real analysis - 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?

Answer

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)|

$$\begin{align*} \int\chi_{E_{n}}|f|\uparrow\int|f|, \end{align*}$$

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