I'm having some trouble with the following exercise:

Let $(\mathbb{R}, \mathcal{B})$ denote the real line with the Borel $\sigma$-algebra and let $X=(\mathbb{R}, \mathcal{B})\times(\mathbb{R}, \mathcal{B})$ have the product $\sigma$-algebra. Show that

A = { $(x,y)\in X:|x-y|<1$ }

is a measurable set.

I know that

$\mathcal{B}(\mathbb{R})\times\mathcal{B}(\mathbb{R})$ = <{$U\times V: U,V \in\mathcal{B}(\mathbb{R})$}> = <{$U\times V: U,V \subset \mathbb{R}$ open}>

So I think I have to write A as a product of two open sets in $\mathbb{R}$, but I'm not really sure how to do this.

Thanks for your help!

  • $\begingroup$ $\mathcal B(\mathbb R\times\mathbb R)$ is generated by the products of open sets. Therefore a $\mathcal B(\mathbb R\times\mathbb R)$-measurable set need not be the product of two open sets, as indeed your set $A$ is not. $\endgroup$
    – AndreasT
    Oct 19, 2013 at 15:26
  • 1
    $\begingroup$ You might try to show that, in general, an open set is measurable in $\mathcal B(\mathbb R\times\mathbb R)$. $\endgroup$
    – AndreasT
    Oct 19, 2013 at 15:27
  • $\begingroup$ Show that any open set in the plane is the union of rectangles with the sides parallel to the axes and corners at points with rational coordinates. $\endgroup$ Oct 19, 2013 at 15:34
  • $\begingroup$ @AndreasT I'm not sure if I understand what you mean. So I could show that any open set X $\subset \mathbb{R^2}$ is in $\mathcal{B}(\mathbb{R})\times \mathcal{B}(\mathbb{R})$ and that A is an open set, to prove it? $\endgroup$
    – caedmon
    Oct 19, 2013 at 15:36
  • 2
    $\begingroup$ Indeed, that is what he must have meant. $A$ is indeed open. But you could also follow my suggestion above, substituting $A$ for “any open set”. $\endgroup$ Oct 19, 2013 at 15:43

1 Answer 1


$f(x, y) = |x - y|$ is continuous, so the preimage of $(-\infty, 1)$ is open, hence Borel.


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