# Show that a set is measurable with respect to Borel product $\sigma$-algebra

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.

• $\mathcal B(\mathbb R\times\mathbb R)$ 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. Oct 19, 2013 at 15:26
• You might try to show that, in general, an open set is measurable in $\mathcal B(\mathbb R\times\mathbb R)$. Oct 19, 2013 at 15:27
• @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? Oct 19, 2013 at 15:36
• 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”. Oct 19, 2013 at 15:43
$f(x, y) = |x - y|$ is continuous, so the preimage of $(-\infty, 1)$ is open, hence Borel.