The Cantor Function

Prove that the Cantor function $f$ is well-defined (the ambiguity of ternary expansions does not change its value), nondecreasing, and continuous on $[0,1]$ . For instance, the two expansions $0.0222\ldots_3 = 0.1_3$ must both give $f = 1/2$ .

0.0222\ldots_3:\ N = \infty,\ b = 0.0111\ldots_2 = \tfrac12 \qquad\text{vs}\qquad 0.1_3:\ N = 1,\ b_1 = 1 \Rightarrow f = \tfrac12

x < y,\ \text{first differing digit } m \Rightarrow f(x) \le S + 2^{-m} \le f(y)

|f - f_n| \le 2^{-n}\ \text{uniformly},\ f_n\ \text{piecewise linear} \Rightarrow f\ \text{continuous}

Stage 3:

C_3

has 8 intervals (=

2^3

) of total length

(2/3)^3 = 8/27 \approx 0.29630

— the staircase is already the finished

f

, with the flat step

f = 1/2

over

(1/3,\,2/3)

visible.

Show proof

Ternary expansions are unique except at points $k/3^m$ , which have exactly two: the terminating one $0.c_1\ldots c_{m-1}1$ and the repeating- $2$ one $0.c_1\ldots c_{m-1}0222\ldots_3$ . If some $c_k = 1$ with $k < m$ , both expansions see the first $1$ at the same position, so the $b$ -sequences agree. Otherwise the terminating expansion has $N = m$ with $b_m = 1$ , while the repeating- $2$ expansion has no $1$ at all, giving $b_m = 0$ and $b_k = 1$ for $k > m$ — but $\sum_{k>m} 2^{-k} = 2^{-m}$ , so the totals agree. Example: $0.0222\ldots_3$ has $N = \infty$ , hence $b = 0.0111\ldots_2 = 1/2$ ; $0.1_3$ has $N = 1$ , hence $b_1 = 1$ and $f = 1/2$ . So $f$ is well-defined.
The function $f$ is nondecreasing. Let $x < y$ and let $m$ be the first index where the (chosen) ternary expansions differ; then $(c_m(x),\, c_m(y))$ is one of $(0,1)$ , $(0,2)$ , $(1,2)$ . Write $S$ for the shared contribution of the first $m-1$ digits. Case $(0,1)$ : $x$ has $b_m = 0$ and tail at most $\sum_{k>m} 2^{-k} = 2^{-m}$ , so $f(x) \le S + 2^{-m}$ ; $y$ sees its first $1$ at $m$ , so $f(y) = S + 2^{-m}$ . Case $(0,2)$ : $f(x) \le S + 2^{-m}$ while $y$ has $b_m = 1$ and tail $\ge 0$ , so $f(y) \ge S + 2^{-m}$ . Case $(1,2)$ : $f(x) = S + 2^{-m}$ and $f(y) \ge S + 2^{-m}$ . Endpoint ambiguity was already settled by well-definedness, so $f(x) \le f(y)$ .
The function $f$ is continuous. On each closed rank- $n$ ternary interval $I$ , $f$ varies by at most $2^{-n}$ : if a $1$ occurs among the $n$ digits the tails vanish and $f$ is exactly constant on $I$ ; otherwise $f = \sum_{k\le n}(c_k/2)2^{-k} + \text{tail}$ with the tail between $0$ and $2^{-n}$ . Let $f_n$ be the continuous piecewise-linear interpolant of $f$ at the endpoints $k/3^n$ . For $x$ in a rank- $n$ interval $[l,r]$ , both $f(x)$ and $f_n(x)$ lie in $[f(l),\,f(r)]$ (by monotonicity), an interval of length at most $2^{-n}$ . Hence $f_n \to f$ uniformly, and a uniform limit of continuous functions is continuous, so $f$ is continuous on $[0,1]$ . Note $f(0) = 0$ and $f(1) = 1$ (reading $1 = 0.222\ldots_3$ ).

Prove that $f$ is constant on every open interval complementary to the Cantor set. Conclude that $f' = 0$ at every point of the complement, which has total length $1$ .

(a,b)\ \text{removed at stage } m:\quad a = 0.c_1\ldots c_{m-1}1_3,\ b = 0.c_1\ldots c_{m-1}2_3

x \in (a,b) \Rightarrow c_m(x) = 1 \Rightarrow f(x) = \sum_{k<m}\tfrac{c_k}{2}2^{-k} + 2^{-m}\ \text{(constant)}

\sum_{m \ge 1} 2^{m-1} 3^{-m} = \tfrac13\sum_{m\ge 1}\left(\tfrac23\right)^{m-1} = 1

Stage 5:

C_5

has 32 intervals (=

2^5

) of total length

(2/3)^5 = 32/243 \approx 0.13169

— the bars thin out, but the flat steps of the staircase (e.g.

f = 1/2

over

(1/3,\,2/3)

) stay put.

Show proof

The complement of the Cantor set is the disjoint union of the removed open middle thirds. Each interval removed at stage $m$ has the form $(a,b)$ with $a = 0.c_1\ldots c_{m-1}1_3$ (terminating) and $b = 0.c_1\ldots c_{m-1}2_3$ , where $c_1,\ldots,c_{m-1}$ contain no $1$ . Let $x \in (a,b)$ . Since $x > a$ , the $m$ -th ternary digit cannot be $0$ (that would give $x \le 0.c_1\ldots c_{m-1}0222\ldots_3 = a$ ); since $x < b$ , it cannot be $2$ (that would give $x \ge 0.c_1\ldots c_{m-1}2000\ldots_3 = b$ ). Hence $c_m(x) = 1$ , and with no $1$ before it, $N(x) = m$ . So $f(x) = \sum_{k<m}(c_k/2)2^{-k} + 2^{-m}$ , a constant independent of $x$ — $f$ is constant on every complementary interval. Numerically: $f \equiv 1/4$ on $(1/9,\,2/9)$ and $f \equiv 1/2$ on $(1/3,\,2/3)$ .
A function constant on an open interval has derivative $0$ at every point of it (the difference quotient is identically $0$ ). Hence $f'(x) = 0$ at every point $x$ of the complement — the staircase is flat "almost everywhere".
The total length of the complement sums the $2^{m-1}$ intervals of length $3^{-m}$ removed at stage $m$ : $\sum_{m\ge 1} 2^{m-1}3^{-m} = \frac13\sum_{m\ge 1}(2/3)^{m-1} = \frac13\cdot\frac{1}{1-2/3} = 1$ . So the set where $f' = 0$ has length $1$ : the staircase climbs a total of $1$ while moving only on the Cantor set, which has length $0$ .

Compute that $f(1) - f(0) = 1$ yet $\int_0^1 f' = 0$ , and prove that $f$ is therefore not absolutely continuous. This is why the Fundamental Theorem of Calculus needs a stronger hypothesis than "differentiable almost everywhere". (The definition of absolute continuity itself may be stated at idea level.)

\text{AC:}\quad \forall\varepsilon > 0\ \exists\delta > 0:\ \sum (y_i - x_i) < \delta \Rightarrow \sum |f(y_i) - f(x_i)| < \varepsilon

f(1) - f(0) = 1,\qquad \int_0^1 f' = 0\ \ (\text{since } f' = 0\ \text{a.e.})

\sum_{C_n\text{ intervals}} (r_i - l_i) = \left(\tfrac23\right)^n,\qquad \sum (f(r_i) - f(l_i)) = 1 \not< \tfrac12

Stage 4:

C_4

has 16 intervals (=

2^4

) of total length

(2/3)^4 = 16/81 \approx 0.19753

—

f

climbs from

0

to

1

on these thin bars alone.

Show proof

By problem 1, $f$ is continuous on $[0,1]$ with $f(0) = 0$ and $f(1) = 1$ , so $f(1) - f(0) = 1$ . On the other hand, Lebesgue's theorem says a monotone function is differentiable almost everywhere, and problem 2 gives $f' = 0$ wherever it exists — so $f' = 0$ a.e., and the (Lebesgue) integral $\int_0^1 f' = 0$ .
If $f$ were absolutely continuous, the Fundamental Theorem of Calculus would give $f(1) - f(0) = \int_0^1 f'$ . But $1 \neq 0$ — a contradiction. Hence $f$ is not absolutely continuous: continuity plus differentiability almost everywhere does not imply the FTC. "Differentiable almost everywhere" is not enough to recover a function by integration; absolute continuity is the sharp hypothesis.
This can also be seen directly. Absolute continuity means: for every $\varepsilon > 0$ there is $\delta > 0$ such that any finite family of disjoint intervals with total length below $\delta$ has total $|f|$ -increment below $\varepsilon$ . Take $\varepsilon = 1/2$ . The $2^n$ intervals of $C_n$ have total length $(2/3)^n \to 0$ , eventually below any $\delta$ . But $f$ is constant on the gaps between them and nondecreasing, so $\sum_i (f(r_i) - f(l_i)) = f(1) - f(0) = 1 \ge \varepsilon$ — every $\delta$ fails. The Cantor set is nearly $0$ in length yet covers all of $[0,1]$ in $f$ -value ( $f$ maps $C$ onto $[0,1]$ ).

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