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.
 A: 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}
