Functions which are $C^n$ everywhere, $D^{n+1}$ nowhere.

Question

For any $n \geq 1$, does there exist a real function which has $n$ continuous derivatives everywhere but which has $n+1$ derivatives nowhere? Does there exist a real function which has $n$ derivatives everywhere but where the $n$th derivative is continuous almost nowhere?

I assume this question has been asked already but was unable to find it myself. If it has already been asked, I will remove my question. My guess for the first part of my question is that if you take the $n$th antiderivative of the Weierstrass function, the resulting set of functions will all possess $n$ continuous derivatives everywhere, but will possess $n+1$ derivatives nowhere.

Answer 1

Let $f\colon\mathbb{R}\to\Bbb{R}$ be arbitrary. Then the set $C_f=\{x\mid f\text{ is continuous at }x\}$ is a $G_\delta$; that is: the intersection of a sequence $(U_n)_{n\in\Bbb{N}}$ of open sets.
On the other hand, a derivative $f$ being a function of first class (i.e. the pointwise limit of a sequence of continuous functions), $C_f$ must be a dense subset of $\Bbb{R}$ (result due to Baire).
Hence for a derivative the set $C_f$ must be a dense $G_\delta$ subset of $\Bbb{R}$.
It happens that the converse is true:

given $A\subset\Bbb{R}$, a dense $G_\delta$, there exists a derivable $F$ such that $C_{F'}=A$.

A proof is given in Andrew Bruckner's Differentiation of real functions (chapter 3).

Now it's possible to construct a dense $G_\delta$ having measure $0$. Consider $(r_k)_{k\in\Bbb{N}}$ an enumeration of $\Bbb{Q}$, and chose some integer $n\geqslant1$.
The set $U_n=\bigcup_{k\geqslant0} \left]r_k-\frac1{n2^k}\mathbin;r_k+\frac1{n2^k}\right[$ is an open set having measure ${}\leqslant\frac4n\cdot$ Also $U_n$ is dense, since $U_n\supset\Bbb{Q}$.
Conclusion: $A=\bigcap_{n\geqslant1}U_n$ is a dense $G_\delta$ having measure $0$.