Measuring the Reals

What is the length of the interval $[a, b]$ ? It is simply $b-a$ . That bit is trivial. But can we assign lengths to meaningful subsets $K \subseteq \mathbb R$ in general? This endeavour requires a lot more effort.

Lemma 1. Let $\mathcal F^0$ be the collection of finite (and without loss of generality, disjoint) unions of intervals of the form $[a, b)$ and $(-\infty, c)$ . Then $\mathcal F^0$ forms an algebra (i.e. it is closed under set complementation and finite unions).

The proof of Lemma 1 is relatively trivial, and it gives us an algebra on $\mathcal F^0$ . Clearly, we want to define $\lambda_0([a,b)) := b-a$ , $\lambda_0((-\infty, c)) = \infty$ . If we can prove that $\lambda_0$ is countably additive, then we can take advantage of the Carathéodory extension theorem to extend it to a proper measure $\lambda$ such that $\lambda|_{\mathcal F^0} = \lambda_0$ . We shall ensure this is the case in Lemma 2.

Lemma 2. Extend the map $\lambda_0 : \mathcal F^0 \to [0, \infty]$ by

$\displaystyle \lambda_0\left( \bigsqcup_{i=1}^\infty [a_i, b_i) \right) := \sum_{i=1}^\infty \lambda_0([a_i, b_i)).$

Then $\lambda_0$ is countably additive: for any pairwise disjoint $\{K_j\} \subseteq \mathcal F^0$ ,

$\displaystyle \lambda_0 \left( \bigsqcup_{j=1}^\infty K_j \right) = \sum_{j=1}^\infty \lambda_0(K_j),\quad \bigsqcup_{j=1}^\infty K_j \in \mathcal F^0.$

Proof. Firstly, for any $n$ , $\bigsqcup_{j=1}^n K_j \in \mathcal F^0$ . By monotonicity,

$\displaystyle \sum_{j=1}^n \lambda_0(K_j) = \lambda_0 \left( \bigsqcup_{j=1}^n K_j \right) \leq \lambda_0 \left( \bigsqcup_{j=1}^\infty K_j \right).$

Taking $n \to \infty$ , we have

$\displaystyle \lambda_0 \left( \bigsqcup_{j=1}^\infty K_j \right) \geq \sum_{j=1}^\infty \lambda_0(K_j).$

It remains to prove that $\lambda_0$ is countably subadditive:

$\displaystyle \lambda_0 \left( \bigsqcup_{j=1}^\infty K_j \right) \leq \sum_{j=1}^\infty \lambda_0(K_j) =: M.$

If $M = \infty$ , then we are done. Suppose $M < \infty$ . Under the assumption that $K := \bigsqcup_{j=1}^\infty K_j \in \mathcal F^0$ , $K$ is a finite union of intervals of the form $[a, b)$ . Assume $K = [a, b)$ then, so that

$\displaystyle \bigsqcup_{j=1}^\infty [a_j, b_j) = [a, b).$

Fix $\epsilon > 0$ . We observe that $\{ (a_j - \epsilon / 2^j, b_j + \epsilon / 2^j) \}$ forms an open cover for the compact space $[a, b]$ , and without loss of generality admits a finite sub-cover $\{ (a_j - \epsilon / 2^j , b_j + \epsilon / 2^j ) : j = 1, \dots, n\}$ . Hence,

$\displaystyle \bigsqcup_{j=1}^\infty [a_j, b_j) \subseteq \bigsqcup_{j=1}^n [a_j - \epsilon / 2^j , b_j + \epsilon / 2^j).$

By monotonicity,

$\begin{aligned} \lambda_0\left( \bigsqcup_{j=1}^\infty [a_j, b_j)\right) &\leq \sum_{j=1}^n \lambda_0([a_j - \epsilon / 2^j , b_j + \epsilon / 2^j)) \\ &= \sum_{j=1}^n \left( \lambda_0([a_j, b_j)) + \frac{\epsilon}{2^{j-1}} \right) \\ &= \sum_{j=1}^n \lambda_0([a_j, b_j)) + \epsilon \cdot \sum_{j=1}^n \frac{1}{2^{j-1}} \\ &\leq \sum_{j=1}^\infty \lambda_0([a_j, b_j)) + \epsilon \cdot \sum_{j=1}^\infty \frac{1}{2^{j-1}} \\ &= M + \epsilon \cdot \frac{1}{1-1/2} \\ &= M + 2\epsilon. \end{aligned}$

Taking $\epsilon \to 0^+$ yields the desired result.

At last, we can define the Lebesgue measure on $\mathbb R$ .

Theorem 1. There exists a $\sigma$ -algebra $\mathcal F \supseteq \frak{B}(\mathbb R) \supseteq \mathcal F^0$ on $\mathbb R$ such that there exists a measure $\lambda : \mathcal F \to [0, \infty]$ , called the Lebesgue measure, such that $\lambda|_{\mathcal F} = \lambda_0$ . In particular, $\lambda([a, b)) = b - a$ .

Proof. Apply Carathéodory’s extension theorem via Lemma 2. All that remains is to check the subset relation. We have

$\displaystyle [a, b) = \bigcap_{k=1}^\infty (a-1/k, b) \in \frak{B}(\mathbb R),\quad (a, b) = \bigcup_{k=1}^\infty [a+1/k, b) \in \sigma(\mathcal F^0).$

Since $\mathcal F$ is a $\sigma$ -algebra containing $\mathcal F^0$ and $\frak{B}(\mathbb R) = \sigma(\mathcal F^0)$ by the above argument, we have $\mathcal F^0 \subseteq \sigma(\mathcal F^0) \subseteq \frak{B}(\mathbb R) \subseteq \mathcal F$ .

Let’s now compute the lengths of various sets.

Example 1. We have the following lengths of the following subsets of $\mathbb R$ .

$\lambda(\{x\}) = 0$ ,
$\lambda((a, b)) = \lambda((a, b]) = \lambda([a, b]) = \lambda([a, b))$ ,
$\lambda(\mathbb Q) = 0$ ,
$\lambda(\mathbb R) = \infty$ .

Proof. For the first result, we use the outer measure formulation. For any $\epsilon > 0$ , $[x-\epsilon, x+\epsilon) \supseteq \{x\}$ . Therefore, $\lambda^*(\{x\}) \leq (x+\epsilon) - (x-\epsilon) = 2\epsilon$ . Taking $\epsilon \to 0^+$ , $\lambda(\{x\}) =\lambda^*(\{x\}) = 0$ .

For the second result, we use countable additivity to obtain

$\lambda ([a, b)) = \lambda (\{a\} \cup (a, b)) = \lambda(\{a\}) + \lambda((a, b)) = 0 + \lambda ((a, b)) = \lambda ((a, b)).$

For the third result, we first let $f : \mathbb N \to \mathbb Q$ denote a bijection (since the latter is countable), and denote $q_i := f(i)$ . Fix $\epsilon > 0$ . For each $i$ , $\lambda(\{ q_i \}) < \epsilon/2^i$ . Therefore, by countable additivity,

$\displaystyle 0\leq \lambda(\mathbb Q) = \lambda \left(\bigcup_{i=1}^\infty \{q_i\} \right) = \sum_{i=1}^\infty \lambda(\{q_i\}) \leq \sum_{i=1}^\infty \frac{\epsilon}{2^i} = \epsilon.$

Hence, $\lambda (\mathbb Q) = 0$ .

For the fourth result, for any $n \in \mathbb N$ , $\mathbb R \supseteq [0, n]$ so that monotonicity yields

$\lambda(\mathbb R) \geq \lambda([0, n]) = n.$

Taking $n \to \infty$ yields $\lambda(\mathbb R) = \infty$ .

The computation $\lambda(\mathbb Q) = 0$ is the motivation for the convention $0 \cdot \infty = 0$ , since we are interpreting $\infty$ in this context to mean countable infinity. If we adopted this convention, the computation simplifies to

$\displaystyle \lambda(\mathbb Q) = \sum_{i=1}^\infty \lambda(\{q_i\}) = \sum_{i=1}^\infty 0 = \infty \cdot 0 = 0 \cdot \infty = 0.$

What’s with all this hard work? Couldn’t we just define $\lambda$ for all subsets of $\mathbb R$ ? Sadly, we cannot.

Lemma 3 (Translational Invariance). For any $K \subseteq \mathbb R$ and $\alpha \in \mathbb R$ , define $\alpha + K := \{\alpha + x : x \in K\}$ . If $K \in \mathcal F$ , then $\alpha + K \in \mathcal F$ and $\lambda (\alpha + K) = \lambda(K)$ .

Proof. Fix $\epsilon > 0$ . By definition of $\lambda^* = \lambda$ on $\mathcal F$ , there exists a countable cover $\{K_i\} \equiv \{[a_i,b_i)\}$ of $K$ such that

$\displaystyle \sum_{i=1}^\infty \lambda^*(K_i) < \lambda(K) + \epsilon.$

We observe that $\{\alpha + K_i\}$ is a countable cover of $\alpha + K$ . Furthermore,

$\lambda (\alpha + K_i) = \lambda ([\alpha + a_i, \alpha + b_i)) = \lambda([a_i, b_i)) = \lambda (K_i).$

Therefore,

$\displaystyle \lambda^*(\alpha + K) \leq \sum_{i=1}^\infty \lambda^*(\alpha + K_i) = \sum_{i=1}^\infty \lambda^*(K_i) < \lambda^* (K) + \epsilon.$

Taking $\epsilon \to 0^+$ , we have $\lambda^*(\alpha + K) \leq \lambda^*(\alpha + K) \leq \lambda^*(K)$ . For the reverse inequality,

$\lambda^*(K) = \lambda^*(-\alpha + (\alpha + K)) \leq \lambda^*(\alpha + K).$

Therefore, $\lambda^*(\alpha + K) = \lambda^*(K) = \lambda(K)$ . It is not hard then to verify that $\alpha + K \in \mathcal F$ , so that

$\lambda(\alpha + K) = \lambda^*(\alpha + K) = \lambda(K).$

Theorem 2. There exists some set $V \subseteq \mathbb R$ , called a Vitali set, such that $\lambda(V)$ is undefined.

Proof. Define the equivalence relation $\sim$ on $\mathbb R$ by $x \sim y$ if and only if $x - y \in \mathbb Q$ . Then we obtain the quotient set $\mathbb R/\mathbb Q := \mathbb R/{\sim}$ whose take the form $x + \mathbb Q$ . Furthermore, for each $x \in \mathbb R$ , $(x + \mathbb Q) \cap [0, 1) \neq \emptyset$ . By the axiom of choice, select $\tilde x \in (x + \mathbb Q) \cap [0, 1)$ . Define the Vitali set by

$\displaystyle V := \bigcup_{x \in \mathbb R} \{\tilde x\}.$

Suppose for a contradiction that $V \in \mathcal F$ and $\lambda(V) \in [0, \infty]$ . Let $f : \mathbb N \to \mathbb Q \cap [-1, 1] =: \mathbb Q_{[-1,1]}$ be an enumeration of $\mathbb Q$ defined by $q_k := f(k)$ (as per the countability of $\mathbb Q$ ). Defining $V_k := q_k + V \in \mathcal F$ , it is not hard to check that $\{V_k\}$ is pairwise disjoint. Hence,

$\displaystyle \bigsqcup_{k \in \mathbb Q_{[-1,1]}} V_k \in \mathcal F.$

We leave it as an exercise to verify that

$\displaystyle [0, 1] \subseteq \bigsqcup_{k \in \mathbb Q_{[-1,1]}} V_k \subseteq [-1, 2].$

By monotonicity,

$\displaystyle 1 = \lambda([0, 1]) \leq \sum_{k \in \mathbb Q_{[-1,1]}} \lambda(V_k) \leq \lambda ([-1, 2]) = 3.$

Since $\lambda(V_k) = \lambda(q_k + V) = \lambda(V)$ by translational invariance, $1 \leq \infty \cdot \lambda(V) \leq 3$ . However, $\infty \cdot \lambda(V) \in \{0, \infty \}$ , a contradiction.

Theorem 2 is one key motivator for all of this measure theoretic language—so that we can still discuss integration and measure without running into contradictions. All of the sets we care about are still present in $\frak{B}(\mathbb R)$ and some of them appear in $\mathcal F$ as well, meaning that we have the correct logical basis to discuss these ideas in further detail.

—Joel Kindiak, 11 Jul 25, 1219H

KindiakMath

Measuring the Reals

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Measuring the Reals

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