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.


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 $$\lim_{n \to \infty} \frac{1}{n} \sum_{k = 0}^{n-1} \varepsilon_k g(x_k) = \int_{[0, 1]} g(x) \ df(x)?$$

  • 2
    $\begingroup$ 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 $a\in[0,1]$. $\endgroup$ Jun 3, 2021 at 12:49
  • $\begingroup$ 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). $\endgroup$ Jun 3, 2021 at 14:48
  • $\begingroup$ 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. $\endgroup$
    – Nate River
    Jun 3, 2021 at 15:59

1 Answer 1


$\newcommand{\ep}{\varepsilon}$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 $0\le a<b\le1$ 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!


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