Let F and G be two σ-algebras of subsets of Ω. Is $F\cup G$, the collection of subsets of Ω lying in either F or G a σ-algebra?

I am able to prove this using an example:-

Let F={$\varnothing$,$\Sigma$}

and G=$2^\Sigma$

So their union would be the bigger set G,which is a $\sigma$ algebra already.
based on this can I argue that whenever I am union-ing two $\sigma$ algebras , one will be the subset of the other and so the result would be the bigger set which is already a $\sigma$ algebra?

  • $\begingroup$ I think you swapped $\Sigma$ and $\Omega$ partway through . . . $\endgroup$ – Noah Schweber Aug 29 '16 at 17:22

Your reasoning is incorrect. In fact, you haven't really provided any reasoning! How do you go from an example of one $\sigma$-algebra being contained in another, to the claim that for every pair of $\sigma$-algebras, one is contained in the other? There's no reason to believe that the example you give is somehow "the usual case."

HINT: For each element $a\in\Omega$, the set $\Sigma_a=\{\emptyset, \Omega, \{a\}, \Omega\setminus\{a\}\}$ is a $\sigma$-algebra (exercise). For $a, b\in\Omega$ with $a\not=b$, what can you say about $\Sigma_a$ versus $\Sigma_b$? Does either contain the other?

Thinking more about this example, do you see how to answer your main question?

  • $\begingroup$ I thought of it like this:- Since every σ-algebra must contain Ω, so their union contains Ω. Now since $Ω^c$ is ∅ it must also be present(by σ-algebra property) in the union. Now $Σ_a$ already has {a} and ${a}^c$,same for $Σ_b$. So their union will contain both. hence every property is fulfilled. $\endgroup$ – Sukanta Chatterjee Aug 29 '16 at 18:46
  • $\begingroup$ @SukantaChatterjee I think you need to re-examine the definition of $\sigma$-algebra. A collection of sets, $C$, is a $\sigma$-algebra iff it is closed under complement, countable unions, and countable intersections. Is the set $\Sigma_a\cup\Sigma_b$ really a $\sigma$-algebra? (HINT: can you find two elements $X, Y\in\Sigma_a\cup\Sigma_b$ such that $X\cup Y$ is not in $\Sigma_a\cup\Sigma_b$?) $\endgroup$ – Noah Schweber Aug 29 '16 at 18:57
  • $\begingroup$ Also, you still haven't addressed the question of whether either of $\Sigma_a, \Sigma_b$ is contained in the other. Do you see what the answer to this is? $\endgroup$ – Noah Schweber Aug 29 '16 at 18:58
  • $\begingroup$ No $Σ_a$ and $Σ_b$ are not contained in one another.So my previous assumption was wrong.Also X={a} and Y={b} then XUY ={a,b} $\notin$ $Σ_a∪Σ_b$. So it is not closed under union, hence not a σ-algebra. $\endgroup$ – Sukanta Chatterjee Aug 29 '16 at 19:06
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    $\begingroup$ Thanks a lot.@Noah Schweber You actually walked me through the entire proof instead of giving one directly. This felt really good! :) $\endgroup$ – Sukanta Chatterjee Aug 29 '16 at 19:11

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