Lebesgue Measure

Outer measure is the cheapest countable interval cover of a set. A single point — and even all the rationals — cost nothing. Measurable sets and countable additivity need the Carathéodory construction (idea level only).

Part of these learning paths

• Integration

Examples & Exercises

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

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

Show proof

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

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

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

Show proof

1. $\mathbb{Q}\cap[0,1]$ is countable, so enumerate it as $\{q_1, q_2, \dots\}$. Let $\varepsilon > 0$ be arbitrary.
2. 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 $\sum_{n=1}^{\infty} \varepsilon/2^n = \varepsilon$.
3. 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 $A \subseteq B \implies m^*(A) \leq m^*(B)$.

Show proof

1. Let $A \subseteq B$. Any countable interval cover $\{I_n\}$ of $B$ also covers $A$: each $I_n$ is an interval, so $\bigcup_n I_n \supseteq B \supseteq A$.
2. 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 $m^*(A) \le m^*(B)$.
3. 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.

10. Integration