Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is right differentiable function almost everywhere continuous? If not, is it measurable?

share|cite|improve this question
For clarity, could you define "right differentiable" and "almost everywhere continuous"? – Potato Dec 27 '11 at 20:13
I think I retagged it properly. Measure theory was added – user21436 Dec 27 '11 at 20:13
function f(x) is defined on [a,b). It is right differentiable which means for any x in [a,b), lim y->x (f(y)-f(x))/(y-x) exists. – Tim Dec 27 '11 at 20:21
And almost everywhere continuous means the discontinuous points of f(x) constitute a measure-zero set. – Tim Dec 27 '11 at 20:22

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.

share|cite|improve this answer
Very clear and complete. Thanks. I've never been good at decomposing sets. – Tim Dec 28 '11 at 2:52

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.