For non-differentiable functions, does fundamental theorem of calculus give the subgradient?

As we know, if $f:R→R$ is continuous, and $F(x)=\int_0^xf(y)dy$, then $F$ is differentiable and $F'(x)=f(x)$. (Fundamental Theorem of Calculus)

Consider the case that $f$ is not continuous, let $F(x)=\int_0^xf(y)dy$. Assume $f$ is increasing thus $F$ is convex. Now $F$ doesn't have to be differentiable but it has subgradients (subderivatives). Is there any relationship between $f$ and subgradient of $F$?

Is it true? Can we prove this: $c$ is a subgradient of $F(x)$, if for any $y<x,f(y)<c$ and for any $y>x, f(y)>c$.

Here is an example to justify this proposition. Suppose $f(x)=1$ for $x>=0$, $f(x)=-1$ for $x<0$. $F(x)=\int_0^xf(y)dy=|x|$. The proposition holds in this example.