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I have been studying measure theory of late, and i was stuck up in these two things.

BOUNDED CONVERGENCE THEOREM: It states that $f_n$ is a sequence of measurable functions, defined on a measurable set $E$ and if $f_{n} \to f$ pointwise on $E$,and is $f_{n}$ is uniformly bounded, that is if $|f_{n}(x)| \leq M$ for each $x$ and $n \in \mathbb{N}$ then $$ \lim \int f_{n} = \int f$$

Could anyone tell me as to why this will fail if we don't assume the uniformly bounded criterion. Please elaborate.

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closed as too localized by Robin Chapman, Harald Hanche-Olsen, Pete L. Clark, Yemon Choi, S. Carnahan Jul 18 '10 at 14:39

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This isn't true as stated: one needs $E$ to have finite measure. When $E$ has finite measure a standard sort of counterexample with $f_n$ not uniformly bounded is to take say $E=(0,1)$ with Lebesgue measure and $f_n$ supported on $(0,1/n)$ with integral whatever you like. –  Robin Chapman Jul 16 '10 at 11:05
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Just to add one observation: the usual proofs of the bounded convergence theorem (or the dominate convergence theorem as Franklin alluded to in his answer) are fairly straight forward. It is quite easy to identify where each assumption is used in the proof. A good exercise to help in learning the subject is to look at that and try to see how it can fail. In your case the boundedness is used to control the difference of the integrals on sets where the difference of the functions is large. By pointwise convergence the measure of the set is small. So if the functions are bounded, the integrals... –  Willie Wong Jul 16 '10 at 11:31
    
...will be small. (Assuming that E has finite measure.) Flip that around and you have that the errors are not controllable if the integral on those smaller and smaller sets remain approximately constant. Which will then require the function to be larger and larger on those smaller sets. –  Willie Wong Jul 16 '10 at 11:35
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I note that this question has been downvoted twice so far (not by me!) without a corresponding comment having been left. My feeling here is that this is because the answer to your question is not very difficult to look up. For example, your question is answered on the Wikipedia pages for Fatou's lemma and the Dominated Convergence Theorem. Indeed, the latter of these two pages is the first Google hit I get upon searching for "bounded convergence theorem". –  Ian Morris Jul 16 '10 at 13:26
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I would classify this as an (easy) exercise in measure theory, and thus not appropriate for MO. –  Keenan Kidwell Jul 16 '10 at 13:46

2 Answers 2

up vote 5 down vote accepted

I am not sure if the question fits MO standards as it is an elementary measure theory question (and if it's not, expect to be tazered by the MO police). Here goes an answer, anyway.

To expand on Robin Chapman's comment, first, the theorem as stated is false withouth the assumption that $E$ has finite measure. The correct generalization is the Lebesgue dominated convergence where the sequence $f_{n}$ is such that there is an integrable $g$ such that $\|f_n{x}\|\leq g$.

To see why, it fails without the boundedness condition consider the sequence of intervals $E_n= [0, 1/n]$ and take the sequence $(n\chi(E_n))$ where $\chi(E_n)$ is the characteristic function (or indicator functions) of $E_n$. This sequence converges pointwise to $0$ but

$\int n\chi(E_n) = 1$

so that the sequence of integrals does not converge to $0$. What is happening is that you are shrinking the support of the functions but the same time increasing their "amplitude" so that the two cancel each other out and the integral stays constant while the functions themselves converge to zero. The uniform bound on the sequence, prevents their "amplitudes" of running off to infinity and screwing up the integrals.

Examples can be concocted where the convergence is uniform instead of just pointwise. The idea is to do the reverse of the previous example: shrink the amplitude of the functions (to guarantee their uniform convergence) while enlarging their support. This will need a measure space of infinite measure. I will leave that as an exercise.

Regards, G. Rodrigues

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+1 just for the remark about tasers ;o) –  Ian Morris Jul 16 '10 at 12:05

If $f_n$ is the characteristic function of the interval $[n,2n]$ then $f_n\rightarrow0$ but $\int f_n = n$ which does not tend to zero. You need even more than the boundedness. Adding that your space is of finite measure will help is you want to bound with a constant. Alternatively you can bound with an integrable function. Without a bound the example above can be even worst.

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@Ben: Dear Ben, since you are a moderator i request you to remove this question. –  S.C. May 10 '11 at 15:25

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