Constructing a null set and a Lipschitz function nowhere differentiable on it I'm trying to solve the following exercise.

Now, Rademacher's theorem says that locally Lipschitz functions are $\mathcal L^N$-a.e. differentiable, so $E$ must be a null set, and this is clearly the case. The suggested function is Lipschitz by the triangular inequality. I get stuck when I try to identify the "suitable subsets" of the sets $I_{j,n}$. We want select them in such a manner that upper and lower derivatives do not coincide at each point in $E$. For this it is enough that at each point at least one partial derivative does not exist. Does it suffice to fix an $N$ and take the union over $j$ and over $n$ running from $1$ to $N$? i.e.,
$$ E := \cup_{n=1}^N \cup_{j=0}^{2^{n-1}-1} I_{j,n} ?$$
If not, may you provide hints or a full solution?
 A: Notice first that a finite union can not work because $f$ is constant on $I_{j,n}$, and they are isolated segments if they are in a finite number.
1) Prove that the restriction of $f$ to the first axis is not differentiable at any diadic point $q_{j_0,n_0}$. To see this write
$$f(t)=4^{-n_0}dist(t,I_{j_0,n_0})+\sum_{(n,j)\neq (n_0,j_0)}4^{-n}dist(t,I_{j,n})$$
and observe that the first term is not differentiable at $q_{j_0,n_0}$, while the second is convex.
In particular if you don't care about $\mathcal{H}^1(E)$ being finite the set $E=\bigcup_{j,n}I_{j,n}$ will work.
2) If you require $\mathcal{H}^1(E)<\infty$ you have to select from each segment $I_{j,n}$ a set $E_{j,n}$ of small measure, say $\mathcal{H}^1(E_{j,n})=4^{-n}$, to make the total measure converge. The rough idea is to take $E_{j,n}$ as a union of many segments inside $I_{j,n}$, that become finer and finer as $n\to\infty$. Let's say you select inside every $I_{j,n}$ $m(n)$ equidistributed segments each of length $\frac{1}{m(n)}4^{-n}$. Now set $E=\bigcup_{j,n} E_{j,n}$.
If $m(n)$ grows rapidly enough as $n\to\infty$ you can prove that $f|_E$ is not differentiable at any point. This comes from the fact that you can approach every point $x\in E$ both from the left and from the right with a sequence of points $(x_n)\subset E$ such that the directions $x_n-x$ are asymptotically horizontal, and you can conclude from point 1).
