Is right differentiable function almost everywhere continuous? Is right differentiable function almost everywhere continuous?
If not, is it measurable?
 A: A right-differentiable function is easily seen to be right-continuous, so it has only countably many points of discontinuity and is therefore certainly continuous almost everywhere.
To see this, let $$\operatorname{osc}(x)=\limsup_{y_1,y_2\to x}|f(y_1)-f(y_2)|$$ be the oscillation of $f$ at $x$; clearly $f$ is continuous at $x$ iff $\operatorname{osc}(x)=0$. For $n\in\mathbb{N}$ let $D_n=\{x\in\mathbb{R}:\operatorname{osc}(x)>2^{-n}\}$, and let $D=\bigcup_{n\in\mathbb{N}}D_n$; $D$ is the set of points of discontinuity of $f$, so to show that $D$ is countable, it suffices to show that each $D_n$ is countable.
Fix $n\in\mathbb{N}$. For each $x\in\mathbb{R}$, $f$ is right-continuous at $x$, so there is an $\epsilon_x>0$ such that $|f(y)-f(x)|<2^{-n-1}$ whenever $y\in(x,x+\epsilon_x)$. But then $$|f(y_1)-f(y_2)|<2\cdot 2^{-n-1}=2^{-n}$$ whenever $y_1,y_2\in(x,x+\epsilon_x)$, so $(x,x+\epsilon_x)\cap D_n=\varnothing$. 
Now let $x_1,x_2\in D_n$ with $x_1<x_2$; $(x_1,x_1+\epsilon_{x_1})\cap D_n=\varnothing$, so $x_2\ge x_1+\epsilon_{x_1}$, and hence $(x_1,x_1+\epsilon_{x_1})\cap (x_2,x_2+\epsilon_{x_2})=\varnothing$. In other words, the intervals $(x,x+\epsilon_x)$ with $x\in D_n$ are pairwise disjoint, so of course there can be only countably many of them, and $D_n$ is indeed countable.
