Let $g : [0,1] \to \mathbb{R}$ be continuos and of bounded variation such that has the (N) proprety ( i.e. $\mu(E) = 0 \implies \mu(g(E)) = 0$ ), show that $g$ is absolutely continuos.
My attempt :
To show my idea, first suppose that $g$ is injective, then define $\nu(A) := \mu(g(A))$ I have that $\nu << \mu$ so by Radon-Nykodim for every $\epsilon > 0$ there exists $\delta > 0$ such that $\mu(A) < \delta \implies \nu(A) < \epsilon$, but then if $\sum_{k=1}^{\infty}{b_k - a_k} < \delta$ I have $\sum_{k=1}^{\infty}{|g(b_k)-g(a_k)|} \leq \sum_{k=1}^{\infty}{\nu([a_k,b_k])} = \nu(\bigcup_{k=1}^{\infty}{(a_k,b_k)}) < \epsilon$
So indeed $g$ is absolutely continuos.
The key is that if $(a_k,b_k)$ are pairwise disjoint then $g((a_k,b_k))$ are pairwise disjoint, which uses strongly the injectivity.
Therefore I want to ""injectivize"" the function. I had this Idea :
Consider $g^*(x) := (x , g(x))$ and $\nu(A) := \mathcal{H}^1(g^*(A))$
I would like to show that $\mu(A) = 0 \implies \nu(A) = 0$ but I didn't managed to do this, anyway if this was true then by Radon-Nykodim
$\forall \epsilon > 0 \exists \delta > 0 \; : \; \mu(A) < \delta \implies \nu(A) < \epsilon$
From which the claim follows since it's easy to show that $V_{x_1}^{x_2}(g) \lesssim \nu([x_1,x_2])$ (there should be a costant of $\frac{1}{\sqrt{2}}$ or somethin like this )
The problem is that I'm unable to show that $\mu(A) = 0 \implies \nu(A) = 0$