*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?

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

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.