Monoton function proof of discontinuity 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.
 A: 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?
A: 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?)
