I want to mention that a similar questions like this or this was asked before, but I want to ask about specific proof I read in Introduction to Real Analysis by Robert Bartle and Donald Sherber
fist some notation if $f:I \to \mathbb{R} $ is increasing on $I$ and if $c$ is not an endpoint of $I$, we define the jump of $f$
at $c$ to be $j_f (c):= \lim_{x\to c^+}f(x)- \lim_{x\to c^-}f(x)$.
theorem :- Let $I \subseteq \mathbb{R}$ be an interval and let $f:I \to \mathbb{R} $ be monotone on $I$. Then the
set of points $D \subseteq I$ at which $f$ is discontinuous is a countable set.
PROOF
we will assume that $I:=[a,b]$
$D={x\in I :j_f (x)≠0}$ since $f$ is increasing $j_f (x)≥0$ for all $c \in I$ and if $a≤ x_1 < …<x_n ≤ b$ then $f(a)≤ j_f(x_1) + …+j_f(x_n) ≤ f(b)$ and it follow that $j_f(x_1) + …+j_f(x_n) ≤ f(b)-f(a)$ Consequently there can be at most $k$ points in $I:=[a,b]$ where $j_f(x) ≥ \frac{f(b)-f(a)}{k}$
here I couldn't understand why the following is true
there can be at most $k$ points in $I:=[a,b]$ where $j_f(x) ≥ \frac{f(b)-f(a)}{k}$
and if it was true how does this proof that the set $D $ is countable
Best Answer
Firstly, there can be at most that many "jumps" because if there were more than $k$ jumps of size greater than $\frac{f(b)-f(a)}{k}$, then by monotonicity the total increase in the function over the interval would be more than $k\cdot \frac{f(b)-f(a)}{k} = f(b)-f(a)$, which is impossible.
As for how it proves countability, the sequence $\frac{f(b)-f(a)}{k}$ approaches $0$, so since every discontinuity satisfies $j_f(x)>0$, you know there is some $k$ for which $j_f(x)\geq\frac{f(b)-f(a)}{k}$. Therefore if $D_k$ is the set of jump discontinuities where $j_f(x)\geq\frac{f(b)-f(a)}{k}$ is satisfied, then $D=\bigcup_k D_k$, and we just proved each $D_k$ was finite, so we have that $D$ is a countable union of finite sets, and is hence countable.