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Suppose $\{\mathcal{F}_n\}_{n=1}^{\infty}$ is a sequence of $\sigma$-algebras of subsets of $X$ where they are properly nested: $\mathcal{F}_n \subset \mathcal{F}_{n+1}, \mathcal{F}_n \neq \mathcal{F}_{n+1}$. Prove that there is a sequence of disjoint sets $\{A_n\}_{n=1}^{\infty}$ and a subsequence $\{\mathcal{F}_{n_k}\}_{k=1}^{\infty}$ such that $A_k \in \mathcal{F}_{n_{k+1}}-\mathcal{F}_{n_{k}}$.

I found the proof in" A. Broughton and B. W. Huff: A comment on unions of sigma-fields. The American Mathematical Monthly, 84, no. 7 (1977), 553-554" but wondered to know if anyone knows any other proof than this one.

domath
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1 Answers1

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As it has already been mentioned, the title of this post is not related to the question you have asked.

Isn't this essentially the construction of the proof of the Monotone convergence for sets. By disjointification rule, it's always possible to construct such a sequence {$A_n$}. Disjointification principle can be stated as follows.

If $M\subset2^X$ is a $\sigma $-algebra and $G_1,G_2,...$ is a sequence of sets from $M$, then there exists a sequence {$F_n$} of pairwise disjoint sets from $M$ such that $\cup_{k=1}^{n} G_k =\cup_{k=1}^{n} F_k$,for $n$ is either a positive integer or $\infty$

This is possible if we choose $F_n = G_n\backslash \cup_{k=1}^{n-1} G_k$.

In your case, where the given sequence {$G_n$} is increasing so that $F_n = G_n\backslash \cup_{k=1}^{n-1} G_k=G_n\backslash G_{n-1}$.

SL_MathGuy
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