The Cantor Function

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 look at the (chosen) ternary expansions.

Let $k$ be the first index where either expansion has digit $1$ or the two expansions differ (such an index exists since $x \ne y$).

The first $k-1$ digits are shared and contain no $1$, so they lie in $\{0,2\}$; write $S = \sum_{i<k}(c_i/2)2^{-i}$ for their shared contribution.

The pair $(c_k(x),\, c_k(y))$ at index $k$ is one of $(1,1)$, $(0,1)$, $(0,2)$, $(1,2)$.

Case $(1,1)$: both see their first $1$ at $k$, so $f(x) = f(y) = S + 2^{-k}$ (e.g. $x = 0.1000\ldots_3$, $y = 0.1200\ldots_3$ gives $k = 1$ and $f(x) = f(y) = 1/2$).

Case $(0,1)$: $x$ has $b_k = 0$ and tail at most $\sum_{i>k} 2^{-i} = 2^{-k}$, so $f(x) \le S + 2^{-k}$; $y$ sees its first $1$ at $k$, so $f(y) = S + 2^{-k}$.

Case $(0,2)$: $f(x) \le S + 2^{-k}$ while $y$ has $b_k = 1$ and tail $\ge 0$, so $f(y) \ge S + 2^{-k}$.

Case $(1,2)$: $f(x) = S + 2^{-k}$ and $f(y) \ge S + 2^{-k}$.

In every case $f(x) \le f(y)$, and endpoint ambiguity was already settled by well-definedness.

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 $1 = 0.222\ldots_3$).

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 $0$.

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, problem 2 gives only that $f' = 0$ on the complement of the Cantor set.

But that complement has length $1$ and the Cantor set $C$ has measure $0$, so $f' = 0$ a.e. (the values on $C$ do not affect an a.e. statement).

Combined with Lebesgue's theorem that a monotone function is differentiable almost everywhere, the (Lebesgue) integral is $\int_0^1 f' = 0$.

An absolutely continuous function with $f' = 0$ a.e. must be constant, which is the heart of the contradiction below.

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 $[0,1]$).