Why is the Cantor function not absolutely continuous?

Solution 1:

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.

Solution 2:

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 $[0,1]$.

Solution 3:

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.

Solution 4:

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.