Lebesgue Measure

The cheapest interval cover prices a set's size. A single point, and even all the rationals, cost nothing.

Length measurement extends to sets like points and rationals. Among all covers, take the smallest total length. That smallest total is the set's size.

For example, cover a set with pieces of 🧵 string; the shortest total string length is the set's size. A single point needs almost no string, so its size is zero.

Examples & Exercises

Prove the outer measure $m^*(\{x\}) = 0$ .

m^*(\{x\}) \leq \varepsilon

\{0.5\}

— covered by a single interval of length

\varepsilon = 0.1

Show proof

For any $\varepsilon > 0$ , the interval $(x - \varepsilon/2,\, x + \varepsilon/2)$ is a countable interval cover of $\{x\}$ with total length $\varepsilon$ .
Since $m^*$ is the infimum of the total lengths over all covers, $m^*(\{x\}) \le \varepsilon$ .
Since $\varepsilon > 0$ was arbitrary, $m^*(\{x\}) = 0$ . A single point has outer measure zero.

Prove $m^*(\mathbb{Q}\cap[0,1]) = 0$ .

\sum_{n=1}^{\infty} \frac{\varepsilon}{2^n} = \varepsilon

\mathbb{Q}\cap[0,1]

— the

k

-th rational gets an interval of length

\varepsilon/2^{k+1}

(total

<\varepsilon

Show proof

$\mathbb{Q}\cap[0,1]$ is countable, so enumerate it as $\{q_1, q_2, \dots\}$ . Let $\varepsilon > 0$ be arbitrary.
Cover each $q_n$ by an interval $I_n$ of length $\varepsilon/2^n$ centered at $q_n$ . Then $\{I_n\}$ is a countable interval cover of $\mathbb{Q}\cap[0,1]$ with total length $\sum_{n=1}^{\infty} \varepsilon/2^n = \varepsilon$ .
Hence $m^*(\mathbb{Q}\cap[0,1]) \le \varepsilon$ , and since $\varepsilon$ is arbitrary, $m^*(\mathbb{Q}\cap[0,1]) = 0$ . Every countable set has outer measure zero.

Prove $A \subseteq B \implies m^*(A) \leq m^*(B)$ .

\mathbb{Q}\cap[0,1] \subseteq [0,1]

— the smaller set's cover total drops below

\varepsilon = 0.05

Show proof

Let $A \subseteq B$ . Any countable interval cover $\{I_n\}$ of $B$ also covers $A$ : each $I_n$ is an interval, so $\bigcup_n I_n \supseteq B \supseteq A$ .
Hence the family of covers of $B$ is contained in the family of covers of $A$ . The infimum over the larger family is smaller, so $m^*(A) \le m^*(B)$ .
In particular $\mathbb{Q}\cap[0,1] \subseteq [0,1]$ gives $m^*(\mathbb{Q}\cap[0,1]) = 0 \le 1 = m^*([0,1])$ . Outer measure is monotone.

Part of these learning paths

Examples & Exercises