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

I am trying to prove (96) on pg. 324 of baby Rudin.

ie. That is $f$ is Riemann integrable on $[a,b]$ and if $$F(x)=\int_a ^x f(t)dt$$ then $F'=f(x)$ $a.e$ on $[a,b]$

Anything I do seems to just create a circular argument. But so far I know if $f$ is Riemann integrable on $[a,b]$, then $f$ is Lesbesgue integrable on $[a,b]$. It is also possible to create the $U(P,f)$ and $L(P,f)$ as in the Riemann integral definition and get $L=U\ a.e$ thus $$L(t)=f(t)=U(t)$$ And this is where I am lost.

share|cite|improve this question
What happens when you look at $\frac{F(x+h)-F(x)}{h}$? You're integrating on an increasingly small interval so the partitions should work out nicely... – Alex R. Dec 10 '12 at 1:56
I cant believe I didnt even consider looking at the definition differentiation. Thanks! – Andrew Jean Bédard Dec 10 '12 at 4:58
Baby Rudin? :D. – D1X May 30 at 11:36
Funny you commented on this and it's nearly 4 years old!!! Baby Rudin refers to the text: Principles of Mathematical Analysis, because it's the bread and butter of many undergraduate courses in Analysis, and it is comparatively elementary to one of Rudin's other well known text: Real and Complex Analysis. – Andrew Jean Bédard Jun 1 at 19:46
up vote 2 down vote accepted

Following Alex's comment and the above answer. Let's take $\,h>0\,$ for simplicity:



Check that the last expression above approaches zero a.e. when $\,h\to 0\,$ as the discontinuities of $\,f\,$ in the integration interval have measure zero.

share|cite|improve this answer

I think you need the condition that a function is Riemann integrable if and only if it is bounded and the set of discontinuities of $f$ has measure 0. This is the so called "Lesbegue's Criterion". Using this it should not be so difficult to prove the above statement.

A quite related article can be found at here.

share|cite|improve this answer
+1 Nice observation. – copper.hat Dec 10 '12 at 2:01

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.