I'm trying to show that a monotone function on a closed interval can only contain jump discontinuities. Could someone give me a hint as to how I should begin? I am not sure how to start this problem.
Edit: Let $f$ be a increasing function. Then if $x \leq y, f(x) \leq f(y)$.
Best Answer
If you're going to deal first with monotone increasing functions, you should begin like this:
Now you should use the definition of continuity at a point to see what this tells you about $f$.
That pretty well exhausts what we can say just on the basis of the assumption that $f$ is discontinuous at $x_0$, so let's look at what the monotonicity of $f$ can tell us. As $x$ approaches $x_0$ from the left, $f(x)$ is increasing; does it have a limit? Yes, because of the completeness of $\Bbb R$. But if we're going to look at $\lim\limits_{x\to x_0^-}f(x)$, we'd better not let $x_0=a$. In fact, the cases $x_0=a$ and $x_0=b$ both look as if they might require a little special handling, so let's set them aside for the moment.
Now prove that $$L=\lim_{x\to x_0^-}f(x)\;.$$ Go on to prove in similar fashion that $\lim\limits_{x\to x_0^+}f(x)$ exists, and call it $R$, say.
The final step is to ask yourself how $\lim\limits_{x\to x_0}f(x)$ can fail to exist when the one-sided limits $L$ and $R$ both exist?
Then you have to clean up the loose ends when $x_0$ is $a$ or $b$, but that's easy once you have the main part done. You also have to deal with monotone decreasing $f$. You can either repeat the argument above with very minor changes, or you can look at $-f$: if $f$ is decreasing, then $-f$ is increasing, so you already know that it has only jump discontinuities, and from that you should be able to show very quickly that the same is true of $f$.