Question: Suppose $E_n$, $n\in\mathbb{N}$, is a sequence of Lebesgue measurable subsets of $[0,1]$. Let $F$ be the set of all points $x\in[0,1]$ that belong to at least $K$ (some positive number) of the $E_n$'s. Prove that $F$ is Lebesgue measurable and $\sum_{n=1}^\infty m(E_n)\geq Km(F)$.
My Attempt/Idea: First, let's show that $F$ is measurable. Let's consider a function $f=\sum_n\chi_{E_n}$. Then, $f:[0,1]\rightarrow[0,\infty]$ is measurable, and so $f^{-1}([K,\infty])$ is measurable. Since $f^{-1}([K,\infty])$ is precisely the number of points that belong to at least $K$ of the $E_n$'s, we have that $F=f^{-1}([K,\infty])$ is measurable.
Now we want to show the inequality. $\int f=\int\sum_n\chi_{E_n}=\sum_n\int\chi_{E_n}$, since $f$ are nonnegative functions by MCT. Let $G$ be the set of all points $x\in[0,1]$ that don't belong to at least $K$ of the $E_n$'s. Then, $\sum_n\int\chi_{E_n}=\sum_n(\int_F\chi_{E_n}+\int_G\chi_{E_n})$…. but I am not sure if I am on the right track…..
Best Answer
You were almost there. Hint: $\int_0^1 f\, dm \ge \int_F f\,dm.$