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Volterra's function is differentiable everywhere with bounded derivative, but its derivative is not Riemann integrable

Counterexample

There is a differentiable function $W : [0, 1] \to R$ whose derivative is bounded but not Riemann integrable.

Let $S$ be the Smith--Volterra--Cantor set. Put $g (0) = 0$ and $g (t) = t^{2} \sin (1 / t)$ for $t > 0$ . Choose a continuously differentiable cutoff $χ : [0, \infty) \to [0, 1]$ with $χ = 1$ on $[0, 1 / 2]$ and $χ = 0$ on $[1, \infty)$ . For each removed component $(u, v)$ of $[0, 1] ∖ S$ , set $r = (v - u) / 3$ and define

$W (x) = g (x - u) χ ((x - u) / r) + g (v - x) χ ((v - x) / r) (u < x < v),$

and put $W (x) = 0$ for $x \in S$ . The two summands have disjoint interiors of support. Then $W$ is differentiable everywhere, $W^{'} = 0$ on $S$ , and $W^{'}$ is bounded. Nevertheless every interval meeting $S$ has oscillation of $W^{'}$ at least $1 / 2$ , so $W^{'}$ fails the Riemann criterion.

Facts & Assumptions

Given: The fat Cantor set $S$ and the displayed construction.

[L1]

At every stage the Smith--Volterra--Cantor construction removes a nonempty open middle interval from each retained interval, and the retained interval lengths are at most $2^{- n}$ at stage $n$ (The Smith-Volterra-Cantor set: the same construction removing, at stage $n \geq 1$ , an open middle interval of length $4^{- n}$ from each of the $2^{n - 1}$ remaining intervals).

[L2]

The set $S$ is closed, nowhere dense, and every interval cover of $S$ has total length at least $1 / 2$ (The Smith-Volterra-Cantor set is compact, perfect and nowhere dense, and does not have measure zero).

[L3]

The product and chain rules and the derivatives of sine and cosine give $g^{'} (t) = 2 t \sin (1 / t) - \cos (1 / t)$ for $t > 0$ (Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , The derivatives of sine and cosine are cosine and minus sine).

[L4]

Sine and cosine have absolute value at most $1$ ; $π > 0$ , and their values at integer multiples of $π$ alternate by the quarter-turn and shift formulas (Parity and the Pythagorean identity for sine and cosine, Pi as twice the smallest positive zero of cosine, Quarter-turn values and shifts by pi/2 and pi).

[L5]

Reciprocals of positive natural numbers tend to $0$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L6]

Oscillation on a set is the supremum of $∣ f (x) - f (y) ∣$ , and it is monotone under inclusion (The oscillation $ω_{f} (S) = \sup {∣ f (x) - f (y) ∣ : x, y \in S}$ of $f$ on a set and the oscillation $ω_{f} (c) = \inf_{δ > 0} ω_{f} (A \cap N_{δ} (c))$ at a point, both taken in the extended reals).

[L7]

A bounded function is Riemann integrable if and only if it has partitions with arbitrarily small upper-minus-lower sum (Riemann's criterion: a bounded $f$ on $[a, b]$ is Darboux integrable if and only if for every real $ε > 0$ there is a partition $P$ with $U (f, P) - L (f, P) < ε$ ).

Verification

technique · construction

1.1

A concrete cutoff is obtained by taking $χ (s) = 1$ for $s \leq 1 / 2$ , $χ (s) = 1 - 3 (2 s - 1)^{2} + 2 (2 s - 1)^{3}$ for $1 / 2 \leq s \leq 1$ , and $χ (s) = 0$ for $s \geq 1$ ; the values and first derivatives agree at both joins.

constructalgebra

2.1

On a gap $(u, v)$ the two supports lie in $(u, u + r]$ and $[v - r, v)$ and are disjoint because $2 r < v - u$ . Each summand and its derivative vanish at its cutoff join, so $W$ is differentiable throughout every gap.

step 1.1algebra

3.1

By [L4], $∣ g (t) ∣ \leq t^{2}$ . If $x$ lies in a gap, then $∣ W (x) ∣ \leq dist (x, S)^{2}$ ; if $s \in S$ , this is at most $∣ x - s ∣^{2}$ . Hence $W^{'} (s) = 0$ for every $s \in S$ , including $0$ and $1$ .

givenstep 2.1L4

4.1

By [L3]--[L4], $∣ g^{'} (t) ∣ \leq 2 t + 1 \leq 3$ on $0 < t \leq 1$ . Differentiating inside a support gives a sum of $g^{'} χ$ and $g χ^{'} / r$ , and $∣ g (t) ∣ / r \leq t^{2} / r \leq r \leq 1$ on $0 \leq t \leq r$ . Since $χ^{'}$ is bounded on its two polynomial pieces, one constant bounds $W^{'}$ on all gaps and on $S$ .

step 1.1step 3.1L3L4

4.2

Let $I$ be a nondegenerate closed interval meeting $S$ . If $S$ meets the interior of $I$ , choose such a point and then, using the shrinking bound in [L1], a retained interval around it contained in $I$ ; its next-stage middle gap lies in $I$ . If $S$ does not meet the interior, an endpoint of $I$ lies in $S$ and the interior lies in one removed gap, so gap points approach that endpoint from inside $I$ . In either case the endpoint recursion in [L1] keeps the relevant gap endpoint in every later retained stage, hence in $S$ . On its adjacent half-support $χ = 1$ . By [L3]--[L5], the points at distances $1 / (2 π k)$ and $1 / ((2 k + 1) π)$ from that endpoint eventually lie in the half-support and give derivative values $- 1$ and $1$ , whereas step 3.1 gives value $0$ at the endpoint. Thus $ω_{W^{'}} (I) \geq 1 / 2$ .

step 3.1L1L2L3L4L5L6

5.1

Fix any partition and retain its subintervals that meet $S$ . These finitely many closed intervals cover $S$ , so their total lengths are at least $1 / 2$ by [L2]. Step 4.2 therefore gives $U (W^{'}, P) - L (W^{'}, P) \geq (1 / 2) (1 / 2) = 1 / 4$ .

step 4.2L2L6

6.1

The derivative is bounded by step 4.1, but the fixed lower bound in step 5.1 contradicts [L7]. Hence $W^{'}$ is not Riemann integrable.

step 4.1step 5.1L7∎

Depends on

Used by

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Sources

W. Chen, The Cantor Set Before Cantor, Sections 3.2--3.3 (standard reference, not scraped)