Fatou's Lemma and the bounded convergence theorem. - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-05-22T04:02:33Z http://mathoverflow.net/feeds/question/32149 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/32149/fatous-lemma-and-the-bounded-convergence-theorem Fatou's Lemma and the bounded convergence theorem. Chandrasekhar 2010-07-16T10:59:40Z 2010-07-16T12:11:46Z <p>I have been studying measure theory of late, and i was stuck up in these two things.</p> <p>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 $$\lim \int f_{n} = \int f$$</p> <p>Could anyone tell me as to why this will fail if we don't assume the uniformly bounded criterion. Please elaborate.</p> http://mathoverflow.net/questions/32149/fatous-lemma-and-the-bounded-convergence-theorem/32152#32152 Answer by Franklin for Fatou's Lemma and the bounded convergence theorem. Franklin 2010-07-16T11:19:52Z 2010-07-16T11:19:52Z <p>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. </p> http://mathoverflow.net/questions/32149/fatous-lemma-and-the-bounded-convergence-theorem/32156#32156 Answer by G. Rodrigues for Fatou's Lemma and the bounded convergence theorem. G. Rodrigues 2010-07-16T11:40:24Z 2010-07-16T12:11:46Z <p>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.</p> <p>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 $\|f_n{x}\|\leq g$.</p> <p>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</p> <pre><code>$\int n\chi(E_n) = 1$ </code></pre> <p>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.</p> <p>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.</p> <p>Regards, G. Rodrigues</p>