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Let $f:[0,1] \rightarrow \mathbb{R}$ be a continuous function. It is given that it has "no flat regions", i.e. its level sets have empty interiors. My question is, must it mean that each of its level sets has Lebesgue measure zero? In other words, can a continuous function with "no flat regions" have level sets which have positive Lebesgue measure?

Somewhere on this site I read a comment to a question that said a continuous function can have, as one of its level sets, the fat Cantor set, which has empty interior but positive Lebesgue measure. Unfortunately I can't find it now, and the comment also did not give an actual example.

Any example is most appreciated.

Apass.Jack
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Canine360
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    Given any closed set in $[0,1]$ say there is a continuous function that is $0$ precisely only on $F$ and for which the level set of any other point is at most countable - take $[0,1]-F$ union of disjoint at most countably many open intervals and define the function to be $0$ on $F$ and triangular with height $1$ on each interval (start at $0$ at an end go up linearly to $1$ at midpoint and back to $0$); note that you can make $F$ nowhere dense of arbitrary measure less than $1$ – Conrad Dec 24 '22 at 17:04
  • But if $F$ is a union of disjoint at most countably many open intervals, it does not have empty interior. You did mention "you can make F nowhere dense of arbitrary measure less than 1" but I do not know any such construction. I'm very much a beginner in measure theory. Any example would be most helpful. – Canine360 Dec 24 '22 at 17:08
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    The complement of $F$ being open is a disjoint union of open intevals; a fat Cantor set is nowhere dense – Conrad Dec 24 '22 at 17:10
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    Consider your favourite closed set $C\subseteq[0,1]$ with empty interior and positive Lebesgue measure (namely, the complement of the fat rationals). The map $f(x)=\inf_{y\in C}\lvert x-y\rvert$ is zero on $C$ and it takes any specific real value $c\ne 0$ at most twice in every connected component of $[0,1]\setminus C$. Therefore $f^{-1}(c)$ is $C$ for $c=0$ and a countable subset for $c\ne 0$. Either way, its interior will be empty. – Sassatelli Giulio Dec 24 '22 at 17:23
  • @Conrad You can't make the function triangular of height $1$ on every interval, because you still need to preserve global continuity. – Sassatelli Giulio Dec 24 '22 at 17:32
  • Yes something like height $1/n$ should work - realized it – Conrad Dec 24 '22 at 17:33
  • Meaning that if you construct inductively a set $F$ by removing intervals of length going to zero, one makes the height decreasing with the length of the interval excluded – Conrad Dec 24 '22 at 17:42
  • And then if $f_n$ is the obvious function at step $n$ (appropriate height on the intervals excluded up to step $n$ and zero everywhere else), $f_n$ converges uniformly to $f$ since $f_n, f_m$ different only the intervals excluded from step $n$ to $m$ and by definition difference it is at most $1/n$ there – Conrad Dec 24 '22 at 17:49
  • @Canine360 I just wrote an answer that constructs a fat Cantor set explicitly. – Apass.Jack Dec 25 '22 at 19:39
  • Thank you, I have accepted it. – Canine360 Dec 25 '22 at 20:40

2 Answers2

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Take $$\operatorname{dist}(x,{\mathcal F}):=\inf_{y\in\mathcal F}|x-y|$$ where $\operatorname{dist}$ is the distance function and $\mathcal F$ is the fat Cantor set. Since $\mathcal F$ is closed, this function is zero precisely on $\mathcal F$.

EDIT: The construction of a fat Cantor set is outlined here.

Akiva Weinberger
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    If you want (once-)differentiable, you could probably do$$\big({\operatorname{dist}(x,{\mathcal F}{\le x})\cdot\operatorname{dist}(x,{\mathcal F}{\ge x})}\big)^2,$$where ${\mathcal F}_{\le x}={y\in{\mathcal F}\mid y\le x}$ etc, which seems to me like it would smooth things out – Akiva Weinberger Dec 25 '22 at 19:36
  • Thank you so much! I had already accepted ApassJack's answer per his comment in the question thread. But I'm most grateful for both answers. This is a simple and easily understood construction, so that you especially for that. – Canine360 Dec 25 '22 at 20:42
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An idea

Suppose we have a fat Cantor set $\mathcal F\subset[0,1]$, which has empty interior. We can define a function $f$ that is $0$ on $\mathcal F$. The complement of $\mathcal F$ in $[0,1]$ is a union of open intervals. For each interval, let $f$ be a continuous function that looks like "$\land$" on it with $0$ at boundaries.

This idea is also stated by as Sassatelli Giulio.

A fat cantor set

For any $a=\sum _{k=1}^\infty a_{k}2^{-k^2}\in[0,1]$, where $a_k=\lfloor2^{k^2}a-2^{2k-1}\lfloor2^{(k-1)^2}a\rfloor\rfloor\in\{0,1, 2, \cdots, 2^{2k-1}-1\}$. In terms of positional notation system, $$a=a_1a_2a_3a_4\cdots,$$ where the weight of each position is respectively $\frac1{2^1},\frac1{2^4},\frac1{2^9},\frac1{2^{16}},\cdots.$

Consider $\mathcal F=\{\sum _{k=1}^\infty a_{k}2^{-k^2}: a_k\in\{0,1, 2, \cdots, 2^{2k-1}-1\}, a_k\not=2^{2k-2}\}\subset[0,1]$. In plain words, we start with $[0,1]$. At round $k\in\{1,2,\cdots\}$, for each remaining intervals, we split it into $2^{2k-1}$ equal pieces and then take away the interior of one of the pieces in the middle. All the points that have never been taken away form $\mathcal F$.

$\ \mathcal F$ is a fat Cantor set with Lebesgue measure $$\prod_{k=1}^{\infty}\frac{2^{2k-1}-1}{2^{2k-1}}>1-\sum_{k=1}^{\infty}\frac1{2^{2k-1}}=\frac13.$$

A continuous function $f$ with $f^{-1}(0)$ having positive lebsegue measure

Suppose $a\in[0,1], a\notin\mathcal F$. Since $\mathcal F$ is a closed subset of $\Bbb R$, there are open intervals that are disjoint with $\mathcal F$ that contain $a$. Let $(\ell_a,r_a)$ be the biggest one of them.

Define $f:[0,1]\to\Bbb R$, $$f(a)=\begin{cases} 0&\text{if }a\in\mathcal F,\\ (\ell_a+r_a)/2-|a-(\ell_a+r_a)/2|&\text{if }a\notin\mathcal F.\\ \end{cases}$$

$f$ is continuous.
$f^{-1}(a)$ is finite if $a\not=0$.
$f^{-1}(0)=\mathcal F$, a fat Cantor set with Lebseque measure $>\frac13$.
All level sets of $f$ have empty interiors.

KCd
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Apass.Jack
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