We have sequence of measurable functions $f_1, f_2, \dots$ such that $f_n \rightarrow f$ a.e. and $f$ measurable. If we know that $\int |f_n|d\mu < B$ for all $n$ ($B$ is fixed and finite). Also assume that $\mu[\Omega]$ is finite. Is this enough to imply that all the $f_n$'s and $f$ are integrable? How or counterexample?
If the absolute value of each function in a sequence has a Lebesgue integral that is bounded by a fixed number, will the limit function be integrable?
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measure-theory
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5Right. Since you ask a question with the tag (measure-theory), surely you are aware of [Fatou's lemma](http://en.wikipedia.org/wiki/Fatou%27s_lemma). You could try to apply it. – 2011-11-18
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Edited. Note that for Lebesgue integration, a function $g$ is integrable (the integral $\int g\,d\mu$ exists and is finite) if and only if $g\in\mathcal{L}^1(\mu)$, if and only if $\int |g|\,d\mu$ is finite; so it makes no sense to both assume that $|f_n|$ are integrable (which you do when you write $\int |f_n|\,d\mu\lt B$) and to ask whether $f_n$ are integrable; they have to be for your hypothesis to make sense.
Detour Removed
Since $|f_n|\to |f|$ almost everywhere, by Fatou's Lemma $|f|$ is integrable and $\int |f|\,d\mu \leq \liminf_{n\to\infty}\int |f_n|\,d\mu.$ So $|f|$ (hence $f$) is in $\mathcal{L}^1(\mu)$, as desired.
Finite measure of $\Omega$ is not required.
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0@DidierPiau: Been way too long; I should review my notes. I always liked measure theory... Thanks. – 2011-11-20