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I am reading both the proof and the lecture notes of the theorem and I don't understand the following. The proof is in a 1996 book "Probability" by Shiryaev (ISBN 978-1-4757-2541-4).

We have $(A_n)_{n=1}^\infty$ a decreasing sequence of sets, $A_{n+1}\subset A_{n}$, and $\cap^\infty_{n=1}A_n=\emptyset$. Here the $A_n$ are subintervals or finite disjoint unions of subintervals in $\mathbb{R}$ and of the type $(a,b]$. Pick an auxiliary sequence $(B_n)$ such that $\cap^\infty_{n=1}[B_n]\subset\cap^\infty_{n=1}A_n$, where $[B_n]$ is the closure of $B_n$. Assume $A_1\subset [-N,N]$ for $N<\infty$:

$$[-N,N]=[-N,N]-\cap^\infty_{n=1}B_n=\cup^\infty_{n=1}([-N,N]-[B_n])$$

The claim is now that we have a open covering of $[-N,N]$, but I don't see how $[-N,N]-[B_n]$ are always open. The simple case where $[B_n]=[-1,1]$ already gives such an example. The only other information for how $B_n$ is chosen is the requirement that $P(A_n-B_n)=2^{-n}\epsilon$, but this seems unrelated to my question.

Waaal
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1 Answers1

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They are open in the relative topology of $[-N,N]$. So for example $[-1,-1/2)\cup (1/2,1]$ is relatively open in $[-1,1]$, because it is the intersection of $[-1,1]$ with the open (in $\mathbb{R}$) set $(-2,-1/2)\cup (1/2,2)$. I guess the argument uses now compactness, etc.

GReyes
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