This question is related to a previous one.
Suppose that $f:[a,b]\rightarrow\mathbb{R}$ has countably many discontinuous points and $g:\mathbb{R}\rightarrow (-\pi/2,\pi/2)$ is a countinous function ($g=\arctan$). What to say about the points of discontinuity of $g\circ f=\arctan(f)$?
My thoughts are that there are at most as many points of discontinuity as $f$ has. Is it true?
Thanks
If $f$ is continuous at $a$ and $g$ is continuous at $f(a)$, then $g\circ f$ is continuous at $a$.
Since your $g$ is continuous everywhere, it follows that $g \circ f$ is continuous at every point where $f$ is continuous. Said another way, if $g \circ f $ is discontinuous at $a$, then so is $f$. So yes, your conclusion is correct.