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Possible Duplicate:
Showing properties of discontinuous points of a strictly increasing function
How to show that a set of discontinuous points of an increasing function is at most countable

I'm struggling to find an elegant proof of the following problem

Let $f : (a, b) \to \mathbb R$ be non-decreasing, $a, b \in \mathbb R$, then $f$ only has countably many discontinuities.

My intuition was to show by contradiction that the set of discontinuities $N \subseteq (a,b)$ is discrete, i.e. all discontinuities are isolated. From there on it's easy to prove that there is an injective function $N \to \mathbb Q$.

But does my first step make sense? Say we had non-isolated discontinuities like $\epsilon > 0, x_0 \in N$ such that $B_\epsilon(x_0) \subseteq N$ - how could one derive a contradiction?

I've already shown that $\lim_{x \nearrow x_0} f(x) \text{ and } \lim_{x \searrow x_0} f(x)$ exist for all points $x_0 \in (a, b)$ and that $f$ is continuous at $x_0$ iff both limits equal. I just somehow fail to do the final step properly. Any thoughts, please?

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    Since the question was already explained, here is an intuitive explanation of this phenomena: Any monotonic function only has jump discontinuities. If the set of jump discontinuities would be uncountable, you can find uncountably many jump discontinuities in some compact interval $[c,d]$. A simple countability argument shows that there is an n so that infinitely many of "jumps" have to be at least $\frac{1}{n}$, which contradicts the fact that $f$ only changes $|f(d)-f(c)|$ over that compact interval...2011-12-28

2 Answers 2

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No, the set of discontinuities need not be discrete. For example, it's quite possible to have a discontinuity at $0$ and also at $1/n$ for each positive integer $n$.

Hint: for each discontinuity, there are rational numbers that are not in $f((a,b))$.

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    Okay well, thanks for helping me out of this discrete-set dead-end. The solution is much easier without.2011-12-28
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Certainly $\lim_{x \nearrow x_0} f(x) \leq \lim_{x \searrow x_0} f(x)$. Also, $\sum\limits_{x_0\in (a,b)}\lim_{x \searrow x_0} f(x) - \lim_{x \nearrow x_0} f(x)$ is finite. It is not hard to show that since this is finite, all but countably many terms are $0$.

EDIT: When I shoot from the hip I forget that if $f(x)$ is unbounded you need to normalize it as $f(x)/\max\{|f(y)| : y\in (\frac{x+a}{2},\frac{x+b}{2})\}$ which is defined because $f(x)$ is monotonic and has all the discontinuities of $f(x)$.