Absolute continuity of a Borel measure

This is a question from Ph. D Qualifying Exam of real analysis.

Let $$F$$ be an increasing function on $$[0,1]$$ with $$F(0)=0$$ and $$F(1)=1$$. Let $$\mu$$ be a Borel measure defined by $$\mu((a, b))=F(b-)-F(a+)$$ and $$\mu(\{0\})=\mu(\{1\})=0$$. Suppose that the function $$F$$ satisfies a Lipschitz condition $$|F(x)-F(y)|\le A|x-y|$$ for some $$A>0$$. Let $$m$$ be the Lebesgue measure on $$[0,1]$$.

(a) Prove that $$\mu \ll m$$.

(b) Prove that $$\dfrac{d\mu}{dm} \le A$$ a.e.

My attempt: (a) Since $$F$$ is Lipschitz continuous, it is clear that $$F$$ is absolutely continuous and $$F$$ is differentiable a.e. and $$\mu((a, b))=\int_{a}^{b}F'dm$$ by absolute continuity. Since $$\mu$$ is a Borel measure, it extends to $$\mu(E)=\int_E F'dm$$ for every Borel set $$E$$.(I'm not sure for this part. Is there any related theorem or counterexample for this one?)

Therefore, it suffices to show that $$\int_E F' =0$$ whenever $$E$$ is a Borel set and $$m(E)=0$$, and this is obvious from the definition of Lebesgue integral.

(b) From (a), $$\dfrac{d\mu}{dm}=F'$$ and by Lipschitz continuity, $$|F'|\le A$$ a.e. and the result is obvious.

Am I correct? Is there any errors or logical jumps in my attempt?

You don't need the full extension to Borel sets - open sets will suffice. Recall that any open set $$G \subset \mathbb R$$ may be written as a union of disjoint open intervals $$\{(a_k,b_k)\}$$.
Suppose that $$N \subset (0,1)$$ is a Borel set with Lebesgue measure zero. For any $$\epsilon > 0$$ there exists an open set $$G \subset (0,1)$$ with the property that $$N \subset G$$ and $$m(G) < \epsilon$$. Writing $$G = \cup (a_k,b_k)$$ as above you find $$\mu(G) = \sum_k \mu(a_k,b_k) = \sum_k \int_{(a_k,b_k)} F' \, dm = \int_G F' \, dm.$$
The monotonicity of $$\mu$$ and the fact that $$|F'| \le A$$ almost everywhere give you $$\mu(N) \le \mu(G) = \int_G F' \, dm \le A \mu(G) < A \epsilon.$$ Since $$\epsilon > 0$$ is arbitrary you get $$\mu(N) = 0$$.
In general, if $$N \subset [0,1]$$ is a Borel set with Lebesgue measure zero then $$\mu(N) \le \mu(\{0\}) + \mu(N \cap (0,1)) + \mu(\{1\}) = 0$$ so that $$\mu(N) = 0$$ too. Thus $$\mu \ll m$$.
I would suggest a simpler approach: First note that even though $$F$$ was only assumed Borel-measurable, we also assume $$F$$ is Lipschitz and thus continuous. Since $$F$$ is also increasing, we see that in fact $$\mu((a,b))=F(b)-F(a)=|F(b)-F(a)| \leq A |b-a| = A \: m((a,b))$$. It follows immediately that $$\mu \leq A m$$ (in the sense of on every measurable set). Absolute continuity is clear now, and using the Radon-Nikodym theorem on $$\mathbb{R}$$, i.e. $$$$\frac{d \mu}{d m}(x)=\lim_{\epsilon \rightarrow 0} \frac{\mu((x-\epsilon,x+\epsilon))}{m((x-\epsilon,x+\epsilon))}$$$$ For almost all $$x \in \mathbb{R}$$, we can immediatly conclude $$$$\frac{d \mu}{d m}(x) \leq A$$$$