Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

"Lebesgue's Theorem" states that for any bounded $f:[a,b] \to \mathbb{R}$, $f$ is Riemann Integrable iff $m\{x:f \text{ is not continuous at x }\}=0$, and if so Riemann's integral coincides with Lebesgue's. ($m$ is Lebesgue's measure).

Does there exist a generalization of this theorem for higher dimensions? Can I have a proof or a reference please?

share|improve this question
add comment

1 Answer

up vote 1 down vote accepted

First note that the Lebesgue-Riemann theorem states that $f:[a,b]\to \mathbb R$ is Riemann integrable if, and only if, $f$ is bounded and is continuous almost everywhere (if a function is not bounded, it can't be Riemann integrable).

A proof of the more general result for multivariable functions can be found in volume 1 of "Real Analysis" by Duistermaat. A better exercise in fact will be to take this wiki proof and adapt it to $\mathbb R^n$.

share|improve this answer
    
Shouldn't "almost always" be "almost everywhere"? –  Pedro Tamaroff Mar 7 '13 at 22:11
    
@PeterTamaroff corrected, thanks. –  Ittay Weiss Mar 7 '13 at 22:18
    
Could you please state the generalization? –  user1337 Mar 7 '13 at 23:07
add comment

Your Answer

 
discard

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.