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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.

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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
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@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

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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.)

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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

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