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Show that the $\sigma$-algebras generated by the collection of all intervals of the form $[a,b]\subset\Bbb R$ and by the collection of all the intervals of the form $(-\infty,b]\subset\Bbb R$ are equivalent.

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  • $\begingroup$ Where did you come across this question? Why are you interested in the answer? $\endgroup$ Nov 25, 2014 at 0:44
  • $\begingroup$ one of my homework question $\endgroup$
    – usengec
    Nov 25, 2014 at 0:49
  • $\begingroup$ Are you sure you've transcribed everything correctly? $[-\infty,b]$ isn't a subset of $\Bbb R,$ since $-\infty\notin\Bbb R.$ $\endgroup$ Nov 25, 2014 at 1:54
  • $\begingroup$ yes. it is a closed interval. @Cameron Buie $\endgroup$
    – usengec
    Nov 27, 2014 at 2:10
  • $\begingroup$ In that case, your text must have a misprint, or perhaps is using a strange convention $\endgroup$ Nov 27, 2014 at 4:11

1 Answer 1

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Hint: show that you can construct every type of generators from the other, using "allowed operations" in sigma algebras (countable unions, intersections,....).

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  • $\begingroup$ I am new to probability. can you be more specific? $\endgroup$
    – usengec
    Nov 25, 2014 at 0:55
  • $\begingroup$ $[a,b]=\cap_{n \in \mathbb{N}} (a- 1/n,b]$, and $(a- 1/n,b]=(-\infty,b] \cap (-\infty,a- 1/n]^c$, so by definition of sigma algebra we have $\{ [a,b] / a,b \in \mathbb{R} \} \subset \sigma( \{(-\infty,b] / b \in \mathbb{R} \})$. By definition of generated sigma algebra then $\sigma (\{ [a,b] / a,b \in \mathbb{R} \} ) \subset \sigma( \{(-\infty,b] / b \in \mathbb{R} \})$. You can prove the other content with similar arguments and conclude the equality. $\endgroup$
    – somebody
    Nov 25, 2014 at 7:42
  • $\begingroup$ @CameronBuie Thank you, I corrected it. $\endgroup$
    – somebody
    Nov 25, 2014 at 7:44

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