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Is there an easy way to see that the Cantor function is not absolutely continuous that directly uses the definition of absolutely continuous?

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5 Answers 5

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A function $f: E \to \mathbb{R}$ is absolutely continuous on an interval $E$ if for every $\epsilon > 0$ there is a $\delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $E$ satisfies

$$ \sum_{k} |y_{k} - x_{k}| < \delta$$

then

$$\sum_{k} |f(y_{k}) - f(x_{k})| < \epsilon$$

Put in words, an absolutely continuous function does not fluctuate on a set of measure zero. To see that the Cantor function is not absolutely continuous, pick $\epsilon < 1$. Then, for every $\delta > 0$, I can find a collection of intervals $(x_{k},y_{k})$ that cover the Cantor points in $[0,1]$ such that

$$ \sum_{k} |y_{k} - x_{k}| < \delta$$

this is because the Cantor set has measure zero. However, since the Cantor function only changes on the Cantor set,

$$\sum_{k} |f(y_{k}) - f(x_{k})| = 1$$

and absolute continuity is violated.


More generally, note that the Cantor function is singular. It is easy to prove (and feels right intuitively) that an absolutely continuous, singular function must be constant. However, the Cantor function is far from constant.

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    $\begingroup$ Excellent answer $\endgroup$ Mar 28, 2012 at 6:55
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    $\begingroup$ The answer taken is not correct since one can not cover canter set by finite union of sets with arbitrarily small measure in total. One can check Royden (4ed) P120 for reference to this question. And the solution taken will become true (with some modification) after proveing Prob38 on P123. $\endgroup$
    – user49021
    Nov 11, 2012 at 16:33
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    $\begingroup$ The Cantor set $C$ is the intersection of sets $C_n$ constructed iteratively. We know that $C_n \supset C_{n+1}$. Moreover, the total length of the line segments in $C_n$ is $(2/3)^n$. Note that a cover for any $C_n$ is automatically a cover for $C$. Do you not think that I can take $N$ large enough to make $(2/3)^N$ smaller than some fixed $\delta$? $\endgroup$ Nov 11, 2012 at 18:44
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    $\begingroup$ Isaac Solomon: Well, in that case you probably should not write that finding a finite cover of $C$ with measure smaller than $\delta$ is possible because $C$ has measure zero. As you wrote yourself, this is not the case and is rather due to the specific construction of $C$. $\endgroup$
    – balu
    Dec 6, 2015 at 16:43
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    $\begingroup$ @balu It is not difficult to show that the Cantor set is compact. Because of Heine-Borel, it is then okay to say that because C has measure zero, we can find a finite cover of C with measure smaller than $\delta$. $\endgroup$ Jun 22, 2020 at 12:42
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One definition of absolutely continuous functions is that they map sets of measure zero to sets of measure zero. However, the cantor ternary function maps the cantor set (of measure zero) onto $[0,1]$.

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Let $f$ be the Cantor function. $f$ is increasing not negative, $f(0)=0$ and $f(1)=1$. Then $f$ is differentiable a.e. and since $f$ is constant on every interval removed in the construction of the Cantor set, $f^\prime=0$ a.e.

Once we assume that $f$ is absolutely continuous, we have $$\int_0^1 f^\prime=f(1)-f(0),$$ but this says that $0=1$. Thus $f$ can not be absolutely continuous.

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    $\begingroup$ Potential typo: "uniformly" should probably be "absolutely" (both times). $\endgroup$ Jun 11, 2012 at 5:57
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The variation of the Cantor function on each approximation $C_n$ of the Cantor set is $1$. The measure of $C_n$ goes to zero when $n\to\infty$. Ergo.

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A function is absolutely continuous if and only if it satisfies the (Lebesgue) fundamental theorem of calculus. Since $f'=0$ a.e., $f$ satisfies the fundamental theorem calculus if and only if

$$f(1)-f(0) = \int_{0}^{1} f' = 0 $$

which we know is false.

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  • $\begingroup$ This is certainly a short (and clever) treatment, but it can't be said it "directly uses the definition of absolutely continuous". It also repeats a point made in one of the previous answers. $\endgroup$
    – hardmath
    Jan 21, 2016 at 22:46
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    $\begingroup$ Ahh he wanted it direct, missed that. I kind of thought leo's answer didn't have the direct punch that makes it easy to follow for someone just learning the subject, so I decided to share this. $\endgroup$ Jan 21, 2016 at 22:54

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