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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?

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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 $|E_{n_{0}}|\leq 2n_{0}<\infty$.

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  • $\begingroup$ I made a typo in the answer I attempted. Does it make more sense now? $\endgroup$ – TuringTester69 Apr 8 '18 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 '18 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. $\endgroup$ – TuringTester69 Apr 8 '18 at 1:38
  • $\begingroup$ Can you explain the $\uparrow$ notation? Does it mean “converges & increases to”? $\endgroup$ – TuringTester69 Apr 8 '18 at 1:50
  • $\begingroup$ Yes, convergence and the sequence is monotone, so the limit is the supremum. $\endgroup$ – user284331 Apr 8 '18 at 1:51

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