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The complement of the Cantor set in $[0, 1]$ has Lebesgue measure one, computed from the removed intervals

Example

Assume the Axiom of Countable Choice and let $C$ be the Cantor set. Then the open intervals removed in its construction form a countable pairwise disjoint family whose total Lebesgue measure is

$\sum_{n = 0}^{\infty} 2^{n} 3^{- n - 1} = 1,$

so the complement $[0, 1] ∖ C$ has Lebesgue measure one.

Facts & Assumptions

Given: The Axiom of Countable Choice and the Cantor set $C$ with stages $(C_{n})_{n \in N}$ .

[L1]

The Cantor set is an uncountable subset of $R$ of Lebesgue measure zero (The Cantor set is an uncountable subset of $R$ of Lebesgue measure zero).

[F1]

There is a unique family $(C_{n})_{n \in N}$ of subsets of $R$ with $C_{0} = [0, 1]$ and $C_{n + 1} = \frac{1}{3} C_{n} \cup (\frac{2}{3} + \frac{1}{3} C_{n})$ for every $n \in N$ (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds).

[F2]

If $∣ r ∣ < 1$ then the series $\sum r^{k}$ converges (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Assuming countable choice, a box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

Verification

technique · direct

1.1

By induction from [F1], the $n$ -th stage removes exactly $2^{n}$ pairwise disjoint open intervals, each of length $3^{- n - 1}$ ; the case $n = 0$ is the single interval $(\frac{1}{3}, \frac{2}{3})$ .

F1algebra

2.1

By [L2], the union removed at stage $n$ has Lebesgue measure $2^{n} 3^{- n - 1}$ , so the full removed set has measure $\sum_{n = 0}^{\infty} 2^{n} 3^{- n - 1} = \frac{1}{3} \sum_{n = 0}^{\infty} (\frac{2}{3})^{n} = 1$ by [F2].

step 1.1F2L2algebra

3.1

The complement of $C$ in $[0, 1]$ is exactly the union of those removed intervals, so step 2.1 computes $λ_{1} ([0, 1] ∖ C) = 1$ ; this agrees with [L1], which already gives $λ_{1} (C) = 0$ .

step 2.1L1∎

Depends on

Used by

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Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Example 2.14 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.2.9 (standard reference, not scraped)

The complement of the Cantor set in [0,1][0,1] has Lebesgue measure one, computed from the removed intervals