# Writing the Lebesgue–Stieltjes integral as a sum of equidistributed Dirac delta measures

Problem set up:

Let $$f: [0, 1] \to \mathbb R$$ be an absolutely continuous function (thus a fortiori of bounded variation) such that its total variation on any open interval $$(a, b)$$ is $$b-a$$.

We say a sequence $$x_k$$ taking values in $$[0, 1]$$ is equidistributed if for any open interval $$(a, b)$$ it holds that $$\lim_{n \to \infty} \frac{1}{n} \#\{j \mathrel| j < n, x_j \in (a, b)\} = b - a,$$ where $$\#$$ denotes the cardinality of a finite set.

Question:

Let $$x_j \in [0, 1]$$ be an equidistributed sequence. Does it hold that there exists a sequence $$\{\varepsilon_n\}_{n \in \mathbb N}$$ taking values in $$\{-1, 1\}$$ such that for all Riemann integrable functions $$g: [0, 1] \to \mathbb R$$ such that the Lebesgue–Stieltjes integral $$\int_{[0, 1]} g(x) df(x)$$ exists, we have $$\lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n-1} \varepsilon_k g(x_k) = \int_{[0, 1]} g(x) \ df(x)?$$

• I suspect this to be false, but it is probably hard to construct a counterexample. To do that, one can try something like this: let the positive part of the measure $df$ have $1_S$ for the density, where $S$ is the Smith–Volterra–Cantor (fat) set, so that no function coinciding with $1_S$ a.e. be Riemann integrable; let $(x_j)$ be a specific, convenient equidistributed sequence with values in $S^c$; finally, consider $g=1_{[0,a]}$ for all $a\in[0,1]$. Jun 3, 2021 at 12:49
• A more direct description of the $df$s considered here would be to say that $df=s\, dx$, with $s=\pm 1$ (and measurable, obviously). Jun 3, 2021 at 14:48
• I feel like in that scenario, you could still choose the $\varepsilon_k$ wisely so that the sum and integral agree, at least for $g$ of the form $1_{[0, a]}$ - despite the fact that all the $x_j$ lie outside $S$. We should be able to approximate “patches” of $S$ with points $x_j$ in $S^c$ since $S^c$ is dense. And then the “somewhat continuity” of Riemann integrable functions might take care of the rest. Jun 3, 2021 at 15:59

$$\newcommand{\ep}{\varepsilon}$$This is at best a partial answer, with a simple result in the positive direction, which also highlights certain difficulties which may lead to a counterexample.
Let $$p_+$$ and $$p_-$$ denote, respectively, densities of the positive and negative parts of the Lebesgue--Stieltjes measure $$df$$, so that $$p_+p_-=0$$ almost everywhere (a.e.). The condition that the total variation of $$f$$ on any open interval $$(a,b)$$ with $$0\le a is $$b-a$$ means that $$p_++p_-=1$$ a.e. and hence a.e. $$\begin{equation*} p_+=1_S,\quad p_-=1_{S^c} \end{equation*}$$ for some Legesgue-measurable set $$S\subseteq[0,1]$$, where $$S^c:=[0,1]\setminus S$$. So, $$\begin{equation*} I:=\int_{[0,1]}g(x)\,df(x)=\int_0^1 h(x)\,dx, \end{equation*}$$ where $$g$$ is any Riemann-integrable function on $$[0,1]$$ and $$\begin{equation*} h:=g1_S-g1_{S^c}; \end{equation*}$$ then, of course, $$g$$ is Lebesgue integrable and hence so is $$h$$, so that the integral $$I$$ exists.
Suppose now for a moment that, possibly up to the a.e.-equivalence, the density $$p_+=1_S$$ is Riemann integrable. Then, without loss of generality, $$h$$ is Riemann integrable as well. So, letting $$\begin{equation*} \ep_k:= \begin{cases} 1&\text{ if }x_k\in S,\\ -1&\text{ if }x_k\in S^c, \end{cases}\tag{1} \end{equation*}$$ we get $$\ep_k g(x_k)=h(x_k)$$ and hence $$\begin{equation*} \frac1n\sum_{k=0}^{n-1}\ep_k g(x_k) =\frac1n\sum_{k=0}^{n-1}h(x_k) \to\int_0^1 h(x)\,dx=\int_{[0,1]}g(x)\,df(x), \end{equation*}$$ as desired.
This reasoning falls apart if $$I_S$$ is not a.e.-equivalent to a Riemann-integrable function -- which happens, for instance, when $$S$$ is the Smith--Volterra--Cantor (fat) set. Letting now $$(x_j)$$ be a specific, convenient equidistributed sequence with values in $$S^c$$ and probing a large enough collection of Riemann-integrable functions $$g$$, one might find a counterexample to the conjecture in general. Indeed, (1) strongly suggests that the $$\ep_k$$'s should mimic the set $$S$$, taking of course the membership of the equidistributed $$x_k$$'s in $$S$$ into account. This mimicking task may be too hard to accomplish if $$I_S$$ is not a.e.-equivalent to a Riemann-integrable function. Therefore, I think in general there is a counterexample -- but it would be an instructive surprise otherwise!