Assume $A_1,A_2,...$ are measurable sets. Let $m\in\mathbb{N}$, and let $E_m$ be the set defined as follows:
$x\in E_m\iff$ $x$ is a member of at least $m$ of the sets $A_k$. Prove that $E_m$ is measurable and that
$$m\lambda(E_m)\le\sum_{k=1}^{\infty}\lambda(A_k)$$
Here everything is in Lebesgue measure. ($\lambda(E_m)$ denotes the Lebesgue measure of $E_m$)
My thoughts: I think the inequality is fairly intuitive since $E_m$ contains the $x$'s that appear at least $m$ times in $A_1,A_2,...$, and so $m$ is a lower bound. But I'm not sure how to prove rigorously.
To show $E_m$ is measurable, let $S$ be the collection of all subsets of $\mathbb{N}$ with at least $m$ elements. I think we can represent $E_m$ as $\cup_{T\in S}\cap_{i\in T}A_i$. Please correct me if I'm wrong. Thank you.
This is not unique to Lebesgue measure and in fact is valid in any measure space. Let $\chi_E(x)$ denote the indicator function of $E$.
Since $$\chi_{A_1}(x) + \chi_{A_2}(x) + \cdots \ge m$$ for all $x \in E_m$ you have $$m \chi_{E_m}(x) \le \sum_{k=1}^\infty \chi_{A_k}(x)$$ for all $x$. The measure inequality is obtained by integrating both sides and applying the monotone convergence theorem.
As for measurability, a point $x$ belongs to exactly the $m$ sets $A_{k_1},\ldots,A_{k_m}$ if and only if $$x \in A_{k_1} \cap \cdots \cap A_{k_m} \cap \left( \bigcap_{j \in \mathbb N \setminus \{k_1,\ldots,k_m\}} A_j^c \right).$$ This means $$E_m = \bigcup_{k_1,\ldots,k_m \in \mathbb N} \left[ A_{k_1} \cap \cdots \cap A_{k_m} \cap \left( \bigcap_{j \in \mathbb N \setminus \{k_1,\ldots,k_m\}} A_j^c \right) \right]$$ where the union is taken over (the countable family of) $m$-tuples of distinct natural numbers.