# 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 $\sigma$-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.

• For the borel algebra: – rmh52 Aug 7 '12 at 2:47
• Why does it have to be closed and empty sets? For example consider the unit interval. The doesnt the collection of closed sets describe the whole interval? Why do we also require the open sets? – rmh52 Aug 7 '12 at 2:49
• If I follow, you're wondering why we need both closed and open sets measurable, when we can cover the unit interval with just closed sets. The intuition would be this: say we give the unit interval measure 1. Then we might expect $[0,1/2]$ to have measure $1/2$. Then shouldn't the rest of the interval also have measure $1/2$? But the remainder, the complement of $[0,1/2],$ is open. – Kevin Carlson Aug 7 '12 at 2:58
• thank you. That is exactly the answer I was looking for. – rmh52 Aug 7 '12 at 3:18
• One more clarification: When is a sigma algebra NOT a borel sigma algebra? – rmh52 Aug 7 '12 at 3:26