# Lebesgue integral of a function that is measurable, finite almost everywhere, and integrable on a set of finite measure

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?

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 $|E_{n_{0}}|\leq 2n_{0}<\infty$.
• 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 $\infty$. Apr 8, 2018 at 1:34
• Can you explain the $\uparrow$ notation? Does it mean “converges & increases to”? Apr 8, 2018 at 1:50