Fundamental theorem of calculus for Lebesgue–Stieltjes integrals?

Note: Throughout, we denote by $$\mathcal L$$ the Lebesgue measure on $$\mathbb R$$.

Let $$g: [0, 1] \to \mathbb R$$ be a continuous function of bounded variation. Denote by $$\mu_g$$ its associated Lebesgue–Stieltjes measure, and $$\lvert\mu_g\rvert$$ its total variation measure.

We say that a function $$f: [0, 1] \to \mathbb R$$ is absolutely continuous with respect to $$g$$ if $$f$$ is continuous, and for every $$\varepsilon > 0$$, there exists a $$\delta > 0$$ such that whenever $$I_k$$, ($$k = 1, \dotsc, n$$) are disjoint open intervals with $$\sum_{i = 1}^n \lvert\mu_g\rvert (I_n) <\delta$$, we have $$\sum \mathcal L(f(I_n)) < \varepsilon$$.

For fixed $$x \in [0, 1]$$, denote by $$E_x$$ the set $$\{y \in [0, 1] \, \mid \, g(x) - g(y) \neq 0\}$$, and consider the limit

$$\lim_{y \to x\, ,\, y \in E_x} \frac{f(x) - f(y)}{g(x) - g(y)}.$$

We shall say that the above limit exists, and denote it by $$\frac{df}{dg}(x)$$ if for every $$r > 0$$, the set $$E_x \cap B_r (x)$$ is nonempty, and the limit along $$E_x$$ exists in the usual sense.

Note that for $$\lvert\mu_g\rvert$$ a.e. $$x \in [0, 1]$$, $$E_x \cap B_r (x)$$ is nonempty for every $$r > 0$$.

Question: Let $$g$$ be a continuous function of bounded variation as above, and $$f$$ a function absolutely continuous with respect to $$g$$. Is it true that the following two statements hold?

1. $$\frac{df}{dg}$$ exists $$\lvert\mu_g\rvert$$-a.e.

2. For every $$x \in [0, 1]$$, we have the following fundamental theorem of calculus style formula:

$$\int_0^x \frac{df}{dg} \, dg = f(x) - f(0).$$

Remark:

The "Riemann FTC" version of the above is true, and not overly difficult to prove — if the limit $$\frac{df}{dg}$$ exists everywhere, then the integral formula holds.

It is thus left to see if the "Lebesgue FTC" version holds — if $$f$$ is absolutely continuous with respect to $$g$$, then $$\frac{df}{dg}$$ exists $$\lvert\mu_g\rvert$$-a.e., and the integral formula holds.

This works, because: (1) Your definition of absolute continuity is equivalent to the other standard definition, namely, the condition that $$|\mu_g|(A)=0$$ implies $$|\mu_f|(A)=0$$ (your condition implies that $$f\in BV$$, so $$\mu_f$$ is well defined).
(2) By a sufficiently general version of the (Lebesgue) differentiation theorem, $$\lim_{h\to 0} \mu_h(x,x+h)/|\mu_g|(x,x+h)$$ exists for $$|\mu_g|$$-a.e. $$x$$ and if $$\mu_h\ll |\mu_g|$$, then the limit computes the Radon-Nikodym derivative $$d\mu_h/d|\mu_g|$$. Thus your quotient converges to $$(d\mu_f/d|\mu_g| )/(d\mu_g/d|\mu_g|)$$; the denominator takes the values $$\pm 1$$, so the division doesn't make any trouble.
In general, a Radon-Nikodym derivative satisfies $$\int f (d\mu/d\nu)\, d\nu = \int f\, d\mu$$. This gives the formula from part (2) of your question.