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The Cantor measure is a singular atomless probability measure concentrated on the Cantor set

Statement

Assume the Axiom of Countable Choice. Let $μ_{c}$ be the Cantor measure of The Cantor measure and let $C \subseteq [0, 1]$ be the Cantor set. Then:

$μ_{c} ([0, 1]) = 1$ , so $μ_{c}$ is a probability measure;
$μ_{c} ({x}) = 0$ for every $x \in R$ , so $μ_{c}$ is atomless;
$μ_{c} (R ∖ C) = 0$ , so $μ_{c}$ is concentrated on the Cantor set;
since $λ (C) = 0$ , the measure $μ_{c}$ is singular with respect to Lebesgue measure.

Facts & Assumptions

Given: The Axiom of Countable Choice, the Cantor function $c$ , its extension $F_{c}$ , the Cantor measure $μ_{c} = μ_{F_{c}}$ , and the Cantor set $C$ .

[L1]

The Cantor function is continuous, nondecreasing, satisfies $c (0) = 0$ and $c (1) = 1$ , and is constant on every complementary gap of $C$ . (The Cantor function is continuous on $[0, 1]$ , The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set)

[L2]

The interval and singleton formulas hold for every Lebesgue-Stieltjes measure. (Interval formulas and atoms for a Lebesgue-Stieltjes measure)

[L3]

Assuming Countable Choice, the Cantor set has Lebesgue measure zero. (The Cantor set is an uncountable subset of $R$ of Lebesgue measure zero)

[L4]

Measures are continuous from below. (Continuity from below for measures)

Proof

technique · direct

1.1

By [L2],

L1L2

$μ_{c} ([0, 1]) = F_{c} (1) - F_{c} (0^{-}) = 1 - 0 = 1.$

Likewise, for every $n \geq 1$ , $μ_{c} ((1, n]) = F_{c} (n) - F_{c} (1) = 0$ and $μ_{c} ((- n, 0)) = F_{c} (0^{-}) - F_{c} (- n) = 0$ . So all mass lies in $[0, 1]$ , and $μ_{c}$ is a probability measure. [L1, L2]

2.1

If $x \in (0, 1]$ , then [L1] makes $F_{c} (x^{-}) = F_{c} (x)$ , so [L2] gives [step 1.1, L1, L2] $μ_{c} ({x}) = 0$ . The same holds for $x < 0$ and $x > 1$ because $F_{c}$ is constant on $(- \infty, 0)$ and on $(1, \infty)$ , and it holds at $x = 0$ because $F_{c} (0) = F_{c} (0^{-}) = 0$ . Hence $μ_{c}$ is atomless.

step 1.1L1L2

3.1

Every point of $[0, 1] ∖ C$ lies in a complementary gap $(u, v)$ of [step 2.1, L1, L2, L4, algebra] $C$ , and [L1] makes $c$ constant on $[u, v]$ . Therefore [L2] gives

$μ_{c} ((u, v)) = F_{c} (v^{-}) - F_{c} (u) = c (v) - c (u) = 0.$

Because $[0, 1] ∖ C$ is the countable union of those disjoint gaps, countable additivity gives $μ_{c} ([0, 1] ∖ C) = 0$ . Together with step 1.1 this yields $μ_{c} (R ∖ C) = 0$ . [step 2.1, L1, L2, L4, algebra]

4.1

By [L3], $λ (C) = 0$ , while step 3.1 shows that $μ_{c}$ is concentrated [step 1.1, step 2.1, step 3.1, L3] on $C$ . That is exactly the statement that $μ_{c}$ is singular with respect to Lebesgue measure.

step 1.1step 2.1step 3.1L3

5.1

Steps 1.1 through 4.1 prove the probability, atomless, concentration, and [step 1.1, step 2.1, step 3.1, step 4.1] singularity claims.

step 1.1step 2.1step 3.1step 4.1∎

Depends on

Used by

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Sources

John K. Hunter, Measure Theory, Example 2.37 (standard reference, not scraped)