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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?

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

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