A measurable set from a sequence of sets

Question

Let (X,M, $\mu$) be a measure space and $E_n$ be a sequence of measurable sets. Let k be a fixed integer and G be the set of all those points that belong to $E_n$ for at least k values of n.

a) Prove that G is measurable.

I am trying to find a way to rewrite G as a countable union of countable intersections. I have come up with the following:

For k = 2, $G_2 = (E_1 \cap E_2) \cup (E_1 \cap E_3) \cup...\cup (E_2 \cap E_3) \cup (E_2 \cap E_4) \cup...\cup (E_k \cap E_{k+1}) \cup (E_k \cup E_{k+1})... $

$G_2 = \bigcup_{n=2}^\infty (E_1 \cap E_n) \cup \bigcup_{n=3}^\infty(E_2 \cap E_n) \cup...\cup \bigcup_{n=k+1}^\infty(E_k\cap E_n) \cup....$

$G_2 = \bigcap_{k=1}^\infty \bigcup_{n=k+1}^\infty (E_k \cap E_n)$

Since $G_2$ is a countable union of intersections it is in M and hence measurable.

I'm having trouble generalizing this approach for an arbitrary K...

b) Show $k\mu(G)\le \sum_{n=1}^\infty \mu(E_n)$

One hint was to look at: $\sum_{n=1}^\infty \int_G \chi_{E_n}d\mu$ however without the correct breakdown in part a) I'm not sure if I can proceed.

Edit: Here's what I have:

$k\mu(G) = \int_G k d\mu \le \int_G \sum_{n=1}^\infty \chi_{E_n}d\mu$ (not confident with the inequality here. Is this because if x is in g, it is in k or more of the En's?)

Continuing by MCT

$\int_G \sum_{n=1}^\infty \chi_{E_n}d\mu = \sum_{n=1}^\infty \int_G \chi_{E_n} d\mu = \sum_{n=1}^\infty \int \chi_G \chi_{E_n}d\mu = \sum_{n=1}^\infty \int \chi_{G \cap E_n} d\mu =\sum_{n=1}^\infty\mu(G\cap E_n) \le \sum_{n=1}^\infty \mu(E_n)$

*I have not learned about Markov inequality. This is a qualifying exam question and Markov inequality is not on the syllabus.

Answer 1

Let $\mathscr F$ be the set of all intersections of $k$ $E_n$'s. Then $\mathscr F$ is countable and $G$ is the union of all elements of $\mathscr F$.

Answer 2

Hint: Express $G$ as an upper level set of the measurable function $$ F:x\mapsto\sum_{n=1}^\infty 1_{E_n}(x). $$ (What does $F(x)$ mean?) For (b), apply Markov's inequality.

EDIT: Since $F(x)$ is the number of $n$ for which $x\in E_n$, we have $$ G = \{x\;|\;F(x)\ge k\} $$ is measurable. Now, observe that $k 1_G(x) \le F(x)$ since $k\le F(x)$ on $G$. Integrate both sides with respect to $\mu$ to get $$ k\mu(G) = \int k1_G(x)d\mu(x) \le \int F(x)d\mu(x) = \sum_{n=1}^\infty \mu(E_n) $$ by monotone convergence thereom.