Sigma algebra of a collection of sets.

Question

Let $\Omega= \{ 1, 2, 3, 4\}$, $\mathcal{F} =$ all subsets of $\Omega$ , and take the collections $\mathcal{A}_1=\{\{1,2\},\{1,4\}\}$ and $\mathcal{A}_2=\{\{2,4\}\}$

What is the smallest sigma algebras? $\sigma(\mathcal{A}_1)$ and $\sigma(\mathcal{A}_2)$?

$\sigma(\mathcal{A}_1) = 2^\Omega $

and $\sigma(\mathcal{A}_2)= \{\emptyset, \Omega,\{2,4\},\{1,3\}\} \subset \sigma(\mathcal{A}_1)$

Is this correct?

Answer 1

Remember that if:

a) $X\in\sigma$ then $\Omega\setminus X\in \sigma$

and

b) $X,Y\in\sigma$ then $X\cup Y\in\sigma$

We have:$$\{1,2\}\in \sigma \implies \{3,4\}\in \sigma$$ $$\{1,4\}\in \sigma \implies \{3,2\}\in \sigma$$ $$\{1,2\},\{1,4\}\in \sigma \implies \{1,2,4\},\{3\}\in \sigma$$ $$\{3,2\},\{3,4\}\in \sigma \implies \{3,2,4\},\{1\}\in \sigma$$ $$\{1\},\{3,4\}\in \sigma \implies \{1,3,4\},\{2\}\in \sigma$$ $$\{3\},\{1,2\}\in \sigma \implies \{1,2,3\},\{4\}\in \sigma$$ $$\{1\},\{3\}\in \sigma \implies \{1,3\},\{2,4\}\in \sigma$$

So $\sigma(\mathcal{A}_1) = 2^\Omega $ as you said.

Answer 2

Instead of giving you the actual solutions, I will give you the general method used find solutions to questions like these in the form of an Algorithm:

Since $\Omega$ is a finite set, Every $\sigma$-algebra on $\Omega$ is just an algebra on $\Omega$, And thus it is enough for you to find the smallest algebra generated by $A_1$ and $A_2$ (as opposed to finding the smallest $\sigma$-algebra generated by $A_1$ and $A_2$).

Or more elaborately, If we denote by $\mathscr{A}_\sigma(A)$ and $\mathscr{A}(A)$ the $\sigma$-algebra and the algebra generated by some set $A\subseteq \mathscr{P}(\Omega)$ respectively, Then we have since $\Omega$ is finite that $\mathscr{A}_\sigma(A)=\mathscr{A}(A)$ and thus in order for you to find $\mathscr{A}_\sigma(A)$ you just have to find $\mathscr{A}(A)$.

Inorder to find $\mathscr{A}(A)$ for a given $A\subseteq\mathscr{P}(\Omega)$ you will have to work with an algorithm:

Set $X = A\cup \{\emptyset,\Omega\}$

While $X$ is not an algebra on $\Omega$

Set $X = X\cup\{Y^c|Y\in X\}$

(In English it says: Add the complements of each set in $X$ into $X$)

Set $X = X\cup\{Y\cup Z| Y,Z\in X\}$

(In English it says: Add the unions of any two sets from $X$ into $X$)

Repeat until $X$ is an algebra on $\Omega$

Eventually, When the algorithm will end (And indeed it will end beacuse we always add subsets of $\Omega$ into $X$ in order to complete it into an algebra on $\Omega$, And because the greatest possible algebra on $\Omega$ is $\mathscr{P}(\Omega)$) you will get $\mathscr{A}(A)$, The smallest algebra generated by $A$.

In order to prove that indeed if $\Omega$ is finite then for any set $A\subseteq\mathscr{P}(\Omega)$ we have $\mathscr{A}_\sigma(A)=\mathscr{A}(A)$ we have to show that:

(1) $K$ is an algebra on $\Omega$ $\iff$ $K$ is a $\sigma$-algebra on $\Omega$

(For the $\Longrightarrow$ direction the proof is straightforward,
For the $\Longleftarrow$ direction we can use the fact from Prove that a countably infinite union of subsets of a finite set, can be expressed as a finite union )

(2) $\mathscr{A}(A)\subseteq \mathscr{A}_{\sigma}(A)$

(For this we need to use (1) on $\mathscr{A}_\sigma(A)$ and use the fact that $\mathscr{A}(A)$ is the smallest algebra on $\Omega$ containing $A$)

(3) $\mathscr{A}_\sigma(A)\subseteq \mathscr{A}(A)$

(For this we need to use (1) on $\mathscr{A}(A)$ and use the fact that $\mathscr{A}_\sigma(A)$ is the minimal $\sigma$-algebra on $\Omega$ containing $A$)

And thus we can conclude by using the Algorithm above that for finite $\Omega$ it gives the $\sigma$-algebra on $\Omega$ generated by $A$ as was required in the question.

Note:

Definition of an algebra on $\Omega$:

$\mathscr{A}\subseteq\mathscr{P}(\Omega)$ is an algebra on $\Omega$ If and only if:

$\emptyset,\Omega\in \mathscr{A}$

$\forall A\in \mathscr{A}, A^c\in \mathscr{A}$

$\forall A,B\in \mathscr{A}, A\cup B\in\mathscr{A}$