Estimate of the measure of the set of elements that belongs at least to n elements of a family of measurable sets.

I've been trying to solve this for some hours it sounds so easy but I can not get the answer, I would appreciate any idea or help.

The problem:

Let $(X,A,\mu)$ be a measure space and $F=\{A_n\}_{n\in \mathbb N}$ a family of measurable sets in $A$. For $m \in \Bbb N$ Let $B_m$ be the set of $x \in X$ such that $x$ belongs to at least $m$ members of $F$ . Show that $B_m$ is measurable and :

$\mu (B_m) \le \frac {1} {m} \sum_1^\infty{\mu (A_n)}$

Well so far i have seen it is measurable since i can see the set as the union of al distinct intersection of k diferent $A_n$ and this union is numerable.

After that i have no clue.

For $\mathbb E\subset X$, let $\chi_E$ be the characteristic function of $E$, that is, $$\chi_E(x) = \begin{cases}1,&x\in E\\0,&x\notin E.\end{cases}$$ Then for each $m$ we have $$B_m = \left\{x\in X: \sum_{n=1}^\infty \chi_{A_n}(x)\geqslant m \right\}= \left(\sum_{n=1}^\infty \chi_{A_n} \right)^{-1}([m,\infty)).$$ Set $\tilde A_n = A_n\setminus\bigcup_{j=1}^{n-1} A_j$ (with $A_0:=\varnothing$), then $\bigcup_{j=1}^n \tilde A_j = \bigcup_{j=1}^n A_j$ for all $n$ so $$\bigcup_{j=1}^\infty \tilde A_j = \bigcup_{j=1}^\infty A_j$$ as both $\{A_j\}$ and $\{\tilde A_j\}$ are increasing sequences of sets. Moreover, the $\tilde A_j$ are pairwise disjoint by construction and $\tilde A_j\subset A_j$ for all $j$, so \begin{align} \mu(B_m) &= \mu\left(\left(\sum_{n=1}^\infty \chi_{A_n} \right)^{-1}([m,\infty))\right)\\ &\leqslant \mu\left(\left(\sum_{n=1}^\infty \chi_{\tilde A_n} \right)^{-1}([m,\infty))\right)\\ &= \mu\left(\left(\frac1m\sum_{n=1}^\infty\chi_{\tilde A_n} \right)^{-1}([1,\infty))\right)\\ &=\frac1m\mu\left(\bigcup_{n=1}^\infty \tilde A_n \right)\\ &=\frac1m\sum_{n=1}^\infty \mu\left(\tilde A_n\right)\\ &\leqslant \frac1m\sum_{n=1}^\infty \mu\left(A_n\right). \end{align}