In Bartle's textbook-"Introduction to Real analysis(4th)", the following theorem is introduced:
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.
As a remark, the above theorem has some useful applications. For example, if $h$ is a monotone function satisfying the functional equation $h(x+y)=h(x)+h(y)$ for all $x,y\in\mathbb{R}$, then $h$ must be continuous on $\mathbb{R}$.
I have already proved the case $h$ is continuous at $x=0$. But, i have a trouble in case of $h$ is monotone, and applying the above theorem.
Give me some advice. Thank you!
I have to admit I don't see where the theorem enters -- this may be due to the fact that the following reasoning may implicitly contain a proof of it, or that I'm missing something...
Clearly $f(0) = f(0)+ f(0) $ so $f(0)=0$.
Assume wlog that $h$ is increasing and you want to show $h$ is right continuous at $x=0$ (i.e. $\lim_{y\rightarrow 0, y>0} f(0+y) = f(0)=0$)
Fix some $y>0$ and note that, by simple induction, $f(\frac{y}{n}) = \frac{1}{n}f(y)$. This implies (as you probably have shown when you demonstrated continuity at $y=0$) that $\lim_{n\rightarrow\infty} f(\frac{y}{n}) = 0$. But then, by monotonicity it follows easily that $\lim_{y\rightarrow0} f(y) = 0$ Similarly you can show continuity from the left hand side at $0$
But then, for any $x$, $$\lim_{z\rightarrow x} f(z) = \lim_{y\rightarrow 0} f(x+y)= \lim_{y\rightarrow 0} (f(x)+f(y)) = f(x) + \lim_{y\rightarrow 0} f(y) = f(x)$$