Determine $\left \{ u\geq a \right \}$ for all $a\in \mathbb{R}$, and is $u$ $\mathcal B(\mathbb{R})/\mathcal B(\mathbb{R})$-measurable?

Question

Let $u:\mathbb{R}\to\mathbb{R}$ be given by $u(x)=\left \lfloor x \right \rfloor$. Determine the set $\left \{ u\geq a \right \}$ for all $a\in \mathbb{R}$. Show that $u$ is $\mathcal B(\mathbb{R})/\mathcal B(\mathbb{R})$-measurable.

Note that the notation of the set can be understood as $$\left \{ u\geq a \right \}=\left \{ x\in\mathbb{R}:u(x)\geq a \right \} =u^{-1}([a,\infty)).$$

I am unable to answer the other problem without the determined set. Help me, please.

Answer 1

to show it's measurable, you just have to take the inverse image of a generating set of the range and show that's a measurable set in the domain. So, [a,b], for real numbers a and b generate the borel sets (Closed intervals. Open intervals work as well). The pre-image of u of this set is going to be all the x values who's floor is between a and b. This you can show is an interval itself, by looking at a few simple cases.

If you still don't have it by tomorrow, I can edit this to finish the details.