Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For a certain $\sigma$-algebra $A$ on the real line, I would like to show that it contains the Borel sets. I can show that $A$ contains the left and right half-line $(a,\infty)$ and $(-\infty,b)$ for any real numbers $a$ and $b$. My question is : can I infer that $A$ contains the Borel sets by only prooving that it contains the left half-line or is it mandatory to show that $A$ contains both half-line? I'm not clear on how the Borel sets are generated from half-line and open intervals.

share|improve this question
3  
Maybe I'm missing something, but for any $a\leq b$, $(a,b)=(a,\infty)\cap(-\infty,b)$. So the $\sigma$-algebra contains the base of open intervals, and contains the $\sigma$-algebra generated by them, hence the Borel sets. –  Vika Feb 19 '12 at 3:54
3  
@Vika yes, but I think Nicolas was asking if it is enough to start with only one of these -- either right OR left -- half-open intervals; i.e. suppose you have $(a, \infty)$ in your $\sigma$-algebra for all $a\in \mathbb R$. –  William DeMeo Feb 19 '12 at 4:03

1 Answer 1

up vote 5 down vote accepted

What is the complement of one of these half-lines? Then consider intersections, unions, etc. In other words, yes, if you can show your $\sigma$-algebra contains $(a, \infty)$ for any $a$, that is enough... but it sounds like it would be a good exercise for you to prove this. Here are some hints:

  1. What is the complement of $(a, \infty)$?

  2. For $a< b$, what is the intersection of $(a, \infty)$ and $(-\infty, b]$?

  3. Show that a $\sigma$-algebra containing all half-open intervals $(a, b]$ contains all Borel sets. In fact, this is sometimes taken as the definition -- i.e. the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the half-open intervals. So, you are either done, or you need to show that this is equivalent to whatever definition you are using. (Hint: at this point you have all the intervals $(a, b-1/n]$, for $n\in \mathbb N$ in your $\sigma$-algebra.)

share|improve this answer
1  
Your trivia is motivating. Here is my attempt. 1. The complement of $(a,\infty)$ is $(-\infty,a]$. 2. The intersection of $(a,\infty)$ and $(-\infty,b]$ is $(a,b]$. 3. Since $(a,b)=\cup_{n=0}^{\infty}(a,b-1/n]$ then the $\sigma$-algebra $A$ contains all open intervals. Since any open set is the union of a countable collection of open intervals, then $A$ contains all the open sets. Since the Borel $\sigma$-algebra is the intersection of all such $\sigma$-algebra, then $A$ contains the Borel $\sigma$-algebra and hence all Borel sets. Thank you. –  Nicolas Essis-Breton Feb 19 '12 at 21:06

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.