It seems reasonable to me to conjecture that the weak derivative is a special case of Radon-Nikodym derivative between measures. Recall that (missing some technical details):

  • Weak derivative: given a function $f:I=(a,b) \rightarrow \mathbb{R}$, $f'$ is called its weak derivative if for every $\varphi \in C^\infty_c(I)$ it holds that $ \int_I f'\varphi dx = - \int_I f \varphi' dx $
  • Radon-Nikodym derivative: given two measures $\mu, \nu$ (on the same $\sigma-$algebra), if $\nu \ll \mu$ ($\nu$ absoultely continuous w.r.t $\mu$), then there exists a unique $f \in L^1(\mu)$ such that $ \nu(E) = \int_E f d\mu $. Such $f$ is called the Radon-Nikodym derivative of $\nu$ w.r.t $\mu$: $f = \frac{d\nu}{d\mu}$.
  • Lebesgue-Stieltjes measure: given a right-continuous, monotone-increasing function $f$, we can define a measure (called the Lebesgue-Stieltjes measure of $f$) $LS_f((a,b]) = f(b) - f(a)$ (and then we have to extend it to all measurable sets as a Radon measure).

My question is: does something like the following hold true?

given $f \in W^{1,1}(I)$ ($f$ Lebesgue integrable, with weak derivative Lebesgue integrable), its weak derivative $f'$ coincides with the Radon-Nikodym derivative of the Lebesgue-Stieltjes measure of $f$ w.r.t. the Lebesgue measure:

$$f' = \frac{d (LS_f)_{ac}}{dx}$$

where $\nu_{ac}$ indicates the absolutely continuous part of $\nu$ w.r.t. the Lebesgue measure.

This seems to me to be true from the Fundamental Theorem of Calculus for $W^{1,1}$, but I have never seen it stated like that.

I would say it looks interesting as a formulation because it would regard the FTC in a more geeral setting.

  • 2
    $\begingroup$ $LS_f$ need not be absolutely continuous w.r.t. Lebesgue measure. $\endgroup$ Commented Sep 25, 2023 at 9:11
  • $\begingroup$ You must require $f$ to be monotone increasing also. Otherwise $LS_f$ makes no sense. In any case I seriously doubt there is any connection here. We say "Radon-Nikodym derivative" to hint at the mnemonic $\nu=f\,d\mu\Rightarrow f=\frac{d\mu}{d\nu}$. That is not really a derivative. More of a mnemonic aid than anything else. $\endgroup$ Commented Sep 25, 2023 at 9:16
  • $\begingroup$ @geetha290krm Thanks for the useful comment. I tried to edit the question (I added the request that we only consider the absolutely continuous part of the measure when we take the R-N derivative). $\endgroup$
    – rod
    Commented Sep 25, 2023 at 9:17
  • $\begingroup$ @GiuseppeNegro I had already edited the monotonicity hypothesis, thanks. I think there is indeed some connection. The Fundamental Theorem of Calculus for weak derivatives seems to say something very similar: $f(x)-f(a) = \int_a^x f'(t)dt$, i.e. you're writing $LS_f = f' dx$, or in other words: $f' = \frac{dLS_f}{dx}$ $\endgroup$
    – rod
    Commented Sep 25, 2023 at 9:21
  • $\begingroup$ @geetha290krm Also, could you provide an example of a function $f$ such that $LS_f$ is not a.c. w.r.t. the Lebesgue measure, but it has a weak derivative? I don't know any. $\endgroup$
    – rod
    Commented Sep 25, 2023 at 9:33

1 Answer 1


There is indeed a connection there, but your require a bit more regularity. You actually need $f\in W^{1,\infty}_{loc}(\mathbb{R})$, i.e. $f$ has to be locally Lipschitz. Then by Rademachers Theorem (see e.g. https://en.wikipedia.org/wiki/Rademacher%27s_theorem) $f$ has a classical derivative almost everwhere. This and monotonicity of $f$ is enough to show, that $LS_f$ is a Radon measure and absolutely continuous w.r.t. Lebesgue measure $\mathcal{L}^1$. Since we are working in $\mathbb{R}$ we can apply the differentiation theorem for Radon measures, see e.g. Leon Simons book about geometric measure theory. It states, that the Radon-Nikodym derivative can be calculated as follows: $$\frac{ d(LS_f)}{d\mathcal{L}^1}(x) = \lim_{r\rightarrow 0} \frac{LS_f(B_r(x))}{\mathcal{L}^1(B_r(x))} = \lim_{r\rightarrow 0}\frac{f(x+r)-f(x-r)}{2r} = f'(x)\ \mathcal{L}^1\mbox{-a.e.}$$ The last equality is by Rademachers Theorem.

EDIT: I did some research and I do not think the added regularity is really needed. For this you have to show, that $f\in W^{1,1}$ is enough for $f$ to be absolutely continuous, see e.g. https://www.math.ucdavis.edu/~hunter/m218a_09/ch3A.pdf Theorem 3.57

This is indeed true, because $f'\in L^1$ and we therefore have $$\forall \varepsilon > 0\ \exists \delta>0:\ \mbox{for all measurable }A\subseteq\mathbb{R}\mbox{ with }\mathcal{L}^1(A)<\delta$$ we have $\int_A|f'|\, dx < \varepsilon$. This is true by e.g. Vitalis convergence theorem. So let us check absolute continuity for $f$: We have to check the following: For all $\varepsilon>0$ exists a $\delta>0$ such that for all $-\infty < a_1 < b_1 < a_2 < \ldots< a_n< b_n< \infty$ with $$\sum_{k=1}^n b_k-a_k < \delta,\mbox{ we have }\sum_{k=1}^nf(b_k) -f(a_k) < \varepsilon.$$ Hence we choose $\delta$ as above for the integral estimate. Then $$\mathcal{L}^1(\bigcup_{k=1}^n [a_k,b_k]) = \sum_{k=1}^n b_k-a_k < \delta$$ and therefore $$\sum_{k=1}^n f(b_k)-f(a_k) = \sum_{k=1}^n\int_{a_k}^{b_k}f'(x)\, dx \leq \int_{\bigcup_{k=1}^n[a_k,b_k]}|f'(x)|\, dx < \varepsilon.$$

  • $\begingroup$ Thank you for your interesting remark. Still, however much relevant, this stretches a connections between the Radon-Nikodym derivative and the strong a.e. derivative, which is a strictly related but different concept from the weak derivative (for example, the Cantor function is differentiable a.e., but it is not weakly differentiable, and the characteristic function of rational numbers is weakly differentiable but not differentiable a.e.)... $\endgroup$
    – rod
    Commented Sep 25, 2023 at 10:06
  • $\begingroup$ Absolutely correct, i got carried away there in the edit and mixed something up. Leaving it here for future reference $\endgroup$ Commented Sep 25, 2023 at 10:08
  • $\begingroup$ Thank you for your reply. In short, you observe that if $f \in W^{1,1}$, then it is absolutely continuous. Therefore, the FTC holds, which implies (is it immediate?) my conjecture. Is that right? $\endgroup$
    – rod
    Commented Sep 25, 2023 at 11:06
  • $\begingroup$ @ rod: Your completely right. If you only have $W^{1,1}$, my argument in the edits shows, that the function is still AC. As far as I can remember, there is a proof of Thm. 3.57, that actually shows your conjecture. Unfortunately I do not have a good reference at hand at the moment. That also was the thing I mixed up earlier. $\endgroup$ Commented Sep 25, 2023 at 11:32

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