# Montone Function only has Jump Discontinuities

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 $x \leq y, f(x) \leq f(y)$.

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Firstly reduce your case to proving only for increasing functions. How do you do this? And, the one-sided limits exist for a monotonic function, what does this tell you? –  user21436 May 1 '12 at 6:08
Huh? Re to your edit: Monotonic functions are not necessarily increasing. –  user21436 May 1 '12 at 6:22
@KannappanSampath: Would it be a bad idea to negate the $\epsilon - \delta$ definition of continuity? –  Student May 1 '12 at 6:27

Hint

1. Reduce the problem to the case of proving for increasing functions by observing that if $f$ is decreasing, $-f$ is increasing.

2. Show that the one sided limits exists and are given by: $$\lim_{t \to c^+}f(t)=\inf\{f(x) \mid x \gt c\}\quad \mbox{and} \quad \lim_{t \to c^-}f(t)=\sup \{f(x) \mid x \lt c\}$$

3. Since one-sided limits exist, the discontinuity is either removable or a jump discontinuity. Can you prove that it cannot be a removable discontinuity?

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If you're going to deal first with monotone increasing functions, you should begin like this:

Assume that $f$ is monotone increasing on $[a,b]$ and discontinuous at $x_0\in[a,b]$.

Now you should use the definition of continuity at a point to see what this tells you about $f$.

We know that $f(x_0)$ exists, so it must be the case that $$\lim_{x\to x_0}f(x)\ne f(x_0)\;.$$

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.

Assume for now that $a<x_0<b$. Since $f$ is increasing, $f(x)\le f(x_0)$ whenever $a\le x<x_0$, so $\{f(x):a\le x<x_0\}$ is bounded above by $f(x_0)$ and therefore has a least upper bound, say $L$.

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 $f$.

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