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The Cantor set has measure zero, yet the Cantor function maps it onto all of $[0, 1]$ : a null set can have image an interval of length $1$

Example

Let $C$ be the Cantor set (The Cantor middle-thirds set as the intersection of the sets $C_{n}$ obtained by removing open middle thirds) and let $c : [0, 1] \to R$ be the Cantor function (The Cantor function on $[0, 1]$ , defined on the Cantor set through ternary digits and extended constantly across each removed interval). Then:

$C$ has measure zero (Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover));
$c [C] = [0, 1]$ : the Cantor function maps the Cantor set onto the whole of $[0, 1]$ (Injection, surjection, bijection);
$[0, 1]$ does not have measure zero (A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero).

So a continuous function can carry a set of measure zero onto a set that is not of measure zero, and indeed onto an interval of length $1$ : being null is not preserved by continuous images.

Facts & Assumptions

Given: The Cantor set $C$ and the Cantor function $c : [0, 1] \to R$ .

[L1]

$C$ has content zero and therefore 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, claim 2, Measure zero (a countable cover by intervals of total length below every $ε$ ) and content zero (a finite such cover)).

[L2]

$c$ is surjective onto $[0, 1]$ as a function on $[0, 1]$ , that is $c [[0, 1]] = [0, 1]$ ; and $c$ is constant on $[u, v]$ whenever $u < v$ lie in $C$ with $(u, v) \cap C = \emptyset$ , while every point of $[0, 1] ∖ C$ lies in the open interval $(u, v)$ of such a pair (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, claims 3 and 4).

[L3]

No set containing a bounded interval with two distinct endpoints has measure zero (A sequence of intervals covering $[a, b]$ has total length at least $b - a$ , so no interval of positive length has measure zero, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

$c$ is continuous on $[0, 1]$ (The Cantor function is continuous on $[0, 1]$ ).

Verification

technique · direct

1.1

Claim 1 is claim 2 of the Cantor set theorem.

1.2

Claim 3 is the nondegenerate-interval lemma applied to $[0, 1]$ , whose endpoints $0$ and $1$ are distinct.

1.3

$c [C] \subseteq [0, 1]$ , since $c [[0, 1]] = [0, 1]$ and $C \subseteq [0, 1]$ .

1.4

$[0, 1] \subseteq c [C]$ : let $y \in [0, 1]$ and take $x \in [0, 1]$ with $c (x) = y$ . If $x \in C$ we are done. Otherwise $x \in [0, 1] ∖ C$ , so $x$ lies in the open interval $(u, v)$ of a pair $u < v$ of points of $C$ with $(u, v) \cap C = \emptyset$ , and $c$ is constant on $[u, v]$ ; hence $y = c (x) = c (u)$ with $u \in C$ , so $y \in c [C]$ .

2.1

Claim 2 follows from steps 1.3 and 1.4: $c [C] = [0, 1]$ . With claims 1 and 3 this says that the null set $C$ has image the set $[0, 1]$ , which is not null, under the continuous function $c$ .

step 1.1step 1.2step 1.3step 1.4L4∎

Remarks

Nothing here contradicts any theorem about null sets. Measure zero is preserved by countable unions (A countable union of measure-zero sets has measure zero, by countable choice) and by passing to subsets, both of which are statements about covers. What this example shows is that it is not preserved by continuous images, and that is not a gap in any theorem above: no result in this library asserts that a continuous image of a null set is null.
The image is as large as it could possibly be. $c$ takes values in $[0, 1]$ (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), so $c [C] \subseteq [0, 1]$ always; claim 2 says the inclusion is an equality. So $C$ , which is null and nowhere dense (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), surjects onto an interval of length $1$ ; that $C$ is uncountable is proved independently as claim 4 of 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.
Where the increase happens. The remark The Cantor function is continuous and nondecreasing, climbs from $0$ to $1$ , and is constant on every interval removed in the construction of the Cantor set, so all of its increase happens on a set of measure zero records the complementary fact: $c$ is locally constant off $C$ , so all of its climb from $0$ to $1$ takes place on the null set $C$ , and this example says that the climb is complete.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

63 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Cantor function (Wikipedia) (standard reference, not scraped)
Cantor set (Wikipedia) (standard reference, not scraped)
The Cantor Function and the Cantor Set (University of Melbourne) (standard reference, not scraped)

The Cantor set has measure zero, yet the Cantor function maps it onto all of [0,1][0,1]: a null set can have image an interval of length 11