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.
Let $B$ be the Borel measure on the real line, what is a good strategy to prove $B_{2}=B\times B$?
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real-analysis
1 Answers
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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.
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0This 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. – 2012-12-03
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0I 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. – 2012-12-03
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0Yes 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. – 2012-12-03
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0You 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 – 2012-12-03
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0I 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. – 2012-12-03
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0At first sight, I don't see how you would guarantee that your balls will end up lying inside $A$. – 2012-12-03
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0I cover $A_{n}$ with finitely many balls $B_{j}$. I choose the ones contained inside of $A$. Then for the rest of the balls I take its intersection with $A$, which give me a new open set. Then I repeat the process with balls of half the width. Eventually I should be able to reduce the diameter of the open set to arbitrarily small. – 2012-12-03
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0I found this construction is actually quite useful, because later I was asked to prove if two measure coincide on retangles, then it coincides on the whole $\sigma$-algebra. – 2012-12-03