Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Just having a little problem solving this, however it probably is pretty easy and I am just being dumb.

Suppose you have a lebesgue integrable function $f$. The goal is, for any $ \epsilon > 0 $, to find a set $C$ with $ \mu (C) < \infty$ such that $\int_{C^c} |f| d \mu < \epsilon$.

Any ideas on how to construct this set? I think it has to do with the Dominated Convergence Theorem, but I don't see it.

share|improve this question
    
I think you must have your quantifiers out of order; you probably want to show that for all $\epsilon>0$ there is a set $C$ such that $\int_{C^{c}} |f|\, d\mu < \epsilon$. Otherwise, the function $f(x) = e^{-x^2}$ is a counterexample. –  Quinn Culver Nov 19 '12 at 4:14
    
Yep, thats what I meant. Sorry. –  Dengcho Nov 19 '12 at 4:21

3 Answers 3

Let $f_n=f\,1_{[-n,n]}$. Then $|f_n|\nearrow |f|$. By Monotone Convergence (you can use Dominated Convergence also), $$ \int|f|\,d\mu=\lim_n\int |f_n|\,d\mu=\lim_n\int_{[-n,n]}|f|\,d\mu. $$ So $$ \lim_n\int_{[-n,n]^c}|f|\,d\mu=0. $$ Taking $n$ big enough, you can take $C=[-n,n]$ and you get your result.

share|improve this answer

Try approximating |f| by the sequence $f_n := f \chi_{B_n}$ where $B_n$ is a ball of radius n centered at the origin.

share|improve this answer

Let $f_n = \lvert f\rvert1_{B(0,n)^c}$. Then $f_n \searrow 0$ so applying the dominated convergence theorem does the job!

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.