I'm learning measure theory, and I don't quite get how to prove or disprove if a set which is closed at infinity is Lebesgue measurable or not. Is there a difference from open sets?
All the properties in the book I'm studying are stated on $\mathbb{R}$ and it throws me off.
Is the set $[0,\infty]$ Lebesgue Measurable
lebesgue-measuremeasure-theory
Best Answer
The extended real numbers $\overline{\mathbb{R}}$ are defined as the set $\mathbb{R}\cup\{-\infty,\infty\}$, where the symbols $-\infty,\infty$ extend the order of $\mathbb{R}$ by setting $-\infty\le x$ for all $x\in\mathbb{R}$ and similarly for $\infty$. In this way, the new order relation defines a topology on $\overline{R}$ by considering the subbase consisting of all rays $\{x : x\lt a \}$ and $\{ x : x \gt b \}$ for all $a,b\in\mathbb{R}$. With this subbase, all open (in the sense of the order) intervals are open (in the sense of the topology). This topology generates a sigma-algebra, the Borel sigma-algebra of $\overline{\mathbb{R}}$. From here, you can try writing your interval $[0,\infty]$ as union/intersection/complement of elements in the set of generators of your sigma-algebra to see if it is measurable.