# Let $B$ be the Borel measure on the real line, what is a good strategy to prove $B_{2}=B\times B$?

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.

-

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.

-
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