Find the smallest $\sigma$-algebra containing specific sets

measure-theoryprobability theory

Given $\Omega = \{0, 1, 2, 3\}$, my task is to find the smallest $\sigma$-algebra containing $\{0\}$ and $\{0,2\}$.

Here is my attempt:

I know that a $\sigma$-algebra is a collection of sets of $\Omega$ that is closed under countable union, countable intersection, and complements. With this definition in mind and letting $A = \{\{0\}, \{0,2\} \}$, I think I have found the smallest $\sigma$-algebra of $A$, denoted $\sigma(A)$ to be
$$ \sigma(A) = \{ \emptyset, \Omega, \{0\}, \{0,2\}, \{1,2,3\}, \{1,3\} \} $$

Is this correct? My thought process was to just take complements of the provided sets and then take unions of all elements generated through that process, and then finally taking the intersection of all such elements.

Best Answer

Suppose $\Omega$ is any set and that $E_1, \ldots, E_k$ are disjoint subsets of $\Omega$ which form a partition (i.e. their union is $\Omega$). We want to find $\mathscr{F} = \sigma(\{E_1, \ldots, E_k\}).$ We propose to demonstrate that $\mathscr{F} = \left\{ \bigcup\limits_{j \in F} E_j \middle| F \subset \{1, \ldots, k\} \text{ is finite}\right\}.$ As a corollary, $\mathrm{card}(\mathscr{F}) = 2^k.$

Then,

  • $\mathscr{F}$ is clearly closed under countable unions;
  • $\varnothing \in \mathscr{F}$ (take $F = \varnothing$);
  • $A \in \mathscr{F} \implies A^\complement \in \mathscr{F}$ for if $A = \bigcup\limits_{j \in F} E_j$ then $A^\complement = \bigcup\limits_{j \in F^\complement} E_j$ (here $F^\complement$ is $\{1,\ldots,k\}\setminus F$).

Thus, the claim has been demonstrated.

In your exercise, it is rather easy to see that $\sigma(A) = \sigma(\{\{0\},\{2\},\{1,3\}\}).$ Hence, you should have $2^3 = 8$ elements.

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