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I am confused when learning about Borel $\sigma$-Algebra, the source of confusion is that the Borel $\sigma$-Algebra is on the Real numbers...

Confusion arises from the fact that i began learning about $\sigma$-Algebra's and how it can work with sets like the "Power Set", ie discrete cases.

So how can I come to understand that [2,6] or (π,9) is a Borel Set ?

Brofessor
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1 Answers1

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The Borel $\sigma$-algebra on the reals is the smallest $\sigma $-algebra that contains all the open sets. It's a very large class of subsets of reals and it is very difficult to understand in its entirety. Luckily, you don't need to understand each and every Borel set. It suffices to be able to reason about certain sets that interest you. So, the interval $(\pi, 9)$ is an open set, and therefore is an element of any $\sigma $-algebra which contains the open sets. In other words, $(\pi, 9)$ is a Borel set. As for $[2,6]$, remember that any $\sigma$-algebra is closed under taking complements. So, if you can show that $(\infty, 2)\cup (6,\infty )$ is a Borel set, then it would follow that so if its complement, i.e., $[2,6]$. Can you do that now?

Ittay Weiss
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  • Hope this message gets you. I am impressed with the friendly and simple answer you gave in here. I posted a question of related subject in this link: http://math.stackexchange.com/questions/1135884/proving-two-measures-of-borel-sigma-algebra-are-equal , I am wondering if you could apply your analysis to mine. A caveat though: Mine comes from an introductory book by Richard Bass, so any solution has to be without big tools like Dynkin's. There're also some in my post wanting to see solution with no big tools. Thank you for your time! – A.Magnus Feb 07 '15 at 17:29