# Proving Borel set

How to prove that set $$\{(x;y): 0\leq x \leq1, 0\leq y \leq 1, x+y\leq 1\}$$ belongs to Borel $$\sigma$$-algebra $$\mathcal{B}(\mathbb{R}^2)$$ ?

I tried to depict given conditions on coordinate plate and I got a triangle. I know that to prove it we need to get rectangle. So how should I do it?

• What is your definition of elements of $\mathcal B(\mathbb R^2)$, i.e. Borel sets in $\mathbb R^2$? – drhab Nov 23 '18 at 11:21
• yes, it is a Borel set in $\mathbb{R}^2$ – Atstovas Nov 23 '18 at 11:23

In general if $$X$$ is a topological space equipped with topology $$\tau$$ then it induces the $$\sigma$$-algebra generated by $$\tau$$ (i.e. the smallest $$\sigma$$-algebra that contains $$\tau$$ as a subcollection).

This is the so-called Borel $$\sigma$$-algebra on $$X$$ and is often denoted as $$\mathcal B(X)$$.

Since $$\mathcal B(X)$$ contains all open sets and is closed under complements it also contains all closed sets.

If $$\mathbb R^2$$ is equipped with its common topology then the set mentioned in your question is evidently a closed subset of $$\mathbb R^2$$ hence is an element of $$\mathcal B(\mathbb R^2)$$.

P.S.

In your question you mention "rectangles" and this indicates that you are interpreting $$\mathcal B(\mathbb R^2)$$ as $$\mathcal B(\mathbb R)\times\mathcal B(\mathbb R)$$.

Fortunately there is a theorem that states that: $$\mathcal B(\mathbb R^2)=\mathcal B(\mathbb R)\times\mathcal B(\mathbb R)$$.

It is based on the fact that the common topology on $$\mathbb R^2$$ coincides with the product topology on $$\mathbb R\times\mathbb R$$ where $$\mathbb R$$ is equipped with its common topology.

See here for a proof.