Strictly increasing function and its derivative We have "$f'(x)>0$ in interval $I\implies f(x)$ strictly increasing in interval $I$".
However, the converse is not true.
A frequently cited example is the function $f(x)=x^3$, which is strictly increasing but $f'(0)=0$.
Here comes my question:  

What is the necessary and sufficient condition of $f'(x)$ to make $f(x)$ strictly increasing? 

My guess is:  

$f'(x)>0$ almost everywhere.  

This is clearly a sufficient condition, as we can use Lebesgue integration and "drop the word 'almost'".  
But is this a necessary condition? Is there a weaker condition? A full solution or a little hint will both be appreciated. Thank you in advance.
 A: Assume $f$ is differentiable on an interval $I$ and $f'(x)\ge 0$ on $I.$ Let $Z = \{x \in I: f'(x)=0\}.$ Then $f$ is strictly increasing on $I$ iff $Z$ contains no interval. (Here "interval" means "interval of positive length".)
Proof: Suppose $f$ is strictly increasing on $I.$ If $Z$ contained an interval $J,$ then $f'\equiv 0$ on $J,$ which implies $f$ is constant on $J$ by the mean value theorem. Thus $f$ is not strictly increasing, contradiction.
Suppose $Z$ contains no interval. Let $x,y\in I$ with $x<y.$ Because $f'\ge 0,$ we have $f(x)\le f(z) \le f(y)$ for all $z\in [x,y].$ If $f(y) = f(x),$ then $f = f(x)$ on $[x,y].$ This implies $f'\equiv 0$ on $[x,y],$ hence $Z$ contains $[x,y],$ contradiction. The contradiction shows $f(x) < f(y),$ hence $f$ is strictly increasing.
A: If $f(x)$ satisfies $f'(x)\ge 0$ for all $x\in \mathbb R$ and if for every $x_0\in \mathbb R$ with $f'(x_0)=0$, there is a neighborhood around $x_0$ , such that $f'(x)\ne 0$ within this neighborhood (except $x_0$) (in other words : isolated roots of the derivate ) , then $f(x)$ is strictly increasing on $\mathbb R$.
Not sure, whether we have a weaker sufficient condition, but an interval $[a,b]$ with $a<b$ and $f'(x)=0$ for all $x\in [a,b]$ is obviously impossible.
