The question is the following: given $f:\mathbb{R}^n\rightarrow\mathbb{R}$ a Lebesgue medible function. We define the superlevel sets as:
$$E_n = \left\{x\in E: \left|f\right|\geq n\right\}$$
Then the following are true:
- If $f$ is Lebesgue integrable, then $\sum_n |E_n| < \infty$.
- If $E$ is of finite measure, and $\sum_n |E_n| < \infty$, then $f$ is Lebesgue integrable
What have I tried?
To be honest not much, I'm pretty much on the first item.
I've been trying to use that $$\sum_n |E_n| = \sum_n\int_E \chi_{E_n}$$ and $\chi_{E_n}\leq f_n=\frac{|f|}{n}\chi_{E_n}\leq |f|$ (To use dominated convergence or simply bounding each term).
However, since $\frac{1}{n}$ is not summable, this idea isn't really working (I can't see how the decreasing set might help give us a more rapid convergence than $\frac{1}{n}$ either).
I've also thought about making the sets disjoint using $|E_n| = |E_n\setminus E_{n+1}| + |E_{n+1}|$ with perhaps more terms but I don't really see that idea taking any serious direction.
Best Answer
In case someone is interested later, I think my first idea was in the right direction. Consider $$U_n=\left\{x\in E: n\leq |f(x)| < n+1\right\}$$ And then the functions $$g_n(x)=\sum_{k\leq n} \chi_{E_k}\quad\quad\quad g = \lim_{n\rightarrow\infty}g_n = \sum_{k\in\mathbb{N}} \chi_{E_k}$$ which are well defined, given that for any $x\in U_n\implies g(x)= n$, as it appears in only the first $n$ superlevels $E_n$. Consequently, we have that $g_n\leq g\leq |f|$ with $|f|$ integrable. Then by dominated convergence (D.C.T.): $$\sum_k |E_k| = \sum_k\int_E \chi_{E_k} = \lim_{n\rightarrow\infty} \sum_{k\leq n}\int_E\chi_{E_k} \stackrel{D.C.T.}{=}\int_E g \leq \int_E |f| <\infty $$
Now the second item is analogous. We want to consider the function $$h_n(x)=g_n(x)+1\quad\quad h=\lim_{n\rightarrow\infty} h_n = g+1$$ With that simple modification we have that $|f|\leq h$
By Fatou's Lemma (F.L.): $$\int_E |f|\leq\int_E h \stackrel{F.L}{\leq} \liminf_{n} \int_E g_n = \lim_n \left(\int_E 1+\sum_{k\leq n}\int_E\chi_{E_k}\right) = |E|+\sum_{k\in\mathbb{N}} |E_k|<\infty$$