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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1The 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. – 2011-11-18
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0...and hence not integrable – 2011-11-18
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0Let's add that the $f_n$'s and $f$ are measurable functions then. – 2011-11-18
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0Is the convergence pointwise convergence? – 2011-11-18
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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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0That's correct, thanks. I suppose we can go from here to that $\int |f_n| d\mu \rightarrow \int |f| d\mu$ as well. – 2011-11-19
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0Arturo, 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. – 2011-11-20
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0@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. – 2011-11-20
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0Precisely. 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). – 2011-11-20
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0@DidierPiau: Been way too long; I should review my notes. I always liked measure theory... Thanks. – 2011-11-20