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The Cantor function is continuous, has derivative $0$ off a null set, and still rises from $0$ to $1$

Counterexample

Let $c : [0, 1] \to [0, 1]$ be the Cantor function. It is continuous, $c (0) = 0$ , and $c (1) = 1$ . At every point outside the Cantor set $C$ , the function is constant on a neighbourhood and hence $c^{'} = 0$ . Since $C$ has measure zero, $c^{'} = 0$ almost everywhere, but

$c (1) - c (0) = 1.$

Thus even continuity of the primitive and integrability of the zero function do not make an almost-everywhere derivative identity sufficient for Newton--Leibniz.

Facts & Assumptions

Given: The Cantor set $C$ and Cantor function $c$ .

[L1]

The Cantor function is continuous and satisfies $c (0) = 0$ , $c (1) = 1$ (The Cantor function is continuous on $[0, 1]$ , The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set).

[L2]

Every $x \in [0, 1] ∖ C$ lies in a neighbourhood on which $c$ is constant (The Cantor function is well defined, satisfies $c (x) \leq c (y)$ whenever $x \leq y$ , is surjective onto $[0, 1]$ , and is constant on every interval removed from the Cantor set).

[L3]

The Cantor set has measure zero (The Cantor set is compact, perfect, uncountable, nowhere dense and of measure zero, and it contains no interval of positive length, so its only nonempty connected subsets are single points).

[L4]

The derivative of a locally constant function is $0$ directly from the difference quotient (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set).

Verification

technique · direct

1.1L2L4

If $x \notin C$ , [L2] makes $c$ constant near $x$ , so every sufficiently local difference quotient is $0$ and [L4] gives $c^{'} (x) = 0$ .

2.1step 1.1L3

The exceptional set is contained in $C$ , which has measure zero by [L3]; therefore $c^{'} = 0$ almost everywhere.

3.1step 2.1L1

By [L1], $c (1) - c (0) = 1 - 0 = 1$ , despite step 2.1.

4.1step 3.1∎

Hence the implication from an almost-everywhere zero derivative to zero endpoint change is false without further regularity.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources

Cantor function (standard reference, not scraped)

The Cantor function is continuous, has derivative 00 off a null set, and still rises from 00 to 11