Equivalence between Lusin (N) property and absolute continuity. Let $F\colon [a,b] \to \Bbb R$ be continuous and increasing. Then:

$F$ satisfies the Lusin (N) property if and only if it is absolutely continuous.

Here, Lusin (N) property means that for all $N\subseteq [a,b]$ with $\mathfrak{m}(N)=0$ we have $\mathfrak{m}(F[N])=0$, where $\mathfrak{m}$ of course denotes Lebesgue measure. 
That absolute continuity implies the Lusin (N) property is not hard to check directly using the definitions. I also have an outline of proof for the direct implication, which I can understand. This question is about fixing my first attempt for said implication (I feel I was very close).
Assume by contradiction that $F$ is not absolutely continuous, and take $\epsilon > 0$ to witness that. Then, for every $k>0$ there is a disjoint collection of open intervals $I_{k,1}, \ldots, I_{k,n_k}\subseteq [a,b]$ such that $$\sum_{j=1}^{n_k} \mathfrak{m}(I_{k,j})< \frac{1}{k} \qquad \mbox{but}\qquad \sum_{j=1}^{n_k} \mathfrak{m}(F[I_{k,j}]) \geq \epsilon.$$(To write the negation of absolute continuity as above I am using that $F$ is increasing to take intervals into intervals, and that $F$ -- hence $\mu_F$ -- is continuous to disregard the endpoints of the $F[I_{k,j}]$ in the second bound). I want to somehow the take the limit of these collections of intervals. This is my motivation for defining $$N \doteq \bigcap_{k >0} \bigcup_{j=1}^{n_k} I_{k,j}.$$It is clear that $\mathfrak{m}(N)=0$. I want to use the second estimate to contradict the Lusin (N) property. I thought of using the Lusin (N) property to say that $\mathfrak{m}(F[N]) = 0$, and use outer regularity of $\mathfrak{m}$ to get an open set $O\supseteq F[N]$ with $\mathfrak{m}(O) = \mathfrak{m}(O\setminus F[N]) < \epsilon$, and do something with the open set $F^{-1}[O]$.

Question: How can I conclude the proof from here? If not, is any of this salvageable?

 A: You have to modify your choice of the intervals, because the property "absolute continuous" can fail in 'many points'. This makes possible that $\bigcup_{j=1}^{n_k} I_{k,j}$ and $\bigcup_{j=1}^{n_{l}} I_{k,j}$ are disjoint for some $k \ne j$. Thus your construction can lead to $N = \emptyset$! This is the reason why you cannot deduce a contradiction in your argument.
One major point is that we need to ensure that $f$ is not absolute continuous on $\bigcup_{j=1}^{n_k} I_{k,j}$. This allows us to choice the next intervals (in the $k+1$-step of the induction argument) to be a subset of $\bigcup_{j=1}^{n_{l}} I_{k,j}$ and thus
$$\lambda(F(N)) = \lambda (\bigcap_{n=1}^\infty \bigcup_{j=1}^{n_k} F(I_{k,j}) = \liminf_{n \rightarrow \infty} \lambda( \bigcup_{j=1}^{n_k} F(I_{k,j})) \ge \varepsilon.$$
We used in the first step that $f$ is injective. (This is true, because $f$ is increasing!)
Finally, let us justify that we can choose the next intervals as subsets of the previous one. If for some $\delta >0$ and for all choices of disjoint intervals $I_1, \ldots, I_n \subset [a,b]$ with $\lambda(I_n) <\delta$ the function $f$ is absolute continious, then it is on $[a,b]$.
Just take $[a,b] = I_1 \cup \ldots I_n$ with $ \delta/2 < \lambda(I_n) < \delta.$ Then there are $\delta_l>0$ such that for any disjoint $J_1,\ldots,J_m \subset I_l$ with $\sum_{i=1}^m \lambda(J_i) < \delta_n$ we have $\sum_{i=1}^m \lambda(f(J_i)) < \varepsilon/(2n)$. 
Let $\delta' := \min(\delta_1,\ldots,\delta_n,\delta/4)$. For any disjoint intervals with $H_1,\ldots,H_m \subset [a,b]$ with $\sum_{k=1}^m \lambda(H_k) < \delta'$, we know that $H_k$ has non-trivial intersection with at most two sets $I_i$ and $I_j$. The intervals $H_i \cap I_j$ have lenght less than $\delta_i$. Thus $\sum_{k=1}^m \lambda(f(H_k \cap I_j)) <\varepsilon/(2n)$. All in all, we find
$$ \sum_{k=1}^m \lambda(f(H_k)) \leq 2 \sum_{l=1}^n \sum_{k=1}^m \lambda(f(H_k \cap I_j)) < \varepsilon.$$
