Construction of Monotone function which is differentiable on the given set Given a set $A \subset \mathbb{R}$ of measure $0$, is it possible to construct a monotone function whose set of non differentiable points is $A$ ?
 A: This is not possible for all $A\subseteq\mathbb{R}$ of measure $0$. 

Define $\mathbb{D}[f]$ to be the set of non-differentiable points of $f$. The following result is due to Z. Zahorski, exhibited in this $1946$ paper:

Theorem. Given $X\subseteq \mathbb{R}$, there exists $f$ such that $\mathbb{D}[f]=X$ if and only if $X=\mathcal{A}\cup\mathcal{B}$ where $\mathcal{A}$ is a $\mathrm{G}_{\delta}$ set and $\mathcal{B}$ is a $\mathrm{G}_{\delta\sigma}$ set of measure $0$.

The difficult part in the proof is actually dealing with $\mathcal{A}$. Since you require $X$ to be of measure $0$, we can happily impose $\mathcal{A}=\emptyset$. As is adressed in Zahorski's paper, the resulting function is Lipschitz with constant $\leqslant 1$: you can therefore make it monotone by adding on a linear function.
Given that the paper is in French, I will briefly summarise the two results relevant to your question 

Claim. Given a $\mathrm{G}_{ \delta\sigma}$ set $X$ of measure $0$ there exists $f:\mathbb{R}\to\mathbb{R}$ such that $\mathbb{D}[f]=X$.

The following approach is simpler than that taken by Zahorski, and is due to T. Fowler and D. Preiss, details in this $2008$ paper.
Proof sketch: write $X=\bigcup_{k\geqslant 0}X_k$ where $\{X_k\}$ are $\mathrm{G}_{\delta}$ null-sets. A preliminary result by Zahorski is that there are Lipschitz functions $\phi_k:X_k\to\mathbb{R}$ with constant $1$ and $\mathbb{D}[\phi_k]=X_k$ such that: 
$$\limsup_{x\to\alpha}\,\big[\phi_k(x)-\phi_k(\alpha)\big](x-\alpha)^{-1}=1>-1=\liminf_{x\to\alpha}\,\big[\phi_k(x)-\phi_k(\alpha)\big](x-\alpha)^{-1}$$
$\forall \alpha\in X_k$. It is then shown that $3^{-k}\phi_k(x)$ is bounded, Lipschitz with constant $3^{-k}$ on $\mathbb{R}$. Define:
$$f(x)=\sum_{k\geqslant 0}3^{-k}\phi_k(x)$$
For arbitrary $\epsilon>0$ and $x\not\in X, \;\phi'(x)$ exists and $\lVert\phi'(x)\rVert\leqslant 1$. Through simple manipulations:
$$\limsup_{h\to 0}\left\lVert h^{-1}\big[f(x+h)-f(x)\big]-\sum_{k\geqslant 0} 3^{-k}\phi_k'(x)\right\rVert\leqslant 2\epsilon$$
And hence $f'$ exists on $\mathbb{R}\setminus X$ (and is finite -  observe $\lVert f'\rVert\leqslant 2^{-1}$). Conversely if $\alpha\in X,$ letting $m=\min\{n:\,\alpha\in X_n\}$, and using the preliminary result, they arrive at:
$$\big(\limsup_{x\to\alpha}-\liminf_{x\to\alpha}\big)\big[f(x)-f(\alpha)\big](x-\alpha)^{-1}\geqslant 3^{-m}\;\Rightarrow\;\nexists f'(\alpha)$$

Claim. If $\mathbb{D}[f]=X$ is null, it must be $\mathrm{G}_{ \delta\sigma}$.

Back to Zahorski's paper.
Proof sketch: this is essentially a concatenation of known results. The idea is to write $\mathbb{D}[f]$ as: $$\mathbb{D}[f]=\left(\mathcal{M}=\{x:\;\lVert f'(x)\rVert=\infty\}\right)\bigcup\left(\mathcal{N}=\{x:\;\nexists f'(x)\}\right)$$
and further let $\mathscr{E}=\{x\in \mathbb{R}:\;\lVert f_+'(x)\rVert\;\text{or}\;\lVert f_{-}'(x)\rVert=\infty\}$. By using the the Denjoy-Young-Saks theorem concerning Dini derivatives, and a result by Hausdorff from his Mengenlehre ($1927$) which implies that both $\mathcal{M},\mathcal{N}$ are $\mathrm{G}_{ \delta\sigma}$, it only takes a couple lines to decompose them into a union as described in the initial theorem, and hence $X=\mathcal{A}\cup\mathcal{B}$ as above. Unfortunately I am unable to find the exact statement Hausdorff made. 
Applying the  measure-$0$ condition allows us to disregard $\mathcal{A}$.

PS. an explicit example of an $A$ which fails these conditions is the set of non-normal numbers.
