Basic help with sigma algebras and borel sets In non-rigorous, intuitive terms, can someone briefly define:
(i) a measurable set
(ii) a borel set
(iii) a sigma algebra
(iv) a borel sigma algebra
Im studying these concepts independently in preparation for a course in the fall and want to make sure I have a functional intuitive idea before learning them rigorously.  Im not looking for references to textbooks, or textbook definitions, just a quick intuitive description from someone who is familiar. 
 A: A $\sigma$-algebra is, like a topology, a set of subsets of some space $X$. It's both bigger and smaller than a topology, though: smaller, because it's only required to be closed under countable unions, instead of all unions, but bigger, because it's also closed under complementation, and thus by de Morgan's law, countable intersections. So it's both the open and closed sets you'd get starting from the base of a topology if you only took countable unions but also allowed countable intersections.
A measurable set is just an element of some $\sigma$-algebra on $X$. The content comes in when you define measures, which are functions from the $\sigma$-algebra to $[0,\infty]$ that satisfy a few obvious properties of a generalization of length.
The Borel algebra on some topological space $X$ is the $\sigma$-algebra generated by its topology: take all the closed and open sets, countable unions and intersections of those, complements of those, countable unions and intersections of those, and so on. A Borel set is just an element of the Borel algebra. 
Note in this case what I said about a $\sigma$-algebra being smaller than a topology does not hold at all! The Borel algebra is important specifically because it's the smallest $\sigma$-algebra containing the topology: we obviously would like to have open and closed sets of reals measurable, and to accomplish that we've got to let at least all the Borel sets be measurable as well. You can think of the Borel sets as every reasonable set; in particular the rationals are Borel in $\mathbb{R}$ with the standard topology, though they're neither closed nor open.
