# Sigma Algebras generated by two classes of subsets

If $A_1$ and $A_2$ are two collection of subsets in $\Omega$ (Sample Space), I need to prove that $$\sigma(A_1) \subseteq \sigma(A_2).$$ I understand that there exist minimal unique $\sigma$-algebras generated by $A_1$ & $A_2$ respectively. However, I am not sure what needs to be demonstrated mathematically, in order to prove the subset status.

I tried to construct an example for this. Let A1={1,2} , A2={1,2,3} , Ω={1,2,3,4}

Then,

σ(A1)={∅,Ω,{1,2},{3,4}}

σ(A2)={∅,Ω,{1,2,3},{4}}

How can I proceed beyond this. I am confused as how to interpret the subsets as opposed to elements.

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Where is the problem from? – Jonas Meyer Sep 15 '12 at 6:21
Your slashes are the wrong one by the way. You should use the backslash. – aviness Sep 15 '12 at 6:24
Probably add the hypothesis $A_1\subset A_2$. After this is done, remind us how $\sigma(A)$ was introduced to you. – Did Sep 15 '12 at 6:25
If by "with the equality" you mean that $\sigma(A_1)=\sigma(A_2)$ might hold, you might want to use $\subseteq$ (produced by \subseteq) instead of $\subset$ (produced by \subset). Also note that you can get subscripts in $A_1$ and $A_2$ like this: A_1, A_2. – joriki Sep 15 '12 at 6:26
What did means is that you need some condition in order to say something. – AD. Sep 15 '12 at 8:38

You may forget the hypothesis $A_1\subset A_2$ or another one which is equivalent to it.

Then we know that $A_2$ is in $\sigma(A_2)$, then $A_1$ is in $\sigma(A_2)$. Since $\sigma(A_1)$ is the smallest $\sigma$-algebra which contains $A_1$, then $\sigma(A_1)\subset \sigma(A_2)$.

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In view of the Edit, I suggest to solve the exercise below.

Exercise Let $A$ and $B$ denote two subsets of a nonempty set $\Omega$. Let $\mathcal F$ denote the smallest sigma-algebra containing $A$ and $\mathcal G$ the smallest sigma-algebra containing $B$.

• Assume that $\mathcal F\subset \mathcal G$ and $\mathcal F\ne \mathcal G$. Then, either $A=\varnothing$ or $A=\Omega$.
• Assume that $\mathcal F=\mathcal G$. Then, either $A=B$ or $A=\Omega\setminus B$.
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I am sorry, but it's not clear to me yet. Could you please explain further. – Pichchamal Sep 15 '12 at 19:59
?? Did you try the exercise? – Did Sep 15 '12 at 20:08