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A problem in Friedmans Analysis is to show that a $\sigma$-ring is closed under countable intersections. Let the $\mathcal{R}$ be the $\sigma$-ring. I tried to solve it as follows:

Since $A - B \in \mathcal{R}$ for $A,B \in \mathcal{R}$, $\quad$ $A \cap B = A - (A - B) \in \mathcal{R}$

By induction, $\bigcap_{n = 1}^k E_n \in \mathcal{R}$ for $E_n \in \mathcal{R}$. Letting $k \to \infty$, the result follows

However, letting $k \to \infty$ did feel a bit fishy so i search around and found this solution. In it, the fact that $A-B \in \mathcal{R}$ is used but also that $\mathcal{R}$ is closed under countable unions.

Could anyone explain the problem/fault in taking the limit $k \to \infty$ as i did?

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    $\begingroup$ Your induction only gives that it is closed under any finite intersection. No matter how large $k$ gets, you're still only considering large finite intersections, not infinite ones. $\endgroup$
    – qaphla
    Commented Jul 1, 2014 at 16:44

1 Answer 1

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The first problem with letting $k\to\infty$ is what does that mean?! At first, a $\sigma$-algebra does not have any metric, topology, etc, so that we may speak of limits. If you look for it, you will probably find this notion of limit of a sequence $(A_n)$ of sets: Let $\limsup A_n=\bigcap_{n\in\mathbb{N}}\bigcup_{m=n}^\infty A_n$ and $\liminf A_n=\bigcup_{n\in\mathbb{N}}\bigcap_{m=n}^\infty A_n$ (this is based on $\limsup$ and $\liminf$ in the real line). We say that $(A_n)$ converges if $\limsup A_n=\liminf A_n$, and we define $\lim A_n=\limsup A_n$. Notice, however, that in the very definition of $\lim A_n$ we use countable unions and intersections.

A counter example would be the following: Let $\mathscr{Q}=\left\{A\subseteq\mathbb{R}:A=\varnothing\text{ or }\mathbb{R}-A\text{ is finite}\right\}$. Then everything you wrote (except for the limit) is still valid, because $\mathscr{Q}$ is a ring, but $\mathscr{Q}$ is not closed by countable intersections (each set $\mathbb{R}-\left\{0,1,\ldots,n\right\}$ belongs to $\mathbb{Q}$ but their intersection $\mathbb{R}-\mathbb{N}$ does not).

The usual proof of the result you stated is as follows:

By definition, a $\sigma$ ring must be closed by differences and countable unions. So, suppose that $\left\{A_1,A_2,\ldots\right\}$ is a countable family belonging to a $\sigma$-ring $\mathscr{R}$. Since $\bigcap_{n=1}^\infty A_n\subseteq A_1$, then $$\bigcap_{n=1}^\infty A_n=A_1-\left(A_1-\left(\bigcap_{n=1}^\infty A_n\right)\right)=A_1-\left(\bigcup_{n=1}^\infty (A_1-A_n)\right)$$ where the seocnd equality follows by de Morgan Laws. Since $\mathscr{R}$ is a $\sigma$-ring, then each $A_1-A_n\in\mathscr{R}$ (closed by differences), so $\bigcup_{n=1}^\infty(A_1-A_n)\in\mathscr{R}$ (closed by countable unions), and hence $\bigcap_{n=1}^\infty A_n=A_1-\left(\bigcap_{n=1}^\infty (A_1-A_n)\right)\in\mathscr{R}$, as we wanted.

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  • $\begingroup$ I was thinking that since $k$ could be taken as big as possible, $k \to \infty$ made sense. Your answer makes a valid point though, so thank you for clearing that up! $\endgroup$
    – user128779
    Commented Jul 1, 2014 at 18:26

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