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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.

Appreciate your comments. Thank you.

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    Your question does not make sense. If $\sigma(A_1) \subseteq \sigma(A_2)$ is true, then the opposite inclusion is also true(because you don't make any assumption on $A_1$ and $A_2$ other than they are collections of subsets of $\Omega$). So the both are equal. This is clearly not true in general. Your example does not make sense, either. $A_1$ and $A_2$ should be collections of subsets of $\Omega$.2012-09-15

2 Answers 2

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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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    ?? Did you try the exercise?2012-09-15