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

Let $f_n :[a,b] \to R$ be a sequence of Riemann-integrable functions that converges pointwise to a Riemann-integrable function $f$. Suppose there exists a positive number $M$ such that $\left|f_n(x)\right| < M $ for every $n$ and $x$. I would like to prove that

$$ \mathop {\lim }\limits_{n \to \infty } \int\limits_a^b {f_n \left( x \right)dx = } \int\limits_a^b {f\left( x \right)dx } $$

Remark: Thanks for talking me about the "Dominated convergence theorem " but I donĀ“t know, how to prove it )= , and other thing I do not know measure theory, this problem is from a calculus course, did not even know the definition of Lebesgue integral, and I can not use it for the problem )=

EDITED: Because I writed my question wrong. Thanks and sorry for tell me.

share|cite|improve this question
This sounds like the dominated convergence theorem. – Dimitrije Kostic Nov 14 '11 at 4:12
We must be talking Lebesgue measure here, right? (for Riemann integrable you can't guarantee integrability of $f$). Then this is a special case of the dominated convergence theorem often called the bounded convergence theorem. – t.b. Nov 14 '11 at 4:12
You'll want $|f_n(x)|<M$; that is, the absolute value sign is crucial. – Byron Schmuland Nov 14 '11 at 4:19
By searching in the web, this is the Arzelà's bounded convergence theorem, and it seems is not easy to prove. – leo Nov 14 '11 at 7:36
This looks like just what you're looking for. It doesn't use any measure theory. – Dane Nov 14 '11 at 8:14

Your Answer


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

Browse other questions tagged or ask your own question.