Measure Theory process Let $(X,S,\mu)$ be a measure space and let $\ E_1,\ E_2,\ ...,\ E_n\in S$. For fixed $m\in\{1,2,...,n\}$, lets define $C_m=\{x\in X:x\in E_j$ for exactly m indices $j\in \{1,2,...,n\}\}$
Thus, the next follows:


*

*$C_m\in S,\ m\in \{1,2,...,n\}$

*$\sum\limits_{m=1}^n\mu(E_m)= \sum\limits_{m=1}^nm\mu(C_m)$
So,  for the first step I managed to do this:
For each fixed $m\in \{1,2,...,n\}$ we have that $x\in C_m$ iff $x$ is in exactly m sets of $\{E_1,E_2,...,E_n\}$ iff $x$ is in m sets of $\{E_1,E_2,...,E_n\}$ but not in the remaining $(n-m)$ sets. From here I got stuck trying to write that $x$ is in a finite, and disjoint, union of $\binom{n}{m}$ sets of the type  $\cap_{k=1}^m E_{j_k} \setminus \cup_{l=m+1}^n E_{j_l}$
Any ideas on how to write such disjoint union? or another idea to prove both points.
 A: Hint: Integrate the function $\sum_{m=1}^n {\bf 1}_{E_m}=\sum_{m=1}^n m {\bf 1}_{C_m}.$
A: You almost have proved (1). First we simplify notations as follows. Put $[n]=\{1,\dots, n\}$ and $[n]^m=\{A\subset [n]:|A|=m\}$. For each $A\subset [n]$ put $E(A)=\bigcap \{E_i: i\in A\}\setminus  \bigcup \{E_j: j\in [n]\setminus A\}$. Since $S$ is a $\sigma$-algebra, $E(A)\in S$ for each $A\in [n]^m$, and so $C_m=\bigcup\{E(A): A\in [n]^m \} \in S$. 
(2) We claim that  if $A, A’$ are distinct subsets of $[n]$ then the sets $E(A)$ and $E(A’)$ are disjoint. Indeed, suppose to the contrary that there exists an element $x\in E(A)\cap E(A’)$. Since sets $A$ and $A’$, witout loss of generality we may suppose that 
there exists an index $i\in A\setminus A’$. Then $x\in E(A)\subset E_i$ and $x\in E(A’)\subset X\setminus E_i$, a contradiction. 
Thus 
$$\sum_{m=1}^n\mu(E_m)= \sum_{m=1}^n\sum_{ m\in A\subset [n]} \mu(E(A)) =\sum_{A\subset [n]} \sum_{m\in A} \mu(E(A))=$$ $$\sum_{A\subset [n]} |A|\mu(E(A))=\sum_{k=0}^n  \sum_{A\in [n]^k} k\mu(E(A))= \sum_{k=0}^n  k \mu(C_k).$$
