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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?

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The same example that was already mentioned on the site works: take $f_n=\mathbf 1_A-\mathbf 1_C$ for every $n$, with $A$ and $C$ non measurable and disjoint but $A\cup C$ measurable and with finite measure. Then none of the $f_n$'s nor $f$ is measurable. – Did Nov 18 '11 at 16:32
...and hence not integrable – Ilya Nov 18 '11 at 17:08
Let's add that the $f_n$'s and $f$ are measurable functions then. – user18115 Nov 18 '11 at 19:33
Is the convergence pointwise convergence? – Arturo Magidin Nov 18 '11 at 19:52
Right. Since you ask a question with the tag (measure-theory), surely you are aware of Fatou's lemma. You could try to apply it. – Did Nov 18 '11 at 19:54
up vote 2 down vote accepted

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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That's correct, thanks. I suppose we can go from here to that $\int |f_n| d\mu \rightarrow \int |f| d\mu$ as well. – user18115 Nov 19 '11 at 18:21
Arturo, I do not understand your proof that if $h$ is measurable and $|h|$ is integrable, then $h$ is integrable. This is the definition of being integrable. – Did Nov 20 '11 at 2:13
@DidierPiau: Hmmm... I am taking "integrable" to mean that the integral $\int f\,d\mu$ exists and is finite; of course, this is equivalent to $f\in\mathcal{L}^1(\mu)$, which is turn is equivalent to $\int|f|\,d\mu$ exists and is finite. But since the OP uses $\int|f_n|\,d\mu$ and yet still asks whether "the $f_n$ are integrable", I assumed he is not aware of this equivalence. – Arturo Magidin Nov 20 '11 at 3:06
Precisely. As you know (and contrary to what the OP seems to believe), there is no such thing as semi-convergent integrals à la Riemann in Lebesgue's theory. Hence I fail to see the point of the decomposition of $|f_n|$ at the beginning of your post (except to confirm a possible misconception by the OP). – Did Nov 20 '11 at 8:16
@DidierPiau: Been way too long; I should review my notes. I always liked measure theory... Thanks. – Arturo Magidin Nov 20 '11 at 21:50

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