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I need to show that these generated $\sigma$-algebras are the same:

$F_1 = \sigma(\{[a,b) : -\infty<a<b<\infty \}) $

$F_2 = \sigma(\{(-\infty , x] :x \in \mathbb{R} \})$

I am not sure, but my idea thus far:

  1. Show that every interval $[a,b)$ is an element of $\{(-\infty , x) :x \in \mathbb{R} \}$.
  2. Show that every interval $(-\infty , x]$ is an element of $\{[a,b) : -\infty<a<b<\infty \}$.

If my idea is right (please tell me if I'm wrong), I am having trouble trying to express some things. For example, I want to show that I can choose an $a$ and a $b$ such that I get the open interval $(x,\infty)$, a complement of an element expressed in $\{(-\infty , x) :x \in \mathbb{R} \}$. Does it make sense to say something like $a=x+\epsilon : \epsilon \rightarrow 0$ and $lim_{n\rightarrow{}\infty}b_n = \infty$? I am not sure if I am expressing that $a$ approaches a number $x$, and that $b$ approaches $\infty$!


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You need to add $\sigma$ in your two statements to make them correct! Note the difference between a collection of sets such as $\{(-\infty,x]:x\in\Bbb R\}$ and the $\sigma$-algebra that they generate. –  Per Manne Sep 24 '12 at 8:54
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1 Answer

up vote 0 down vote accepted

To answer your last question, you cannot use limits directly in the fashion that you seem to be suggesting, but you can take countable unions, which $-$ if done right $-$ can have the same effect.

Neither of your statements (1) and (2) is correct, so both will be rather hard to prove! Try instead to show that each interval $[a,b)\in F_2$ and each interval $(\leftarrow,a]\in F_1$: that immediately implies that $F_1=F_2$.

Here’s a sequence of steps that will take care of one direction; remember that a $\sigma$-algebra is closed under both countable unions and complementation.

  1. Show that each interval $(\leftarrow,a)\in F_2$.
  2. Show that each interval $[a,\to)\in F_2$.
  3. Show that each interval $[a,b)\in F_2$.

For the other direction:

  1. Show that each interval $[a,b]$ with $a<b$ is in $F_1$.
  2. Show that each interval $(\leftarrow,a]\in F_1$.
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Oh, yes... I guess this is where I employ my newly learned "up-arrows" and "down-arrows": $a_n\downarrow{}a$? –  ialm Sep 24 '12 at 7:28
@iakl: Sounds good to me. –  Brian M. Scott Sep 24 '12 at 7:29
Also, for your first point under "For the other direction", why do I have to show that the interval $[a,b]$ is in $F_1$? –  ialm Sep 24 '12 at 7:32
@iakl: The first way that I saw to show that $(\leftarrow,a]\in F_1$ is to show first that all non-trivial intervals of the form $[a,b]$ are in $F_1$. There are other ways, but that was what immediately occurred to me. –  Brian M. Scott Sep 24 '12 at 7:38
Thank you very much for your help, I think I am starting to get the idea. –  ialm Sep 24 '12 at 7:40
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