Does "continuous + only at countable many points not differentiable (with bounded derivative)" imply Lipschitz-continuity? Let $f$ be continuous on $\mathbb R$ and differentiable with derivative $f'$ on $\mathbb R \setminus \{t_0, t_1, \dots \}$. Let $\sup | f'(t) | < \infty$, then $f$ is Lipschitz continuous with $L=\sup |f'(t)|$.
Does this hold? How could one prove it?
 A: Since $-L \leq f'(x) \leq L $ except on a set of only measure zero, we may integrate this between $x_1$ and $x_2$ and the desired $ -L(x_2-x_1) \leq f(x_2) - f(x_1) \leq L(x_2-x_1) $ pops right out.

Note to all: The following is what was my intial answer, but it is faulty.
First do the problem for each segment that the function is differentiable (so $(-\infty,t_0), (t_0,t_1) $ etc): By the mean value theorem, for $x_1 , x_2 \in (t_k, t_{k+1}) $ we get $$ \frac{f(x_1)-f(x_2)}{x_1-x_2} = f'(c) $$ for some $c\in (x_1,x_2)$. 
This means for any $x_1,x_2 \in \mathbb{R}$ we have in $(t_k, t_{k+1})$ that $$ |f(x_1)-f(x_2) | \leq |f'(c)||x_1-x_2| \leq L|x_1-x_2|. $$
which makes it Lipschitz continuous in each segment with Lipschitz constant $L$. Can you see how to prove that if we stitch together Lipschitz continuous functions like this, it remains so?
A: I try a correct fix of @Ragib's answer, then we can continue to discuss here
Let $A=\{t_0, t_1, \dots \}$. Define 
$$\tilde{f}'(x) = \begin{cases} 0 & x \in A\\ f'(x) & \text{else}
\end{cases}$$ 
$A$ has measure 0 (in other words $f$ almost everywhere differentiable).
Then $\sup|\tilde{f}'(t)| \equiv \sup |f(t)| =L$.
$\tilde{f}'(t)$ has an anti-derivative $\forall x \in R$ and is Lebesgue-integrable, hence with the Fundamental Theorem of Analysis for Lebesgue-integral:
$\int_{x_1}^{x_2} \tilde{f}'(x)=f(x_2)-f(x_1)$

In Wikipedia under generalizations it says "Part II of the theorem is true for any Lebesgue integrable function ƒ which has an antiderivative F (not all integrable functions do, though)."
