Elements in a sigma algebra generated by a partition Suppose you have a set $X$. Consider the $\sigma$-algebra $\mathcal{A}:= \sigma\{A_1,A_2,\cdots,A_n\}$, where $\{A_1,A_2,\cdots,A_n\}$ form a partition on $X$. Is it then true that every element $A \in \mathcal{A}$ can be written as $$A = \bigcup_{i\in I} A_i, $$where $I$ is a subset of $\{1,2,\cdots,n\}$?
It seems very trivial to me, but as I've learned along the years I should not always trust my instincts.
 A: Yes, that is correct. Let us write
$$
\mathcal{A} = \left\{ \bigcup_{i \in I} A_i : I \subseteq \{1, \ldots,n\}\right\},
$$
we will show that $\mathcal{A} = \sigma \{A_1, \ldots, A_n\}$ by showing that


*

*$\mathcal{A}$ is a $\sigma$-algebra,

*any $\sigma$-algebra containing $\{A_1, \ldots, A_n\}$ must contain $\mathcal{A}$,
because then $\mathcal{A}$ is indeed the smallest $\sigma$-algebra containing $\{A_1, \ldots, A_n\}$. Every step is actually quite straightforward, but I found that when I doubt something that seems trivial it helps to write it out completely to take away any doubts I have.
The second point is actually the easiest to check: a $\sigma$-algebra is closed under countable unions and must thus certainly contain all the unions of the elements in $\{A_1, \ldots, A_n\}$, since those are at most finite unions.
The first point just involves checking the three properties of the definition of a $\sigma$-algebra.


*

*$X$ is in $\mathcal{A}$, this checks out because $X = \bigcup_{i \in \{1, \ldots,n\}} A_i \in \mathcal{A}$. Here we used that the $\{A_1, \ldots, A_n\}$ forms a partition of $X$.

*$\mathcal{A}$ is closed under complements, let $A \in \mathcal{A}$, then $A = \bigcup_{i \in I} A_i$ for some $I \subseteq \{1, \ldots, n\}$. Define $J = \{1, \ldots, n\} - I$, then $B := \bigcup_{j \in J} A_j$ is an element of $\mathcal{A}$. Again, using that the $\{A_1, \ldots, A_n\}$ form a partition of $X$ we have that $B = X - A$, and so we see that $\mathcal{A}$ is closed under complements.

*$\mathcal{A}$ is closed under countable unions, in fact $\mathcal{A}$ is finite, so we only need to check that it is closed under finite unions. Let $B_1, \ldots, B_k \in \mathcal{A}$. Then we have $J_1, \ldots, J_k \subseteq \{1, \ldots, n\}$ such that $B_i = \bigcup_{j \in J_i} A_j$ for all $1 \leq i \leq k$. Set $J = \bigcup_{1 \leq i \leq k} J_i$, then clearly
$$
\bigcup_{1 \leq i \leq k} B_i = \bigcup_{j \in J} A_j \in \mathcal{A},
$$
as required.


And so we can conclude that $\mathcal{A}$ is indeed a $\sigma$-algebra.
Note that the above proof essentially goes through as well if $\{A_1, A_2, \ldots\}$ would be a countable partition of $X$.
