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What conditions will it be true that:

$$\int_a^bf(x)dx = \lim_{n\to \infty}\int_a^bf_n(x)dx$$ where $f_n$ is a sequence of functions. I know that just having convergence of $f_n(x)$ to $f(x)$ isn't enough, but I don't know which assumptions will make this equality hold.

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See if Arzela's dominated convergence theorem helps in my post. – Frank Science Mar 2 '13 at 3:02
up vote 4 down vote accepted

The key is uniform convergence of $f_n$. If $f_n$ converges uniformly to an integrable function $f$, then you can interchange the limit and the integral.

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Thanks, This was a lot more straightforward than I thought. – MITjanitor Mar 2 '13 at 0:41

This depends on what type of integral you want (Riemann or Lebesgue) and what you know about $f$:

If you want a Riemann integral, and you don't already know that $f$ is Riemann integrable, then you'd better show that it's Riemann integrable first. Uniform convergence will do the trick (it will also prove the equality) but it is not always the case that the functions converge uniformly (and then you need to use other means to prove the equality).

If you want a Lebesgue integral, then you may prove the limit using many powerful convergence theorems of Lebesgue integrals, such as the dominated convergence theorem or the monotone convergence theorem.

If you want a Riemann integral, don't have uniform convergence, and already know that $f$ is Riemann integrable, then you can use a basic result that says Riemann-integrable functions are Lebesgue-integrable and both integrals are equal, and use the powerful convergence theorems I mentioned above.

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If you are on a set of finite measure, uniform integrability can be your friend. Check it out!

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