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How to find a function whose continuity set and discontinuity set are both uncountable.

Riemann function has continuity set $\Bbb Q^c$, uncountable, but discontinuity set $\Bbb Q$, countable!

Dirichlet function has continuity set $\emptyset$, countable, but discontinuity set $\Bbb R$, uncountable!

Oh, What to do? Some function which is immediate between Riemann and Dirichlet?

xldd
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2 Answers2

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Let $A\subset [0, 1]$ be any set.

$\mathbf{1}_A:[0,1]\to\Bbb{R}$ defined by

$\mathbf{1}_A(x)=\begin{cases}1&x\in A\\ 0& x\notin A\end{cases}$

$\mathbf{1}_A$ is called indicator function of $A$

Set of discontinuity of $\mathbf{1}_A=\partial(A)$

where $\partial(A)$ denote the set of all boundary points of $A$

Particular example: (Using my favorite Cantor set $\mathcal{C}$)

$\mathbf{1}_{\mathcal{C}}(x)=\begin{cases}1&x\in \mathcal{C}\\ 0& x\notin \mathcal{C}\end{cases}$

Then set of discontinuity $:=D_{{\mathbf{1}}_{\mathcal{C}}}= \partial(A)=\mathcal{C}$

The Cantor set is an uncountable set and it's complement $[0, 1]\setminus \mathcal{C}$ has positive measure $ ( m([0, 1]\setminus \mathcal{C})=1) $, is also uncountable.

Now you can produce more examples using indicator functions $ :) $

Probably helpful post.

Sourav Ghosh
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  • Among other possibilities, this answer shows that the continuity set can have full Lebesgue measure and be co-meager in every interval (much stronger than $c$-many points in every interval, or even Hausdorff dimension $1$ in every interval) and the discontinuity set can have Hausdorff dimension $1$ in every interval (much stronger than $c$-many points in every interval). – Dave L. Renfro Jul 17 '22 at 14:39
  • Great. I will definitely see. As usual you and your answer both are awesome. I like the Cantor set. Whenever I get a chance to use it , I can't resist myself. – Sourav Ghosh Jul 17 '22 at 14:51
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Use $f(x)=1$ if $x\,\in\,\mathbb{Q}$, $x=0$ if $x$ is irrational on $[0,1)$ and

$x=1$ on $[1,2]$. Then the two conditions are satisfied!!