Checking when a real function is continuous in an interval where it is monotonic [closed]

I was reviewing for my Real Analysis 1 exam and I found this theorem:

Let $$f : I \to \mathbb R$$ be a function, where $$I\subseteq\mathbb R$$ is an interval. Suppose that $$f$$ is monotonic in $$I$$, then the following statements are equivalent:

• $$f$$ is continuous in $$I$$
• $$f(I)$$ is an interval

I am almost finished with this exam but I never used this equivalence in practice, my question is: can you give me some examples where studying the image of an interval is easier or more useful than directly checking the limits of the function, to know when a function is continuous?

• Note that the implication "$f$ continuous on $I$ $\implies f(I)$ is an interval" is true without the monotone condition. Commented Aug 17, 2020 at 16:39
• FYI, for real valued functions defined on an interval, the "$f(I)$ is an interval" property is equivalent to the Darboux property (= intermediate value property), which all continuous functions (and all derivatives, even when not continuous) satisfy and which many discontinuous functions also satisfy. Commented Aug 17, 2020 at 19:02
• @DaveL.Renfro a function which satisfies the Darboux property and is monotonic in an interval is automatically continuous in that interval then? Commented Aug 17, 2020 at 19:33
• Yes. The main idea behind this is that each discontinuity of a monotone function has to be a jump discontinuity (requires careful proof), and each function with a jump discontinuity will not satisfy the intermediate value property (in the vicinity of the domain point at which the jump discontinuity occurs). So if we assume $f$ is monotone, then: "not continuous implies not Darboux", which is equivalent to "Darboux implies continuous". This isn't all you have to prove for that exam question, but it's probably the most difficult part of it. ("Probably", because I don't know what you can use.) Commented Aug 17, 2020 at 19:50
• @DaveL.Renfro I see, that's very interesting. I'll try to study this concepts more in depth. Thanks for your contribution! Commented Aug 17, 2020 at 20:09

One way to show that the Cantor function is continuous is by showing that it is non-decreasing and has image $$[0,1]$$.