Proving $y=\lfloor x\rfloor$ doesn't have a primitive function Prove that $y=\lfloor x\rfloor$ doesn't have a primitive function. 


I have the proof from a book here but I don't understand it:


  Suppose that there is $F(x)$ such that $F'(x)=\lfloor x\rfloor$. We
  have that $y=\lfloor x\rfloor$ has a jump discontinuity (or type 1)
  but $F'(x)$ has only an essential discontinuity (or type 2), a
  contradiction.


I don't understand, why does $F'(x)$ has an essential discontinuity?
The proof is silently appealing to Darboux’s theorem:


  If $I$ is an open interval, and $f:I\to\Bbb R$ is a differentiable function, then $f'$ has the intermediate value property. That is, if $a,b\in I$, $a&lt;b$, and $y$ is between $f'(a)$ and $f'(b)$, then there is an $x\in[a,b]$ such that $f'(x)=y$.


This means that a derivative cannot have a jump discontinuity and therefore cannot be the floor function. In particular, suppose that $F'(x)=\lfloor x\rfloor$, and take $I$ to be the interval $(-1,2)$. Then $0,1\in I$, and 

$$F'(0)=\lfloor 0\rfloor=0&lt;\frac12&lt;1=\lfloor 1\rfloor=F'(1)\;,$$ 

but there is no $x\in[0,1]$ such that $F'(x)=\frac12$, contradicting Darboux’s theorem.
A fairly 'wild' answer would be for $x&gt;0$:
$$F(x)=\sum_{n=1}^{\infty}  \delta(x-n)$$