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I am talking abount the product sigma algebra of two algebras. I think we had the statement in the lecture so it should be true. I can not remember the proof completely though and need some help to find it.

Let $\mathcal A$ be the set of all countable unions of disjoint rectangles with measurable sides. It suffices to show that $\mathcal A$ is a $\sigma$-algebra. The step that I can not get right though is showing that $\mathcal A$ is closed under complementation. As far as I remember the proof went like this: Let $A=\bigcup_iR_i$ be a countable union of disjoint rectangles. Then $A^\complement=\bigcap_iR_i^\complement$. I can show that $R_i^\complement\in\mathcal A$ and that finite intersections of them are in $\mathcal A$ as well. But I don't know why countable intersections of the $R_i^\complement$s are supposed to be in $\mathcal A$.

I am not 100% sure if the proof really went like this though.

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I believe it to be false. The line $y=x$ in $\mathbb{R}^2$ is Borel, but it not a countable union of disjoint measurable rectangles.

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  • $\begingroup$ Holy shit thas was fast, thanks! And the line is in the product $\sigma$-algebra, since $B(\mathbb R^2)=B(\mathbb R)\times B(\mathbb R)$, right? Strange, I guess I have to ask my teacher about that, since I am pretty sure that we 'proved' that in the lecture and used it later a lot $\endgroup$ – aiw7 Mar 26 '16 at 19:59
  • $\begingroup$ Yes, the Borel $\sigma$-algebra of $\mathbb{R}^2$ is the same as the product $\sigma$-algebra on $\mathbb{R}$, however you can even be more explicit: The complement of $y = x$ is open in $\mathbb{R}^2$ and so is a countable union of open balls (this is a standard construction using the density of $\mathbb{Q}$) a slight modification on this will give you the complement of $x=y$ explicitly as a countable union of rectangles (actual products of intervals), this tells you the line $x=y$ is product-measurable, $\endgroup$ – James Mar 26 '16 at 22:21
  • $\begingroup$ ah nice, so this can be seen as a direct example to why I had to get stuck with the proof at that point. The complement of the line is in what I called $\mathcal A$ but the line itself is not. $\endgroup$ – aiw7 Mar 27 '16 at 8:27

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