Could you prove that the sigma algebra σ generated by the classes $A_1$ and $A_2$ satisfy the condition:$ σ(A_1\cup A_2)=σ(σ(A_1)\cup\sigma(A_2))$ I've been trying to do this exercise but I'm not been able
2026-03-29 11:42:30.1774784550
Sigma algebra properties exercise
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$\sigma (A_1\cup A_2)$ is a $\sigma $ algebra that contain $A_1\cup A_2$. In particular, it contains $A_1$, and thus $\sigma (A_1)$ and $A_2$, therefore $\sigma (A_2)$ as well. Therefore $\sigma (A_1)\cup\sigma (A_2)\subset \sigma (A_1\cup A_2)$ and thus $\sigma (\sigma (A_1)\cup \sigma (A_2))\subset \sigma (A_1\cup A_2)$ since $\sigma (\sigma (A_1)\cup \sigma (A_2))$ is the smallest $\sigma -$algebra that contains $\sigma (A_1)\cup \sigma (A_2)$.
For the converse inclusion, obviously $A_1\cup A_2\subset \sigma (\sigma (A_1)\cup \sigma (A_2))$, and thus $\sigma (A_1\cup A_2)\subset \sigma (\sigma (A_1)\cup \sigma (A_2))$, since $\sigma (A_1\cup A_2)$ is the smallest $\sigma -$algebra that contains $A_1\cup A_2$.