Prove that the set $B_m$ is measurable

Question

Let $(X, \mathcal{A}, \mu)$ be a measure space and $\{A_n\}$ a family of measurable sets. For $m \in \mathbb{N}$ let $B_m$ be the set of points $x \in X$ that belong to at least $m$ of the sets $A_n$. Prove that $B_m$ is measurable and that

$\mu (B_m) \leq \dfrac{1}{m} \sum_{n=1}^{\infty} \mu (A_n)$

I have been trying using the theorem that said that $\mu (\cup_{n=1}^\infty C_n) \leq \sum_{n=1}^{\infty} \mu (C_n)$ for a family of measurable sets $\{C_n\}$ and I'm looking for the family that could help me, but I don't know how to use the factor $\dfrac{1}{m}$.

I'm also trying to see the set $B_m$ as the intersection of other measurable sets that I know.

I tried with induction over $m$ too.

Answer 1

Note that $x\in B_m$ if and only if to every $k \ge m$, there corresponds a function $f : \{1,2,\ldots, k\} \to \Bbb N$ such that $x\in A_{f(j)}$ for $j = 1,2,\ldots, k$. Thus

$$B_m = \bigcap_{k\ge m} \bigcup_{f : \{1,2,\ldots, k\}\to \mathbb N} \bigcap_{j = 1}^k A_{f(j)}\in \mathcal{A}$$

Now $\sum\limits_{n = 1}^\infty 1_{A_n}(x) \ge m1_{B_m}(x)$ for all $x\in X$. Integrating over $X$ yields $$\int_X \sum\limits_{n = 1}^\infty 1_{A_n}(x) \ge m\mu(B_m)$$ By the monotone convergence theorem, $$\int_X \sum_{n = 1}^\infty 1_{A_n}\, d\mu = \sum_{n = 1}^\infty \int_X 1_{A_n}\, d\mu = \sum_{n = 1}^\infty \mu(A_n)$$ Hence $$\frac{1}{m}\sum_{n = 1}^\infty \mu(A_n) \ge \mu(B_m)$$ as desired.

Answer 2

For the first question: The function $f = \sum_n \chi_{A_n}$ is a measurable function from $X$ to $[0,\infty].$ Therefore $f^{-1}([m,\infty])$ is measurable. The latter set is precisely $B_m.$