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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

1.1F1algebra

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 $(\frac{1}{3}, \frac{2}{3})$ .

2.1step 1.1F2L2algebra

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].

3.1step 2.1L1∎

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 $λ_{1} (C) = 0$ .

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

39 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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