Measurability of the floor function

Question

Let $u(x)=⌊x⌋$, i.e the largest integer not greater than $x$ . Determine $\{u≥a\}$ for all $a\in \mathbb{R}$. Show that $u$ is Borel-measurable. Can anyone help me with this problem?

Answer 1

An idea to show that $u$ is measurable is that you can express a floor function as a simple one (a limit of simple functions)

$$ u(x) = \sum_{n=-\infty}^\infty n \, \chi_{[n,n+1)}(x)$$

alternatively $u^{-1}[a,\infty) = \{ x \mid a \leq \lfloor x \rfloor \} = [ \lfloor a \rfloor +1 , \infty)$

EDIT: it is not right $u^{-1}[a,\infty) = [\lceil a \rceil,\infty)$

EDIT:Mainly intuition... i know that if there is an expression for it, it must be an interval written in terms of $a$ , and some type of roof or ceil function. For the proof

$$ a \leq \lfloor x \rfloor \implies \lceil a \rceil \leq \lfloor x \rfloor \leq x$$

$$ \lceil a \rceil \leq x \implies a \leq \lceil a \rceil \leq \lfloor x \rfloor $$