1
$\begingroup$

Let $f$ be measurable, finite a.e, and Lebesgue integrable on $\mathbb{R}$ . Show that for all $\epsilon > 0$ there exists a measurable set $E$ (i.e $\mu(E) < \infty$) such that:

$\int_{E} |f| > \int_{\mathbb{R}} |f| - \epsilon$

My attempt

Suppose that for all $\epsilon > 0$ there exists set of finite measure $E$ such that:

$\int_{E} |f| \leq \int_{\mathbb{R}} |f| - \epsilon$

Then:

$ \int_{\mathbb{R}\setminus{E}} |f| \geq \epsilon$

Since $\epsilon > 0$ was arbitrary then we have:

$ \int_{\mathbb{R}\setminus{E}} |f| = \infty$

But $\mu(\mathbb{R}\setminus{E}) \geq \mu(\mathbb{R}) - \mu(E) = \infty$

This contradicts the fact that $g$ is finite a.e.

I don’t think my answer is correct because it seems pretty hand-wavy. Any thoughts on a better way to do this?

$\endgroup$

1 Answer 1

1
$\begingroup$

You have assumed the result, which is not a valid proof.

Rather, given $\epsilon>0$, since $E_{n}=\{x\in{\bf{R}}:|x|\leq n,|f(x)|<n\}$ is such that $\chi_{E_{n}}|f|\uparrow|f|$ a.e. since $f$ is finite a.e., then Monotone Convergence Theorem gives \begin{align*} \int\chi_{E_{n}}|f|\uparrow\int|f|, \end{align*} so there exists some $n_{0}$ such that \begin{align*} \int\chi_{E_{n_{0}}}|f|>\int|f|-\epsilon. \end{align*} Note that \begin{align*} \int\chi_{E_{n_{0}}}|f|=\int_{E_{n_{0}}}|f|, \end{align*} and $|E_{n_{0}}|\leq 2n_{0}<\infty$.

$\endgroup$
9
  • $\begingroup$ I made a typo in the answer I attempted. Does it make more sense now? $\endgroup$ Apr 8, 2018 at 1:32
  • $\begingroup$ The negation of the statement is there exists some $\epsilon>0$ such that the negation holds, so it is not that $\epsilon>0$ being arbitrary that you can claim the integral is then $\infty$. $\endgroup$
    – user284331
    Apr 8, 2018 at 1:34
  • $\begingroup$ hm yes I realize the negation of the statement is what you are saying but I thought it would be okay since I was using contradiction not contrapositive. But now I see that it’s not correct. Thanks. $\endgroup$ Apr 8, 2018 at 1:38
  • $\begingroup$ Can you explain the $\uparrow$ notation? Does it mean “converges & increases to”? $\endgroup$ Apr 8, 2018 at 1:50
  • $\begingroup$ Yes, convergence and the sequence is monotone, so the limit is the supremum. $\endgroup$
    – user284331
    Apr 8, 2018 at 1:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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