The question is to prove that if union of two sigma algebras is an algebra then it must be an sigma algebra.

My approach to show it is that if $A_1,A_2\in M$ then it must be true that $A_1\cup A_2\in M_1$ and also $A_1 \cup A_2\in M_2$ because if not , consider an $A_k\in M_1$ but $A_k\not\in M_2$ then $A_1\cup A_k\not\in M_1$ and $A_1\cup A_k\not\in M_2$ this implies $A_1\cup A_k\not\in M$. But since we know that M is an algebra this has to be true. I am looking to extend this argument via the process of induction to prove that M is also an sigma algebra but I haven't been able to formalise my thoughts properly please help .

  • $\begingroup$ I don't see how that argument works. Why couldn't you have $A_1\in M_1\setminus M_2$ and $A_2\in M_2\setminus M_1$ yet $A_1\cup A_2\in M_1$? I'm actually pretty sure you can have such situations take the usual measurable sets on $[0,1]$ and extend them by adding $\mathbb R\setminus [0,1]$ to each and taking the closure and do the same for $[2,3]$ then you get take any two non-trivial elements which contain the outsides. Their union is everything yet neither is in both. (note the union of these is not an algebra though). $\endgroup$ – DRF Mar 21 '17 at 10:53
  • $\begingroup$ @DRF Yes , I see your point , but I couldn't come up with the case where such the union of $M_1 and M_2 $ forms an algebra . I have predicated my argument on the assumption that it will always work if the union is an algebra $\endgroup$ – Noob101 Mar 21 '17 at 11:15
  • $\begingroup$ @DRF But as I myself point out that this an unverified assumption made on my part , so I will be glad if you could tell me the correct approach to the problem instead. $\endgroup$ – Noob101 Mar 21 '17 at 11:17

Let $(A_n)_n$ be a sequence in $M=M_1\cup M_2$. To prove that $A=\bigcup\limits_n A_n\in M$ we decompose $\bigcup\limits_n A_n$ as $$\bigcup_nA_n=\left(\bigcup_{n;A_n\in M_1}A_n\right)\bigcup\left(\bigcup_{n;A_n\in M_2}A_n\right).$$ Then $A\in M$, because each (countable) union belongs to $M$ (as $M_1$ and $M_2$ are $\sigma$-algebra) and $M$ is an algebra.


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