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Let $f:[a,b] \to \mathbb {R}$ be nondecreasing.

(a) Recall from Thm. 2.1.9 (c) that $f(x-)$ and $f(x+)$ exist at every $x\in [a,b]$, and conclude that $f$ is discontinuous at $x\in S $ iff $f(x-) < f(x+)$. Im assuming this true because the if the $f(x-)$ < the x+ then we would have a constant function not an increasing funtion. the second part is more obvious since f is nondecreasing then $f(x+) \geq f(x-)$ or it would violate the original statement. how would i try and lay this out with $ e$ and $ \delta$'s?

(b) Give an upper bound for the number of points $x$ at which the `jump' $(f(x+)- f(x-))$ is greater than a given $r>0$.

one point?

(c) Prove that the set $S = \{ x\in [a,b]: f \text{ is discontinuous at }x \}$ is at most countable.

Think i got this one figured out, suppose that S is uncountable then S must contain a Limit Point, but f is discontinuous at x so there exists a B(r,x) r>0 s.t x is isolated for all x in S if this wasn't the case then f(x) would be continuous thus S is countable.

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For (a) let $x_n$ be a non decreasing arbitrary sequence converging to $x$. Then $f(x_n)$ is a non decreasing bounded sequence, with Bolzano Weierstrass we also know that there is a convergent subsequence. Now we look at $a_n=x-\frac{1}{n}$, wlog we say that $f(a_n)$ converges (we could take a subsequence to do so). We gonna call this limit $f(x-)$ and obviously $f(x-)\leq f(x)$. We show now that for every non decreasing to $x$ converging sequence $x_n$ $f(x_n)$ converges to $f(x-)$. As $x_n$ converges to $x$ we know that for all $m$ there a $N$ such that for all $n>N$ $|x_n-x| < |x-a_m|$. Hence $$ f(a_m) \leq f(x_n) \leq f(x-).$$ With sandwhiching we get $\lim f(x_n)= f(x-)$.

For (b) we gonna denote $n$ as the number of possible jumps. As $f$ is not monotone decreasing we have that $n\cdot r \leq f(b)-f(a)$. Hence an upper bound will be \[ n \leq \frac{f(b)-f(a)}{r}\]

As $f$ is non decreasing we surely now that $f(x)\leq f(b)$ for all $x$. A non decreasing function can only have jump discontinuities. Let $y_i=f(x_i+)-f(x_i-)$ for all $i \in I$. With $I$ I mean the indexing set of our discontinuites. The $y_i$ measures the jumps of the function. Furthermore we know that $$f(b) \geq f(a) + \sum_{i\in I} y_i. $$ As $f(b)$ is a real number it is finite and hence $$\sum_{i \in I} y_i$$ must converge. This implies that $I$ must be countable, cause an uncountable sum of positive numbers must diverge (why?)

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C1 = constant C1>C2 $C1 \geq C2 + \sum$ if the sum was uncountably infinite it would diverge to infinity and that violates the cute little proof about there being a biggest number in the reals =) – Faust7 May 16 '13 at 16:05

You have that $S=\bigcup_{n\in \mathbb{N}} \{x\in [a,b]\mid f(x+)-f(x-)<1/n\}$. What can you say about the sets in the union?

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there all subsets of the n=1 case? – Faust7 May 16 '13 at 5:18
No, you could have discontinuities where the jump in the function is much smaller than $1$. Since you care about countability, you should try to think about the size of these sets for each $n$... – Edvard Fagerholm May 16 '13 at 5:21

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