Let $g(x) :\mathbb{R}\to\mathbb{R}$ be a non-decreasing function in the sense that $g(y)\geq g(x) \; \forall y \geq x$. Can $g(x)$ be nowhere continuous? My guess is that any such function that is non-decreasing must have at most countably infinite number of discontinuities. Otherwise, it will cease to be discontinuous due to non-decreasing nature. However, I couldn't come up with a rigorous proof of this. Am I right in my guess?
My attempt was to assume that for each $x_0 \in \mathbb{R}$ there exists a fixed $\varepsilon_{x_0}>0$ such that for all $x \in (x_0,\infty)$ we have the relation that $f(x)>f(x_0)+\varepsilon_{x_0}$ by non-decreasing nature because for each arbitrary $\delta$ neighborhood of $x_0$ there is at least one $x'$ in it that satisfies the relation (hence all $x>x'$[non-decreasing] works but $x'$ can be made arbitrarily close to $x_0$). My intuition is that such a strictly increasing relation can only be forced for countable number of points. Is there a nice way to show it? Or I am on the wrong track?
Nvm: This proof strategy is wrong. Take the Floor function.