# Question about Lebesgue Measurable functions and sets.

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.

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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

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