Show that any element of a sigma algebra is the union of disjoint sets Let $\mathscr{M}$ be a $\sigma$-algebra on $X$ generated by a finite family of sets. Prove that there exists a partition of $X$ into disjoint sets $E_1, E_2, \ldots, E_n$ such that $A$ is an element of $\mathscr{M}$ if and only if $A$ is the union of some sets $E_1, E_2,\ldots, E_n$.
My work so far is
Suppose $\mathscr{M}$ is generated by a finite collection say $\{B_i\}$ $i = 1,\ldots,n$.
Then I define a partition $\mathscr{P}$ by 
$$\mathscr{P} = \left\{\bigcap C_i: C_i = B_i  \text{ or } B_i^c, i = 1,\ldots,n\right\}$$
Let $\mathscr{L}$ be a collection of the arbitrary union of the sets in $\mathscr{P}$.  I claim that $\mathscr{L}$ is a $\sigma$-algebra.  If I can show that $\mathscr{L} = \mathscr{M}$, is this enough to prove the proposition?
Thank you.
Sorry about the typsetting.  I don't know how to type in the symbols.
 A: Yes, your approach is essentially correct. Here is a proof, based on your approach.
Suppose $M$ is the $\sigma$-algebra generated by $B_1,\dots,B_m$.  If $B_1,\dots,B_m$ does not cover $X$, define $B_0=X-\bigcup_{i=1}^m B_i$. 
Then we have $X\subseteq \bigcup_{i=0}^m B_i$, and note that $M$ is also the $\sigma$-algebra generated by $B_0,\dots,B_m$.
Consider the family $\mathcal E= \{\bigcap_{i=0}^m C_i \,| \, \textrm{ para todo } i\in \{0,\dots,m\}, C_i=B_i \textrm{ or } C_i=B_i^c\}$. Then we have: 


*

*$\mathcal E$ is finite (its cardinality is at most $2^{m+1}$);

*If $E,F\in  \mathcal E$ and $E\neq F$ then $E\cap F= \emptyset$

*$X\subseteq\bigcup_{E\in  \mathcal E} E$


So $\mathcal E$ is a finite partition of $X$.
Note that, since $\mathcal E$ is a finite collection, any union of sets in $\mathcal E$ is, in fact, a finite union of sets in $\mathcal E$
Let $\mathcal H$ be the class of finite union of sets in $\mathcal E$. 
Since, $\mathcal E\subseteq M$ and $M$ is a $\sigma$-algebra, we have  $$\mathcal H\subseteq M \tag{1}$$ 
On the other hand, it is easy to prove that $\mathcal H$ is a $\sigma$-algebra. (In fact, any countable union of sets in $\mathcal H$ reduces to finite union of sets in $\mathcal E$). 
Since, given any $r\in \{0,\dots,m\}$, we have 
$$B_r=\bigcup \left \{\bigcap_{i=0}^m C_i \in \mathcal E\,| \, C_r=B_r \textrm{ and, para todo } i\in \{0,\dots,m\}-\{r\}\,,\, C_i=B_i \textrm{ or } C_i=B_i^c\right \}$$ 
So, $\{B_0,\dots,B_m\}\subseteq \mathcal H$. So, 
$$M=\sigma(\{B_0,\dots,B_m\})\subseteq \mathcal H \tag{2}$$
From $(1)$ and $(2)$, we have $M=\mathcal H$. 
