Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Here $B_{2}$ is the $\sigma$-algebra generated by open sets on the plane, $B$ is the $\sigma$-algebra generated by open sets on the real line. I need to prove the product measure coincides with the given measure.

share|cite|improve this question
up vote 1 down vote accepted

Open sets in the plane are generated by open balls (well, discs, but this arguments apply not only to $\mathbb R^2$ but to $\mathbb R^n$). The product topology is generated by open rectangles.

So what you want to prove is that every ball contains a rectangle, and that every rectangle contains a ball.

share|cite|improve this answer
This is not suffice, because I need to prove any open set in the plane is generated by open balls. I do not know how to construct a "countable union" type of argument. – Bombyx mori Dec 3 '12 at 0:55
I see your point. I remember seing a proof many years ago of how to fill a circle with countably many rectangles or viceversa. But here you don't need that. The Borel $\sigma$-algebra is the smallest $\sigma$-algebra that contains the given topology. Here both topologies are the same, so the two Borel $\sigma$-algebras have to be the same. – Martin Argerami Dec 3 '12 at 1:04
Yes I am quite annoyed with the process. I managed to prove every open set is a countable union of open balls, which is quite ugly. Thanks for the hint. – Bombyx mori Dec 3 '12 at 1:07
You are welcome. Off the top of my head, I don't see how to make the countable thing work. That's why it's nice to be able to avoid it! :D – Martin Argerami Dec 3 '12 at 1:12
I made an ugly argument. Suppose we have an open set $A$, then its intersection with balls of radius $n$ constitute a sequence of open sets $A_{n}$. Each of them has a closure which is compact because it is bounded. Then I cover them repeatably by balls of radius $\frac{1}{2^{i}}$, which means I cover the parts left by first time with smaller balls. So eventually I reach a countable union of open balls by this process. – Bombyx mori Dec 3 '12 at 1:15

Your Answer


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.