Show that given $\epsilon>0$ there exists $E\subset\mathbb{R}$ measurable so that $\int_{\mathbb{E}}f\,dm > \int_{\mathbb{R}}f\,dm-\epsilon$

Question

Let $f\colon \mathbb{R}\to [0,\infty]$ be measurable with $\int_{\mathbb{R}}f\,dm<\infty$. Show that given $\epsilon>0$ there exists $E\subset\mathbb{R}$ measurable so that

$$\int_{\mathbb{E}}f\,dm > \int_{\mathbb{R}}f\,dm-\epsilon.$$

Moreover, show that $E$ can be chosen so that $f$ is bounded on $E$

My work.

For $\epsilon>0$ be given. Then we have,

$$\int_{\mathbb{R}}f\,dm = \int_{\mathbb{E}}f\,dm + \int_{\mathbb{R-E}}f\,dm.$$

We also know that, by Fatou's lemma,

$$\int_{\mathbb{R}}f\,dm - \int_{\mathbb{R-E}}f\,dm \leq \liminf\left(\int_{\mathbb{R}}f_{n}\,dm-\int_{\mathbb{E}}f_{n}\,dm\right).$$

Am I going on the correct track? Or Are there any other way solve this one?

Any suggestion would be helpful. Thanks.

Answer 1

What's $f_n$ ?

The problem is to find $E$. Here's the idea.

First observe that integrability implies $f<\infty$ a.e. Next, let $E_n =\{x:f\le n\}$. Then $f$ is bounded on $E_n$, and $f {\bf 1}_{E_n} \nearrow f$ a.e. By monotone convergence

$$\int_{E_n} f = \int f {\bf 1}_{E_n} \nearrow \int_{\mathbb R} f.$$

Thus, for each $\epsilon>0$, $\int_{E_n} f \ge (\int_{\mathbb R} f) -\epsilon$ provided $n$ is large enough.

Now choose $n$ such that $\int_{E_n} f \ge (\int f) - \epsilon/2$.

Let $E_{n,m} = E_n \cap (-m,m)$. Then for each $n,m$, $E_{n,m}$ is a bounded set on which $f$ is bounded. Also, $f {\bf 1}_{E_{n,m}} \underset{m\to\infty} {\nearrow} f {\bf 1}_{E_n}$, so dominated convergence gives,

$$\int_{E_{n,m}} f =\int f {\bf 1}_{E_{n,m}} \underset{m\to\infty}{\nearrow} \int_{E_n}f.$$

Thus for all $m$ large enough, $\int_{E_{n,m}} f >(\int_{E_n} f) -\epsilon/2$, that is

$$ \int_{E_{n,m}} f \ge (\int f) -\epsilon.$$

Answer 2

First, fix an $\epsilon>0$ and let $B_n$ be a metric ball centered around origin, with radius $n$, namely $B_n = B(0,n)$. Note that by monotone convergence theorem, $$ \lim_{n\to\infty}\int_{B_n}f d\mu = \int f d\mu $$ hence, there exists an $N$ such that for every $n\geq N$, we have $$ \int f d\mu -\frac{\epsilon}{2}< \int_{E_n}f d\mu. $$ Next, for this $E_N$, let us define a new family of sets $G_n$ via $$ G_n \triangleq \{x \in E_n : f(x) \leq n\}. $$ Applying MCT one more time, we get $$ \lim_{n\to\infty}\int_{E_n}\mathcal{X}_{G_n}f d\mu = \int_{E_n}f d \mu, $$ hence for an $N^*$, we have $$\int_{E_n}f d\mu - \frac{\epsilon}{2}< \int_{E_n}\mathcal{X}_{G_{N^*}}fd\mu. $$ Combining, and keeping in mind that $G_{N^*} \cap E_n = G_{N^*}$ we get $$ \int f d\mu - \frac{\epsilon}{2}<\int_{G_{N^*}}fd\mu. $$ It is clear that $G_{N^*}\subset E_n$, and hence a bounded set; and furthermore by construction, $f$ is bounded on this set.

Answer 3

Let $f_n = \max(n,f)$. The Dominated convergence theorem shows that $\lim_n \int f_n = \int f$. Define $E_n = \{ x | f(x) \le n \}$. Then $f(x) = f_n(x) $ iff $x \in E_n$ and so $\lim_n \int_{E_n} f = \int f$. Clearly $f$ is bounded on $E_n$.