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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?

Arctic Char
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mathstudent
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    What have you tried, what are your thoughts? What to you know about measurability of monotone functions? – PhoemueX Sep 27 '14 at 21:27
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    @PhoemueX The monotone angle seems the easiest way... one can use the answer of http://math.stackexchange.com/questions/252421/are-monotone-functions-borel-measurable – ttb Jul 04 '16 at 23:49
  • Whenever I'm not wrong, for $v(x) = \lfloor x \rfloor$, $v^{-1}([c, \infty)) = [\lceil c \rceil, \infty)$ and for $u(x) = \lceil c \rceil$, $u^{-1}((c, \infty)) = (\lfloor c \rfloor, \infty)$, for every $c \in \mathbb{R}$. – Blue Tomato Sep 01 '23 at 12:58

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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

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

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

k76u4vkweek547v7
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