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

$F_1 = \sigma(\{[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.

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 $\infty$!

Thanks!

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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.2012-09-24

1 Answers 1

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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 is in $F_1$.
  2. Show that each interval $(\leftarrow,a]\in F_1$.
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    Thank you very much for your help, I think I am starting to get the idea.2012-09-24