# Theorems about functions with uncountable number of discontinuities

I have seen a nice number of theorems that start with "suppose that $f$ is continuous function" or with some equivalent claim and then, with only that, or with some additional assumptions some theorem follows.

But I would like to know about theorems that start with the assumptions like "suppose that $f$ is discontinuous function" and then end with some truth about discontinuous functions.

Suppose that we work with real functions of a real variable.

Because the function can be discontinuous in a finite number of points, in countably infinite number of points and in uncountably infinite number of points let us talk here only about functions that have uncountably infinite number of discontinuities.

So the question is:

Can you give me some examples of theorems that start with the assumption that "$f$ is real function of a real variable which has an uncountable number of discontinuities" (and possibly with some other assumptions) and then some fact about such functions follows?

• What, exactly, do you mean by a discontinuity? Do you mean "there are uncountably many $x$ such that there is $\epsilon$ such that for all $\delta$ there is $y$ with $|y-x| < \delta$ but $|f(y) - f(x)| > \epsilon$"? – Patrick Stevens Apr 29 '16 at 20:24
• @PatrickStevens I mean that the function is discontinuous at a point if it is not continuous at that point. Could I mean something different from that? And that the number of discontinuities is uncountably infinite. – Farewell Apr 29 '16 at 20:27
• You could have had in mind specific kinds of discontinuity, for instance, like jump discontinuities (though it's impossible to have uncountably many of those). – Patrick Stevens Apr 29 '16 at 20:31
• @PatrickStevens Nope, did not have any particular kind of discontinuity in mind, only that there is an uncountable number of them, it could be that some of them are essential, some step, some removable... – Farewell Apr 29 '16 at 20:34
• Seeing as a discontinuous fuctions can have any freaking value and do any freaking thing it wants, there is nothing to conclude as you have no information to conclude anything from. I think Patrick Stevens is wondering if you meant anything more that could be concluded. Example f(x) = x if x is rational and fkx)=0 if x is irrational has uncountable many discontinuities and much can be said about this particular function. But "let f be function" is too unspecific to result in any thing. It's like say "let x be real number" is there a theorem about real numbers. As opposed to what? – fleablood Apr 29 '16 at 20:38

If a function $f : (a,b) \to \Bbb R$ has an uncountable number of discontinuities, then only a countable number of them may be jump discontinuities, the others (uncountably many) being essential discontinuities.

If a function $f : (a,b) \to \Bbb R$ has an uncountable number of discontinuities, then it cannot be monotonic.

• Let me notice that the first statement in my answer is the same as the first paragraph in @DaveL.Renfro's answer. – Alex M. Apr 29 '16 at 21:20
• Ah, I didn't think about the monotonic result. Of course, "monotone" can be replaced with "bounded variation". Also, since any uncountable set of reals has a condensation point (a point such that every neighborhood of the point contains uncountably many points of the set), in fact uncountably many condensation points, if a function has uncountably many discontinuities, then it must have a point (in fact, uncountably many points) at which it does not have bounded variation (a point such that for every neighborhood of the point the function is not bounded variation in that neighborhood). – Dave L. Renfro Apr 29 '16 at 21:46

One result that comes to mind is the fact that at most countably many points can be points at which both the left limit and the right limit exist (finitely or infinitely) and are different from each other and are different from the value of the function at that point. Thus, if $f:{\mathbb R} \rightarrow {\mathbb R}$ has uncountably many discontinuities, then at all but countably many of these discontinuities we must have at least one of the unilateral limits not existing (finitely or infinitely).

In fact, there are much stronger statements that can be made -- see my answer to the mathoverflow question A search for theorems which appear to have very few, if any hypotheses.

• Thank you Dave, surely I believed that there are some known facts, and I do not see any reason for non-existence of some other nontrivial facts that we are not aware of. – Farewell Apr 29 '16 at 21:10

No. There is nothing that can be said about discontinuous fuctions. Nothing at all. Any pairing of real numbers to any other real numbers with no rationale, pattern, predictability, no determination or dependence can be a function, so there is nothing that can be concluded by knowing a function exists.

With one exception...

If f is a function that maps X to Y and $x \in X$ we can conclude there exists an $f (x) \in Y$ but there is noting we can say about what f (x) might be or any relation f (x) might have with any other y and f (y).

But that's the only thing we can say.

• Actually, there are some nontrivial things one can say. See my answer. – Dave L. Renfro Apr 29 '16 at 21:02
• I agree with @DaveL.Renfro: you've rushed a bit with your answer. It's true, though, that the conclusions that follow from uncountable discontinuities are on the technical side. – Alex M. Apr 29 '16 at 21:12
• Hmm... those are valid points. Although, I tend to view a statement "right and left limits exist" as the exception rather than the norm so I'd say "for right and left limits to exist on all be countably many points the function must have at most countably many discontinuities" is a statement about limits, not discontinuity. It's rather a negative property. But it's arguable. And you are right "a function with infinite discontinuities will have at most countably many limits" is a valid theorem but ... that seems ... well, contra, to me. – fleablood Apr 29 '16 at 21:54

One such example is the Lebesgue's Criterion for Riemann Integration.

In context of your question let $f$ be a bounded function defined on a closed and bounded interval such that its set of discontinuities is uncountable and of measure 0, then $f$ is Riemann Integrable.

You can read more about the Lebesgue's Criterion here, and can find an example of such a function here.

• No, the set of discontinuities is not required to be uncountable; what hat did you pull that one from? It is only required to be negligible. Therefore, your "answer" does not address the given question. – Alex M. Apr 29 '16 at 20:57
• Isn't the hypothesis of uncountable discontinuities superfluous once "measure zero" is tacked on? – hardmath Apr 29 '16 at 20:59
• I agree with @hardmath that the set of discontinuities should have a measure 0 and uncountability is not a necessary condition. Answering in context of the question I intended to convey that this result can tell some truth about some special discontinuous functions. – Akul Bansal Apr 30 '16 at 3:10