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Let $n$ be some positive integer. Does there exist a Riemann integrable function $$f: [0,1]^n \to \mathbb{R} $$ with the following property: for every non-empty open subset of $U \subset (0,1)^n$, the set $$ \{x \in U \ \mid \ f \text{ is discontinuous at } x \} $$ is uncountable?

I don't really know how to approach this. I know that a function is Riemann integrable if and only if it is bounded and continuous almost everywhere.

I tried to see if, for instance, I can somehow construct a function like this, such as the characteristic function on the Cantor set in $[0,1]$ (which does not satisfy the above property) and extend it to $[0,1]^n$, but I did not succeed.

S.T.
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  • If you want a Riemann integrable function then it can admits only a finite number of "jump" discontinuities – Gabrielek Feb 27 '21 at 14:09
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    The set of jump discontinuous can be countable, not just finite, right? – S.T. Feb 27 '21 at 14:10
  • https://math.stackexchange.com/questions/2601120/is-this-discontinuous-function-riemann-integrable – Gabrielek Feb 27 '21 at 14:13
  • What you sent me is not an "if and only if statement" – S.T. Feb 27 '21 at 14:14
  • For an example of function that has an infinite countable points of discontinuity but it is not R. integrable take the Dirichlet function defined $1$ on the rationals between $0$ and $1$ and $0$ everywhere else. – Gabrielek Feb 27 '21 at 14:16
  • Take, for instance, the set $[0,1]$, and write it as $$[0,1] = [1/2, 1] \cup \bigcup_{n=2}^\infty \left[ \frac{1}{n+1}, \frac{1}{n} \right), $$ and then take a function $f$ that is constant on all these subintervals and set $f(0) = 0$ (and also assume that this $f$ is bounded). This function is Riemann integrable, and has a countably infinite number of jump discontinuities. – S.T. Feb 27 '21 at 14:16
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    Also, the Dirichlet function is discontinuous in every point. For if you fix $x \in [0,1]$, then if you approach $x$ via the rationals, then the limit of the Dirichlet function is $1$, but if you approach $x$ via the irrationals, the limit is $0$. – S.T. Feb 27 '21 at 14:17
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    Dirichlet function is everywhere discontinuous by just choosing any $\epsilon<1$ on the definition. @S.T. Do you know if the result you posted holds if we just ask ${x\in[0,1] : f \text{ is discontinuous at } x}$ to be uncountable? – R.V.N. Feb 27 '21 at 14:24
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    It should be well known that monotone functions have jump discontinuities only, there are at most a countable number of them and it is Riemann integrable. This is for single variable calculus. Also as noted Dirichlet function is discontinuous everywhere. – Paramanand Singh Feb 27 '21 at 14:26
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    @R.V.N. The characteristic function on the Cantor set is discontinuous on the Cantor set, which is uncountable, but it is still Riemann integrable, as the Cantor set has measure zero. – S.T. Feb 27 '21 at 14:31
  • Between every pair of distinct rationals, construct a Cantor set. Then use the result mentioned in the first comment here. (I do not know if there is something more elementary.) – David Mitra Feb 27 '21 at 15:25
  • How do you define such a set if between any pair of rationals there are other infinite rational numbers? – R.V.N. Feb 27 '21 at 17:12
  • @David Mitra: See the integrability criteria here, let ${I_n: ; n = 1,;2,;\ldots }$ be the rational-endpoint open subintervals of $[0,1],$ let $C_n$ be a Cantor set (or any closed set) of Hausdorff dimension greater than $1-\frac{1}{n}$ contained in $I_n,$ and use the construction in Proof Sketch here (note comment regarding "pairwise disjoint"). This gives a Riemann integrable function whose discontinuities in every open interval has Hausdorff dimension $1$ (MUCH stronger than uncountable). – Dave L. Renfro Feb 27 '21 at 17:50
  • @DavidMitra This is an interesting construction, and I think it's correct. Could you please write it as an answer so that I can accept it? – S.T. Feb 28 '21 at 13:33

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Here is an example for $n=1$:

It is known that if $S$ is a $F_\sigma$ set (a countable union of closed sets) in $\Bbb R$, then there is a function $f:\Bbb R \rightarrow\Bbb R$ whose set of discontinuities is $S$. See this post for a reference to a proof.

So, it suffices to find an $F_\sigma$ in $\Bbb R$ of measure zero with the property that its intersection with any non-degenerate open interval is uncountable. Towards this end, for each pair of distinct rationals, construct a Cantor set between them; then take the (countable) union over all such sets. It is easily verified that this gives what's required.

Also, see Dave Renfro's comments above.

David Mitra
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  • To follow-up on my yesterday's comment (I was out of characters, the comment having 2 or 3 characters left, and didn't feel like continuing with another comment), the reason we get "everywhere of Hausdorff dimension $1$" is that given any nonempty open interval $I,$ there exist infinitely many $n$ such that $I_n\subseteq I,$ and so for every $\epsilon > 0$ and for every nonempty open interval, the intersection of the set with that open interval has Hausdorff dimension greater than $1-\epsilon,$ which implies the intersection of the set with that open interval has Hausdorff dimension $1.$ – Dave L. Renfro Feb 28 '21 at 17:15