Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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
up vote 6 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?

share|cite|improve this answer

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.

share|cite|improve this answer

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 $X''\times Y''$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.