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?
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.
