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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.

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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:

  1. $\mathcal E$ is finite (its cardinality is at most $2^{m+1}$);
  2. If $E,F\in \mathcal E$ and $E\neq F$ then $E\cap F= \emptyset$
  3. $X\subseteq\bigcup_{E\in \mathcal E} E$

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 $M=\mathcal H$.

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