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When reading the book Measures, Integrals and Martingales written by R.L. Schilling, I saw a statement as below: $$\mathcal{A} \times \mathcal{B}:= \{A \times B: A\in \mathcal{A}, B\in \mathcal{B}\}$$ where $\mathcal{A}$ and $\mathcal{B}$ are $\sigma$-algebras, is not $\sigma$-algebra in general.

However, I cannot construct a counterexample. Could anyone offer help here?

Kind regards

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Hint: it is not closed under unions or complements. Any nontrivial $\sigma$-algebras will suffice for a counterexample. –  Nate Eldredge Apr 5 '11 at 1:21

3 Answers 3

up vote 5 down vote accepted

think of the union of two rectangles in $\mathbb{R}^2$, like $[0,1]\times[0,1]\cup[2,3]\times[2,3]$. is this a product?

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Simple example: $A$ and $B$ are both $\mathbb{R}$ and $\mathcal{A}$ and $\mathcal B$ are both Borel $\sigma$-algebra. Then sets like $(-\infty,a)\times(-\infty,b)$ are in $\mathcal A\times \mathcal B$, but not their complements.

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Take the Borel Algebra $B$ in $R$ - generated by the open intervals of the form $(a,b)$. Then, every set of the type $\{a\}$ belongs to $B$, where $a\in R$. Then $(a,a)\in B\times B, \forall a\in R$. But for $a\neq b$ we have that the set $\{(a,a), (b,b)\}$ does not belong to $B\times B$.

In general, $(X\times Y)\cup (X'\times Y')$ is not of the form $X''\times Y''$.

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