TITLE: Show a $\sigma$-algebra contains the Borel sets : with $(a,\infty)$ or $(-\infty,b)$?
QUESTION [6 upvotes]: 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.

REPLY [6 votes]: 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:

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