Continuous right derivative implies differentiability A book of mine says the following is true, and I am having some trouble proving it. (I've considered using the Lebesgue differentiation theorem and absolute continuity, as well as elementary analysis methods.)

Let $f: [0, \infty) \rightarrow \mathbb{R}$ be continuous and have right derivatives at each point in the domain, with the right derivative function being continuous.  Then $f$ is differentiable.

 A: Let us denote the right derivative of $f$ by $g$.

Lemma: Given $a<b$ and $m\le M$, if $m\le g\le M$ on $[a,b]$, then  $$m\le\frac{f(b)-f(a)}{b-a}\le M.$$

Proof: Define 
$$L(a)=g(a)\quad\text{and}\quad L(x)=\frac{f(x)-f(a)}{x-a},\ x\in(a,b].$$
By definition, $L$ is continuous on $[a,b]$, and it suffices to show that for every $\delta>0$, 
$$E_\delta:=\Big\{x\in[a,b]\,\Big|\, m-\delta\le L(y)\le M+\delta, \forall y\in[a,x] \Big\}=[a,b].$$
By definition and the continuity of $L$, we know that $E_\delta=[a,c]$ for some $c\in[a,b]$, and from $m\le g(a)\le M$ we know $c>a$. Then from $c\in E_\delta $ and $m\le g(c)\le M$ it is easy to see that $c<b$ is impossible. Therefore, $c=b$ and the lemma follows. $\quad\square$

Now let us show that $f$ is differentiable for any $x>0$. Since $g$ is continuous, given $0<h<x$, we can define
$$m_h=\min_{y\in[x-h,x]}g(y),\quad M_h=\max_{y\in[x-h,x]}g(y),$$
and we know that
$$\lim_{h\to 0^+}m_h=\lim_{h\to 0^+}M_h=g(y).$$
Due to the lemma, for $a=x-h$, $b=x$, $m=m_h$ and $M=M_h$, we have
$$m_h\le\frac{f(x-h)-f(x)}{-h}\le M_h.$$
Let $h\to 0^+$, it follows that the left derivative of $f$ at $x$ exists and is equal to $g(x)$, i.e. $f$ is differentiable at $x$. $\quad\square$
A: I would have commented but my rep is too low. Concerning the very nice answer of 23rd, I would like to point out the details of the proof.
To prove the lemma it is sufficient to prove (and indeed it is what he's doing) that the sets $E_\delta$ are both closed and open for every $\delta$. The closeness of the sets follows from the fact that the quotient $\frac{f(x)-f(a)}{x-a}$ is a continuous quantity. To prove that the set is open, fix $x\in E_\delta\cap (a,b)$. Since $m\le g(x)\le M$, there exists a point $r'\in [x,b]$ such that
\begin{equation*}
m-\delta\le \frac{f(r)-f(x)}{r-x}\le M+\delta
\end{equation*}
holds for all $r\in [x,r']$. Therefore it follows
\begin{equation*}
f(r)-f(a)=(f(r)-f(x)) + (f(x)-f(a))\le (M+\delta)[(r-x)+(x-a)]=(M+\delta)(r-a)
\end{equation*}
and the converse inequality is analogous. Thus $r'\in E_\delta$ and we are done.
