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

• 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
Commented May 1, 2012 at 6:08
• Huh? Re to your edit: Monotonic functions are not necessarily increasing.
– user21436
Commented May 1, 2012 at 6:22
• @KannappanSampath: Would it be a bad idea to negate the $\epsilon - \delta$ definition of continuity? Commented May 1, 2012 at 6:27

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 $$a. Since $$f$$ is increasing, $$f(x)\le f(x_0)$$ whenever $$a\le x, so $$\{f(x):a\le x is bounded above by $$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 can $$\lim\limits_{x\to x_0}f(x)$$ 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 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 $$f$$.

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?