^\infty a_i = \left( \sum_{i=1}^\infty a_i \right) \cdot c,$

where all sides either converge or diverge. In the case $c = 0$ , the left-hand side converges to $0$ , and so the right-hand side must converge to $0$ as well. In the case $c > 0$ , if one side diverges, so does the other.

Definition 3. For any $a \in [0, \infty]$ , define multiplication as follows:

$\begin{aligned} 0 \cdot a &= a \cdot 0 = 0, \\ \infty \cdot a, &= a \cdot \infty = \infty. \end{aligned}$

Furthermore, if $a \neq \infty$ ,

$\displaystyle \frac{\infty}{a} := \frac 1a \cdot \infty = \infty,\quad a > 0.$

Addition works similarly. For real numbers $a_i, b_i \geq 0$ ,

$\displaystyle \sum_{i=1}^\infty (a_i + b_i) = \sum_{i=1}^\infty a_i + \sum_{i=1}^\infty b_i.$

If both sums on the left-hand side converge, so does the right-hand side. On the other hand, if at least one side diverges, so does the right-hand side. This yields the addition arithmetic for $[0, \infty]$ .

Definition 4. For any $a \in [0, \infty]$ , define addition as follows:

$a + \infty = \infty + a = \infty.$

What about negative numbers? Either we stick to the usual infinite series interpretation, or we extend one of its implications. We do a bit of both as follows. Given $a_i \geq 0$ , $-a_i \leq 0$ , and the equation

$\displaystyle \sum_{i=1}^\infty (-a_i) = (-1) \cdot \sum_{i=1}^\infty a_i$

still makes sense if both sides converge or diverge. As such, we define $-\infty := (-1) \cdot \infty$ . To agree with real number arithmetic, we also agree that

$-(-\infty) = (-1) \cdot ((-1)\cdot \infty) = ((-1) \cdot (-1)) \cdot \infty = \infty.$

Denote $[-\infty, \infty] := \mathbb R \cup \{-\infty, \infty\}$ . Using similar reasoning, we make similar definitions to complete our construction.

Definition 5. Define addition on $[-\infty, \infty]$ as follows:

$a \pm \infty = \infty \pm a = \infty,\quad a \in (-\infty, \infty).$

Furthermore, $(\pm \infty) + (\pm \infty) = \pm \infty$ . For subtraction, define $a - b := a + (-b)$ whenever the right-hand side is well-defined (e.g. we leave $\infty -\infty$ undefined).

For multiplication, we define for $a \in [-\infty, \infty]$ ,

$a \cdot \infty = \infty \cdot a = \begin{cases} \infty, & a \in (0, \infty], \\ -\infty, & a \in [-\infty, 0), \\ 0, & a = 0. \end{cases}$

For division, define $1/(\pm {\infty}) := 0$ and $a/b := a \cdot (1/b)$ whenever $b \neq 0$ and at least one of $a,b$ is finite. It gets too unhelpfully complicated otherwise.

As such, we are not saying that $\infty$ is a real infinity. We are abbreviating our interpretations of the various kinds of sums that arise.

Now that we have properly defined $[0, \infty]$ , we can officially declare the counting measure $|\cdot | : \mathcal P(\mathbb N) \to [0, \infty]$ as a properly meaningful measure, even an infinite one. We can meaningfully say that $|\mathbb N| = \infty$ , and derive many useful properties for measures.

Lemma 3. Let $\mu : \mathcal F \to [0, \infty]$ be a measure on $(\Omega, \mathcal F)$ .

For any $K, L \in \mathcal F$ such that $K \subseteq L$ , $\mu(K) \leq \mu(L)$ . Furthermore, if $\mu(K) < \infty$ , then $\mu(L \backslash K) = \mu(L) - \mu(K)$ .
If $K_1 \subseteq K_2 \subseteq \cdots$ , then $\mu\left( \bigcup_{i=1}^\infty K_i \right) = \lim_{n \to \infty} \mu(K_n)$ .
If $K_1 \supseteq K_2 \supseteq \cdots$ and $\mu(K_1) < \infty$ , then $\mu\left( \bigcap_{i=1}^\infty K_i \right) = \lim_{n \to \infty} \mu(K_n)$ .

Proof. The first claim is immediate from $K \sqcup L \backslash K$ . The second claim comes from the observation that if we defined $L_i$ via

$L_1 = K_1,\quad L_{i+1} = K_{i+1} \backslash K_i,$

then $\bigcup_{i=1}^n K_i = \bigsqcup_{i=1}^n L_i$ , even when $n = \infty$ . The third claim comes from defining $L_i := L_1 \backslash K_i$ . Applying the second result yields

$\displaystyle \mu(L_1) - \mu \left( \bigcap_{i=1}^\infty K_i \right) = \mu \left( L_1 \backslash \bigcap_{i=1}^\infty K_i \right) = \mu\left( \bigcup_{i=1}^\infty L_i \right) = \lim_{n \to \infty} \mu(L_n).$

Therefore,

$\displaystyle \mu \left( \bigcap_{i=1}^\infty K_i \right) = \mu(L_1) - \lim_{n \to \infty} \mu(L_n) = \lim_{n \to \infty} \mu(L_1 \backslash L_n) = \lim_{n \to \infty} \mu(K_n).$

Lemma 4. There does not exist a probability measure on $\mathcal P(\mathbb N)$ with the following property: there exists some $p \in [0, 1]$ such that for any $n \in \mathbb N$ , $\mathbb P(\{n\}) = p$ .

Proof. Fix $p \in (0, 1]$ . For any integer $n > 1/p$ , $\mathbb P(\{0,\dots, n-1\}) = np > 1$ , which implies that

$1 = \mathbb P(\mathbb N) \geq np > 1,$

a contradiction. On the other hand, if $p = 0$ , then

$\displaystyle 1 = \mathbb P(\mathbb N) = \sum_{i=0}^\infty \mathbb P(\{i\}) = \lim_{n \to \infty} \sum_{i=0}^n \mathbb P(\{i\}) = \lim_{n \to \infty} \sum_{i=0}^n 0 = \lim_{n \to \infty} 0 = 0,$

a contradiction.

There are applications of such a definition, even to the more “finite” probability theory. Of course, however, such applications invoke a price for the infinite. We will construct such probability spaces the next time, before we turn to making sense of the length of subsets of $\mathbb R$ .

—Joel Kindiak, 4 Jul 25, 1357H

KindiakMath

Extending the Reals

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Extending the Reals

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