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The graph path $t \mapsto (t, c (t))$ of the Cantor function is rectifiable although its second coordinate is not absolutely continuous

Example

Let $c$ be the Cantor function and define $γ (t) = (t, c (t))$ on $[0, 1]$ . Then $γ$ is a rectifiable path, but its second coordinate $c$ is not absolutely continuous. Thus rectifiability, even together with continuity and coordinatewise monotonicity, does not imply absolute continuity.

Facts & Assumptions

Given: The Cantor set stages and Cantor function.

[L1]

The Cantor function is continuous, nondecreasing, and satisfies $c (0) = 0$ , $c (1) = 1$ (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]

A real function has bounded variation exactly when it is a difference of two nondecreasing functions (Jordan decomposition for functions of bounded variation), and a path is rectifiable exactly when all coordinates have bounded variation (A path in $R^{n}$ is rectifiable exactly when every coordinate has bounded variation).

[L3]

Absolute continuity requires small total input length to force small total endpoint increment over every finite family of intervals with disjoint interiors (Absolute continuity on a compact interval).

[L4]

At stage $n$ , the Cantor construction retains $2^{n}$ disjoint closed intervals of length $3^{- n}$ , indexed by the first $n$ ternary digits in ${0, 2}$ (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds, The Cantor set is exactly the set of $\sum_{k \geq 1} a_{k} 3^{- k}$ with every $a_{k} \in {0, 2}$ , and this gives a bijection with ${0, 1}^{N}$ ).

[L5]

The Cantor function reads those first $n$ ternary digits as binary digits, so its increments over the retained stage- $n$ intervals are all $2^{- n}$ and sum to $1$ (The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval, 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).

[L6]

$(2 / 3)^{n} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L7]

A vector-valued function is continuous exactly when all its coordinate functions are continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

Verification

technique · construction

1.1

The identity coordinate and $c$ are nondecreasing by [L1], hence have bounded variation by [L2]. Their pairing is continuous by componentwise continuity.

givenL1L2L7

1.2

The $2^{n}$ retained stage intervals have pairwise disjoint interiors and total length $2^{n} 3^{- n} = (2 / 3)^{n}$ , which tends to zero by [L4] and [L6].

L4L6

1.3

By [L5], the sum of the absolute increments of $c$ over those same intervals is $2^{n} 2^{- n} = 1$ for every $n$ .

2.1

Therefore [L2] makes $γ$ rectifiable; quantitatively its length is at most the sum of the two coordinate variations, namely $2$ .

step 1.1L2

3.1

Taking $ε = 1 / 2$ , every proposed $δ > 0$ is defeated by a sufficiently large $n$ : step 1.2 makes the total interval length below $δ$ , while step 1.3 leaves total image increment $1$ . This contradicts [L3], so $c$ is not absolutely continuous.

step 1.2step 1.3L3∎

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Sources

Cantor function (standard reference, not scraped)

The graph path t↦(t,c(t))t↦(t,c(t)) of the Cantor function is rectifiable although its second coordinate is not absolutely continuous