Fundamental Theorem of Calculus for functions with one-sided derivative. Let's assume we have a continuous function $F:[0,\infty)\to\Bbb R$ such that its one-sided derivative
$$
f(t):=\lim_{h\searrow 0} \frac {F(t+h)-F(t)}{h}
$$
exists everywhere on $[0,\infty)$.

Does the "Fundamental Theorem of Calculus" hold, i.e. for each $t\in[0,\infty)$ we have
  $$
F(t) = F(0) + \int_0^t f(s)\, ds?
$$

Usually we'd use the Mean Value Theorem to prove the (normal) FCT but I don't know if something similar to the MVT would be provable with only these assumptions.
If the above doesn't hold in general, what kind of condition can we impose on $F$ or $f$ to make the "FTC" holds?
Note: I also think it is possible that the assumptions that $F$ is continuous and that its one-sided derivative exists everywhere might be strong enough to deduce better property of $F$, like the existence of $F'$. If anyone know a result in this direction I'd really love to hear it.
 A: No, that does not give an FTC.
Note that this has nothing to do with the fact that you're talking about one-sided derivatives; in fact the corresponding "normal FTC" for two-sided derivatives is also false!
For example  let $$F(x)=\begin{cases}x^2\sin(1/x^{10}),&(x\ne0),
\\0,&(x=0).\end{cases}$$
Then $F$ is differentiable everywhere but $\int_{-1}^1 F'(t)\,dt$ does not exist (not even as a Lebesgue integral).
I suspect the result is true if you assume in addition that $f$ is continuous.
Edit: Yes, it's true if $f$ is continuous.


Lemma. If $F:\Bbb R\to\Bbb R$ is continuous and the right-hand derivative $D_RF(x)$ exists and equals $0$ for every $x$ then $F$ is constant.


It's enough to prove this:


If $\lambda>0$ then $|F(x)-F(0)|\le\lambda x$ for every $x\ge 0$.


Proof: Let $A$ be the set of $a\ge0$  such that $|F(x)-F(0)|\le \lambda x$ for every $x\in[0,a]$. It's clear that $$A=[0,\alpha] $$for some $\alpha\in[0,\infty]$, and we need only show that $\alpha=\infty$. But if $\alpha<\infty$ then $D_RF(\alpha)=0$ shows that there exists $\delta>0$ with $[\alpha,\alpha+\delta)\subset A$. (Choose $\delta$ so that $|F(\alpha+h)-F(\alpha)|<\frac12\lambda h$ for all $h\in[0,\delta)$.)
And now


Prop. Suppose $F:\Bbb R\to\Bbb R$ is continuous and $f(x)=D_RF(x)$ exists for every $x$. If $f$ is continuous then $$F(x)=F(0)+\int_0^x f$$for every $x>0$.


Proof: Define $G(x)=\int_0^x f(t)\,dt$. Since $f$ is continuous it follows from the standard FTC that $G$ is differentiable and $G'=f$. So $D_R(F-G)=0$, hence $F-G$ is constant.


(Cor. If $f$ is continuous then $F$ is differentiable.)


Alas the question is changing. I suspect it's also true assuming just that $F$ is convex.
Edit: Yes, it's true if $F$ is convex. My version of this if anything seems simpler than the case $f$ continuous, because I saw how to use some high-powered machinery.


If $F:\Bbb R\to\Bbb R$ is convex then $F(x)=F(0)+\int_0^x D_RF$.


Proof. You say you know, and it's not hard to prove, that $F$ is locally Lipschitz. Hence it's locally absolutely continuous, so it's diferentiable almost everywhere and $F(x)-F(0)=\int_0^x F'(t)\,t$ ((where that's a Lebesgue integral).
In case we care it follows that $F(x)-F(0)$ is actually the Riemann integral of $f=D_RF$: Since $f$ is increasing it is continuous almost everywhere, hence the Riemann integral $\int_0^x f$ exists. And it equals the Lebesgue integral of $F'$, since $F'=f$ almost everywhere.
