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I know that the answer to this question is negative, since proving the opposite is an exercise in Terrance Tao's Measure Theory book.

However, it doesn't make sense to me. In another part of the same exercise, we show that the product $\sigma$-algebra of two atomic algebras, is again an atomic algebra. As I understand, the atoms of the product algebra are the products of the atoms of the given atomic algebras.

But then, if $(X,P(X))$ and $(Y,P(Y))$ are two discrete measurable spaces, then the discrete $\sigma$-algebra on $X\times Y$ is $P(X\times Y)$, and the atomic algebra generated by the products of the singletons of $X$ and those of $Y$, $$P(X)\times P(Y) = \left\{\bigcup_{(x,y)\in J}(x,y):J\subset (X\times Y) \right\}=P(X\times Y), $$ as well.

What am I missing?

Yoav Bar Sinai
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    What are you missing? Uncountable unions need not be in the sigma-slgebra. – GEdgar Feb 14 '15 at 12:47
  • So how does the product $\sigma$-algebra atomic? In atomic $\sigma$ algebras, the sets in the algebra are arbitrary unions of the atoms, isn't that so? – Yoav Bar Sinai Feb 14 '15 at 13:38
  • There are non-discrete atomic $\sigma$-algebras. For example, the Borel sets in $\mathbb R$. – GEdgar Feb 14 '15 at 13:49
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    I think there is some confusion in the definition of atomic algebras. On one hand, there is this definition which I have just encountered: http://en.wikipedia.org/wiki/Atom_(measure_theory). This definition involves a specific measure on the space. On the other hand, an algebra in Tao's book is called atomic, if the sets in these algebra are all arbitrary unions of atoms, which are disjoint sets whose union is the whole space. Maybe that is the reason for confusion? – Yoav Bar Sinai Feb 14 '15 at 13:58
  • All the sets of the sigma-algebra are unions of atoms. But perhaps not all unions of atoms are sets of the sigma-algebra. – GEdgar Feb 14 '15 at 17:23
  • So you're saying that a product of two atomic algebras need not be atomic (in the sense that the product algebra need not be closed under arbitrary unions of its atoms). I tend to agree, but then there is a mistake in Tao's book.. – Yoav Bar Sinai Feb 14 '15 at 17:34
  • Tao's two exercises cited are correct. – GEdgar Feb 14 '15 at 17:38
  • Thanks a lot for your help, but I'm sorry, I don't understand how this make sense. The product of two atomic algebras is atomic. If its atoms are the products of the atoms of the given algebras, every arbitrary union of them should be in the product algebra. But then the product of discrete algebras must also be discrete. Where is the mistake? – Yoav Bar Sinai Feb 14 '15 at 17:44
  • See http://math.stackexchange.com/questions/545640/does-mathcal-p-mathbb-r-otimes-mathcal-p-mathbb-r-mathcal-p –  Feb 14 '15 at 20:07

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