# If union of two sigma algebras is an algebra , then it is an sigma algebra

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 .

• 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). – DRF Mar 21 '17 at 10:53
• @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 – Noob101 Mar 21 '17 at 11:15
• @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. – 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.