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According to Terrance Tao's Measure Theory book, a boolean algebra $\mathcal{B}$ on a set $X$ is atomic, if there exist disjoint sets $(A_\alpha)_{\alpha \in I}$ which we refer to as atoms, such that $X=\bigcup_{\alpha \in I}A_\alpha$, and $\mathcal{B}$ consists of all sets $E$ of the form $E=\bigcup_{\alpha \in J}A_\alpha$ for some $J\subset I$. An atomic algebra is obviously a $\sigma$-algebra.

Given two measurabel spaces, $(X,\mathcal{B_X})$ and $(Y,\mathcal{B_Y})$, the product $\sigma$-algebra $\mathcal{B_X} \times \mathcal{B_Y}$ is defined to be the $\sigma$-algebra generated by the sets $E\times F$, where $E\in \mathcal{B_X}$ and $F\in \mathcal{B_Y}$.

Exercise 1.7.19, page 196 in Tao's book:

Show that the product of two atomic $\sigma$-algebras is again atomic.

Let $\mathcal{B_X}$ and $\mathcal{B_Y}$ be two atomic algebras, with atoms $(A_\alpha ^X)_{\alpha \in J_X}$ and $(A_\beta ^Y)_{\beta \in J_Y}$, respectively. The minimal sets with respect to set inclusion in the product $\sigma$-algebra are $A_\alpha ^X \times A_\beta ^Y$ for some $\alpha \in J_X$ and $\beta \in J_Y$. Therefore, if the product algebra is atomic, these are its atoms. Obviously, countable unions of these atoms lie in the product algebra. But what about uncountable ones?

This difficulty and what I wrote here: Is the product of two discrete $\sigma$-algebras necessarily discrete?, indicate that either the product algebra need not be atomic (not very probable), or that I am missing something. I'll be very greatful for your help.

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Yoav Bar Sinai
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2 Answers2

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Counterexample. Let $|X|\gt2^{\aleph_0}$ and let $\mathcal B=\mathcal P(X)$. Then $(X,\mathcal B)$ is an atomic algebra. I claim that $\mathcal B\times\mathcal B$ is not an atomic algebra. Since $\mathcal B\times\mathcal B$ contains all the singletons, this is the same as saying that $\mathcal B\times\mathcal B\ne\mathcal P(X\times X)$. In fact, I will show that the diagonal $\Delta=\{(x,x):x\in X\}$ does not belong to $\mathcal B\times\mathcal B$.

Assume for a contradiction that $\Delta\in\mathcal B\times\mathcal B$. Then $\Delta$ belongs to the $\sigma$-algebra generated by some countable family of rectangles, say $E_1\times F_1,E_2\times F_2,\dots,E_n\times F_n,\dots.$ Since $|X|\gt2^{\aleph_0}$, we can choose two points $x,y\in X,x\ne y$, so that $\{n:x\in F_n\}=\{n:y\in F_n$. Hence, for any $n\in\mathbb N$, we have $(x,x)\in E_n\times F_n\Leftrightarrow(x,y)\in E_n\times F_n$. It follows that, for any set $S$ in the $\sigma$-algebra generated by the rectangles $E_n\times F_n$, we have $(x,x)\in S\Leftrightarrow(x,y)\in S$. Since $(x,x)\in\Delta$ while $(x,y)\notin\Delta$, we have arrived at a contradiction.

See this answer for related problems and results.

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bof
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See this page

Exercise 1.7.19(ii) is not correct as stated and should be deleted.

But see HERE for a more commonly used definition of "atomic", which makes the exercise true. (Perhaps when Prof. Tao put the exercise in, he forgot he was using a non-standard definition?)

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GEdgar
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