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.
Best Answer
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.