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:

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


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.


1 Answer 1


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.

  • $\begingroup$ Can you give a reference to "a sufficiently general version of the (Lebesgue) differentiation theorem"? The usual version of this theorem is based on the Hardy–Littlewood maximal inequality, which seems to be tied to the Lebesgue measure or, more generally, doubling measures. $\endgroup$ Jun 5 at 19:01
  • $\begingroup$ @IosifPinelis: The main trick is to replace the Vitali covering lemma by the Besicovitch covering lemma, then the usual treatment applies to the general situation. See Mattila, Geometry and of sets and measures in Euclidean space, Ch. 2. (I think it's also discussed somewhere in Stein's standard books, maybe in supplementary notes.) $\endgroup$ Jun 5 at 19:11
  • $\begingroup$ @IosifPinelis: The Besicovitch lemma doesn't mention a measure, so it should be plausible right away from its statement that the theory generalizes from Lebesgue measure to general measures. $\endgroup$ Jun 5 at 19:14
  • $\begingroup$ Thank you. I should remember this, especially given that on the real line the Besicovitch covering lemma is easy. $\endgroup$ Jun 5 at 20:23

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