Explicit example of "hyper-discontinuous" function on domain with positive Lebesgue measure?

Question

From this question, suppose $X\subset\mathbb{R}$ and $Y\subset\mathbb{R}$.

Definition: A function $\ f: X \to Y\ $ is hyper-discontinuous if for every $\ x\in X,\ \ \exists\ \delta>0,\ \varepsilon>0\ $ such that $\ y\in X \setminus \{x\},\ \vert y-x \vert < \delta,\ \implies \vert f(x) - f(y) \vert \geq \varepsilon.$

Also, suppose $\lambda^{*}$ is the outer Lebesgue measure, to to define Lebesgue measure $\lambda$ on $X$ measurable in the sense of Caratheodory.

Question: Does there exist a hyper-discontinuous $f$ on $X$, where $\lambda(X)>0$?

If an example doesn't exist, how do we show the example in the attempt doesn't answer the question:

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Attempt:

Consider the function $f:\mathbb{R}\to\mathbb{R}$, where for a base-3 expansion of $x\in\mathbb{R}$, we take the "pseudo-random" iterations of function $\mathscr{F}:\{0,1,2\}\to\{0,1,2\}$ where for the first iteration, replace base-3 digit zero with one; one with two, and two with zero.

In mathematical terms, if $\mathscr{F}:=\left\{(0,1),(1,2),(2,0)\right\}$, such that equation: $$\mathscr{F}^{k}=\underset{k\text{ times}}{\underbrace{\mathscr{F}(\mathscr{F}(\cdots\mathscr{F}(x)))}}$$ is the $k$-th iterations of $\mathscr{F}$, we want $g_{\varepsilon}:\mathbb{R}\to\mathbb{R}$ where:

$$g_{\varepsilon}(x)=\begin{cases} \quad\!\!\,\varepsilon^{x/\varepsilon} & \quad\!\! x<-\varepsilon\\ -1/x & -\varepsilon\le x <0\\ \quad \! 0 & \!\!\quad x=0 \\ \quad\!{1}/{x} & \quad\!\! 0<x<\varepsilon \\ \quad\!\!\,\varepsilon^{-x/\varepsilon} & \quad\!\! x\ge \varepsilon\\ \end{cases}$$

and $\left[\cdot\right]$ rounds to the nearest integer, we define function $z_{\varepsilon}:\mathbb{R}^2\to\mathbb{R}$ where

$$z_{\varepsilon}(x,k)=\big[\left|g_{\varepsilon}(x) k\sin(g_{\varepsilon}(x)k)\right|\big]$$

contains "pseudo-random" outputs of $\mathbb{N}$, such that:

$${f_{\varepsilon}(x)=\left\{\sum\limits_{k=-\infty}^{\infty}{\mathscr{F}^{z_{\varepsilon}(x,k)}(a)}/{3^k}:a\in\left\{0,1,2\right\},x=\sum\limits_{k=-\infty}^{\infty}a/{3^k}\right\}}$$

where for set $X\subset\mathbb{R}$ and $X^{\prime}\subset X$, we want $f\left|_{X}\right.$, such that:

$$\forall(\varepsilon_1>0)\exists(X^{\prime}\subset X)\forall(\varepsilon\in X^{\prime})\exists(M>0)\forall(x\in X)\left(0<\varepsilon\le M\Rightarrow \left|f_{\varepsilon}(x)-f(x)\right|<\varepsilon_{1}\right)$$

How do we show this example is correct/incorrect?

Answer 1

We show that there exists no hyper-discontinuous function on any uncountable set in $\mathbb{R}$ (the proof follows closely the one in Do these "hyper-discontinuous" functions exist?, well more honestly, I carry out the sketch that is presented there). Note that this in particular implies that there exists no hyper-discontinuous function if $X$ has positive Lebesgue measure (as countably many points form a null set Any countable set has measurable zero).

Essentially the proof relies on the fact that the countable union of countable sets is countable (see for example here Prove that the union of countably many countable sets is countable.). Thus, if we can write an uncountable set as a countable union, then one of the sets in the union must be uncountable. The rest is just tedious shuffling of indices.

Let now $X\subseteq \mathbb{R}$ be any uncountable set. For every $x\in X$ we denote by $(\delta(x), \varepsilon(x))$ an admissible choice given by the definition of hyper-discontinuity.

Reduction to some fixed $\varepsilon_0, \delta_0$: Now write

$$ X = \bigcup_{n,m\in \mathbb{N}} \{ x\in X \ : \ \delta(x)\geq 1/n, \varepsilon(x) \geq 1/m \}. $$ As $X$ is uncountable, there exists $\delta_0, \varepsilon_0>0$ such that $$ Z=\{ x\in X \ : \ \delta(x)\geq \delta_0, \varepsilon(x)\geq \varepsilon_0 \} $$ is uncountable.

Finding an interval of length $\delta_0/2$ that contains infinitely many points for which we can choose $\varepsilon_0, \delta_0$: Next, write $$ Z = \bigcup_{n\in \mathbb{Z}} \left(\left[n\frac{\delta_0}{2}, (n+1)\frac{\delta_0}{2}\right) \cap Z\right). $$ As $Z$ is uncountable, there exists $n_0\in \mathbb{Z}$ such that $W:=\left[n_0\frac{\delta_0}{2}, (n_0+1)\frac{\delta_0}{2}\right) \cap Z$ is uncountable.

Reaching a contradiction: However, all points in $W$ have distance less than $\delta_0$ and as $\delta(x)\geq \delta_0$ for all $x\in W$ by construction, we get that there are uncountably many intervals in $\mathbb{R}$ of size $\varepsilon_0$ that are disjoint, which yields a contradiction as shown in the next two paragraphs.

First version: Indeed, we can write $$ W = \bigcup_{m\in \mathbb{N}} \{ x\in W \ : \ (f(x)-\varepsilon_0, f(x)+\varepsilon_0)\subseteq [-m,m] \}. $$ Again, as $W$ is uncountable, there exists $m_0\in \mathbb{N}$ such that $$ \{ x\in W \ : \ (f(x)-\varepsilon_0, f(x)+\varepsilon_0)\subseteq [-m_0,m_0] \} $$ is uncountable. Pick a countably infinite subset $\{ x_j \in W \ : \ j\in \mathbb{N}\}$. As for all $i \neq j$ we have $(f(x_i)-\varepsilon_0, f(x_i)+\varepsilon_0) \cap (f(x_j)-\varepsilon_0, f(x_j)+\varepsilon_0) = \emptyset$ we get by countable additivity of the Lebesgue measure $$ 2m_0=\lambda([-m_0, m_0]) \geq \sum_{j\in \mathbb{N}} \lambda((f(x_j)-\varepsilon_0, f(x_j)+\varepsilon_0)) = \sum_{j\in \mathbb{N}} 2\varepsilon_0 = \infty, $$ which yields a contradiction.

Second version: Alternatively, if one does not like the measure-theoretic argument. We can pick for every $x\in W$ some $q(x)\in \mathbb{Q}\cap (f(x)-\varepsilon_0, f(x)+\varepsilon_0)$. This defines a map $$ q: W \rightarrow \mathbb{Q}, x \mapsto q(x). $$ Note that by construction, the intervals are disjoint and thus $q$ is injective. Therefore, $q(W)$ has the same cardinality as $W$. However, $q(W)\subseteq \mathbb{Q}$ and as subsets of countable sets are countable (see here Subset of a countable set is itself countable), we get that $q(W)$ is countable and hence $W$ is countable. Again we reach a contradiction.

Generalization: This was not really the point of the OP, but we could replace $\mathbb{R}$ by any separable metric space $(X,d_X)$, respectively $(Y,d_Y)$. From now on $B_X(c,r)$ denotes a ball of radius $r$ and center $c$ in the metric $d_X$ and similarly for $B_Y(y,c)$.

We call a map $f: (X, d_X) \rightarrow (Y, d_Y)$ hyper-discontinuous if for all $x\in X$ exists some $\delta>0$ and some $\varepsilon>0$ such that for all $u\in B_X(x,\delta)\setminus \{x\}$ we have $d_Y(f(x), f(u))\geq \varepsilon.$

The claim is now that if both $(X,d_X)$ and $(Y,d_Y)$ are separable metric spaces and $X$ is uncountable, then there exists no hyper-discontinuous map from $(X,d_X)$ to $(Y,d_Y)$.

The first reduction step would remain the same.

For the second step, we fix some countable dense set $A_X\subseteq X$. We write

$$ Z = \bigcup_{a\in A_X} B_X(a, \delta_0/2) \cap Z $$

and define $W=B_X(a_0, \delta_0/2) \cap Z$ for some $a_0\in A_X$ such that $W$ is uncountable.

Again by the definition of hyper-discontinuity (translated to the metric setting), we get that for all $z, w\in W$ with $z\neq w$ we have $B_Y(f(z), \varepsilon_0/2) \cap B_Y(f(w), \varepsilon_0/2) = \emptyset$. Now we use the second version of the contradiction proof. Namely, fix $Q_Y\subseteq Y$ a countable and dense subset and for each $z\in Z$ we pick $q(z)\in Q_Y \cap B_Y(f(z), \varepsilon_0/2)$, which again yields a map

$$ q: Z \rightarrow Q_Y, z \mapsto q(z). $$

However, this map is again injective and $Q_Y$ is countable, thus we get again a contradiction.