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Infinite union of sigma algebras
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[QUOTE="fishturtle1, post: 6330041, member: 606256"] Thanks for the reply. I tried working with the sequence ##(B_n)## where ##B_n## is an element of ##\mathcal{A}_{n+1} \setminus \mathcal{A}_n##. We want to show ##\bigcup_{n} B_n \not\in \mathcal{A}_\infty##. We can observe that ##B_1 \in \mathcal{A}_2\setminus\mathcal{A}_1, B_2 \in \mathcal{A}_3\setminus\mathcal{A_2} \supset \mathcal{A}_3 \setminus \mathcal{A}_1 \dots## and so for all ##n## we have ##B_n \not\in \mathcal{A}_1##. Similarly, for all ##n \ge 2## we have ##B_n \not\in \mathcal{A}_2## and continuing in this way, for all ##n \ge k## we have ##B_n \not\in \mathcal{A}_k##. Assume by contradiction that ##\bigcup_n B_n \in \mathcal{A}_m## for some ##m##. Then, ##B_1, B_2, \dots, B_{m-1} \in \mathcal{A}_m##. Since ##\mathcal{A}_m## is closed under complements and countable intersection, we have ##\bigcup_{k=m}^{\infty}B_k \in \mathcal{A}_m##. By construction, ##B_m \not\in \mathcal{A_m}##. Under these assumptions, can we show ##B_m \in \mathcal{A}_m## to get a contradiction? Thanks for the stack exchange link (I haven't clicked on it yet but maybe if this problem turns out too be too hard I will...) its funny, my homework was too hard so I found this problem in textbook and thought it'd be be a fun one to do as a warm up... if you have time, would you be able to give me a hint, please? [/QUOTE]
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Infinite union of sigma algebras
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