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Let $\left[ {a,b} \right] \subset \mathbb{R}$ and $f,g:\left[ {a,b} \right] \to \mathbb{R}$ be two Riemann-integrable functions.

Let $a = {x_0} < {x_1} < {x_2}... < {x_n} = b$ be a partition of $\left[ {a,b} \right]$ and let $\Delta x = \mathop {\max }\limits_{i = 0}^{n - 1} \left( {{x_{i + 1}} - {x_i}} \right)$.

Let ${t_i} \in \left[ {{x_i},{x_{i + 1}}} \right],\;i = 0,n - 1$ and let $k \in {\mathbb{N}^*}$.

I’d like to prove that

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {{{\left( {{x_{i + 1}} - {x_i}} \right)}^k}g\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

It is obvious for equally spaced partitions ${x_{i + 1}} - {x_i} \equiv \Delta x$

$\mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}f\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta {x^k}g\left( {{t_i}} \right)} }} = \mathop {\lim }\limits_{\Delta x \to 0} \frac{{\sum\limits_{i = 0}^{n - 1} {\Delta xf\left( {{t_i}} \right)} }}{{\sum\limits_{i = 0}^{n - 1} {\Delta xg\left( {{t_i}} \right)} }} = \frac{{\int\limits_a^b {f\left( x \right){\text{d}}x} }}{{\int\limits_a^b {g\left( x \right){\text{d}}x} }}$

But I don’t see how to do it in the general case?

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  • $\begingroup$ Some context would be advisable. Which is your source ? $\endgroup$ Dec 20, 2019 at 18:15
  • $\begingroup$ @TonyPiccolo My question is motivated by my unpublished works on point null hypothesis testing problems such as the Behrens-Fisher problem or the Jeffreys-Lindley paradox. Please see my draft paper A Fully Bayesian Solution to k-Sample Tests for Comparison and the Behrens-Fisher Problem Based on the Henstock-Kurzweil Integral. Problems with discrete parameters are OK. But problems with continuous ones are degenerate. Therefore, I approximate continuous problems by sequences of discrete ones and then I take the limit solution... $\endgroup$ Dec 20, 2019 at 18:22
  • $\begingroup$ @TonyPiccolo This limit is trivial for discrete random variables defined on equipartitions but I still need to prove that it remains the same for arbitrary partitions. Hence my little problem. $\endgroup$ Dec 20, 2019 at 18:24
  • $\begingroup$ @TonyPiccolo Anyway, the main "theorem" of my paper is not satisfactory and I need to fix this. But the main idea should be OK. $\endgroup$ Dec 20, 2019 at 18:37
  • $\begingroup$ @TonyPiccolo Thanks to you, I learned that it is sufficient to consider equipartitions (but arbitrary tagged points) for the Riemann integral. I need to think about it.............. $\endgroup$ Dec 20, 2019 at 18:49

2 Answers 2

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This result can fail to hold with non-uniform partitions. For a counterexample, we look for Riemann integrable functions $f$ and $g$ and a sequence of partitions

$$P_n: a = x_0^{(n)}<x_1^{(n)} < \ldots < x_{n-1}^{(n)} < x_n^{(n)} = b$$

along with a choice of tags $t_j^{n} \in [x_{j},x_{j+1}]$ where

$$\tag{*}\Delta x := \|P_n\| = \underset{0 \leqslant j \leqslant n-1} \max \left(x_{j+1}^{(n)}-x_j^{(n)}\right) \underset{n \to \infty}\longrightarrow 0$$ and, such that for some $k > 1$,

$$\lim_{\Delta x \to 0, \,n \to \infty}\frac{\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k}{\sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k} \neq \frac{\int_a^bf(x) \, dx}{\int_a^b g(x) \, dx}$$

Note that condition (*) is an essential requirement here as it ensures convergence of Riemann sums (with $k=1$) to the respective integrals.

Take $[a,b] = [1,e]$, $f(x) = 1$, $g(x) = x$, $k = 2$, partition points $x_j^{(n)} = e^{j/n}$ and tags $t_j^{(n)} = e^{j/n}$ for $j=0,1,\ldots,n$.

In this case, the $n$th partition is $P_n : 1 < e^{1/n} < e^{2/n} < \ldots < e^{(n-1)/n}< e$, and we have

$$\Delta x = \|P_n\| = \max_{0 \leqslant j \leqslant n-1}(e^{(j+1)/n}-e^{j/n}) = \max_{0 \leqslant j \leqslant n-1}e^{j/n}(e^{1/n}-1) = e^{(n-1)/n}(e^{1/n}-1), $$

where $\Delta x = e^{(n-1)/n}(e^{1/n}-1)\to e\cdot 0 = 0$ as $n \to \infty$.

Note that, with $k=1$,

$$\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \sum_{j=0}^{n-1} 1 \cdot (e^{(j+1)/n}- e^{j/n)})= (e^{1/n} - 1)\sum_{j=0}^{n-1}e^{j/n} = (e^{1/n} - 1)\frac{e-1}{e^{1/n} -1}\\ \sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \sum_{j=0}^{n-1} e^{j/n} \cdot (e^{(j+1)/n}- e^{j/n)})= (e^{1/n} - 1)\sum_{j=0}^{n-1}e^{(2j)/n} = (e^{1/n} - 1)\frac{e^2-1}{e^{2/n} -1} ,$$

and, as we expect for Riemann sums,

$$\lim_{n \to \infty}\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= e-1 = \int_1^e f(x) \, dx\\ \lim_{n \to \infty}\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})= \frac{e^2-1}{2} = \int_1^e g(x) \, dx$$

However, for $k=2$,

$$\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k= \sum_{j=0}^{n-1} 1 \cdot (e^{(j+1)/n}- e^{j/n)})^2= (e^{1/n} - 1)^2\sum_{j=0}^{n-1}e^{(2j)/n} = (e^{1/n} - 1)^2\frac{e^2-1}{e^{2/n} -1}\\ \sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k= \sum_{j=0}^{n-1} e^{j/n} \cdot (e^{(j+1)/n}- e^{j/n)})^2= (e^{1/n} - 1)^2\sum_{j=0}^{n-1}e^{(3j)/n} = (e^{1/n} - 1)^2\frac{e^3-1}{e^{3/n} -1} ,$$

and,

$$\lim_{\Delta x \to 0, \,n \to \infty}\frac{\sum_{j=0}^{n-1} f(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n)})^k}{\sum_{j=0}^{n-1} g(t_j^{(n)})(x_{j+1}^{(n)}- x_j^{(n))})^k} = \lim_{n \to \infty}\frac{e^2-1}{e^3-1}\frac{e^{3/n}-1}{e^{2/n}-1} = \frac{3}{2}\frac{e^2-1}{e^3-1} \\\neq \frac{2}{e+1} = \frac{e-1}{\frac{e^2-1}{2}}= \frac{\int_a^bf(x) \, dx}{\int_a^b g(x) \, dx}$$

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  • $\begingroup$ Thanks a lot RRL. As stated in the comments above, this simple question arises from my research about the Behrens-Fisher problem, which comes in two different variants, discrete and continuous. The situation is as follows: 1) the standard Bayesian solutions in both cases are not correct and satisfactory in many respects 2) the continuous problem has a trivial and totally useless absolute solution because we directly jump into the uncountable 3) the discrete problem has straightforward absolute and relative solutions. Hence it remains to find the relative solution to the continuous problem... $\endgroup$ Dec 1, 2020 at 13:56
  • $\begingroup$ ... by 1) approximating the continuous problem by a sequence of discrete one. This is done by partitioning the parameter space 2) computing the relative solutions to those problem 3) taking the limit of those solutions. But as you shown, this limit solution does not exist. Hence, we conclude that the continuous problem is actually mathematically ill-posed unless we can argue, justify that we should restrict ourselves to particular sequences of partitions, e.g. equipartitions. Therefore, I'm looking for some intuitive principle, typically an invariance principle... $\endgroup$ Dec 1, 2020 at 14:03
  • $\begingroup$ ... that would imply the restriction to e.g. equipartitions. Any idea please? $\endgroup$ Dec 1, 2020 at 14:05
  • $\begingroup$ @FabricePautot: You're welcome. I'll have to delve more deeply into your work and see where passing to a continuous formulation breaks down. $\endgroup$
    – RRL
    Dec 1, 2020 at 17:48
  • $\begingroup$ Thanks. That would be great!!! The beginning of my paper is correct (criticism of the classical solution + solutions to the discrete problem). Only the very end is not (limit relative solution of the continuous problem), I was way too fast because I was not expecting the continuous problem to be ill-posed at all since it is supposed to be well-posed ... according to the standard, but wrong, solution. $\endgroup$ Dec 1, 2020 at 18:33
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I believe this reduces to two copies of a problem which has never been successfully solved in the history of mathematics. The best you could do is conjecture, but you should be able to at least show that your conjecture is reasonable.

As a matter of fact, you can't even prove the result which you claimed was obvious. It's obvious because we accept this definition of the integral. That does not mean that it can be proven unfortunately...

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  • $\begingroup$ Thanks, but I don't get your point. The situation is as follows. We have a problem that comes in two different variants, discrete and continuous. The standard solutions to both problems are not correct for many reasons. The solution to the discrete problem is straightforward. Hence, it remains to solve the continuous one by i) approximating the continuous problem by a sequence of discrete ones ii) solve those discrete problems iii) prove there exists a unique limit solution. But, as shown by Alex above, that is not the case. Hence we conclude that the problem is actually ill-posed unless... $\endgroup$ Nov 26, 2020 at 13:51
  • $\begingroup$ ... we add some well-chosen constraints or conditions that make it well-posed, e.g. restrict ourself to equipartitions. That's the next step... $\endgroup$ Nov 26, 2020 at 13:54
  • $\begingroup$ I don't understand your second paragraph: for equipartitions, the limit is the ratio of both integrals, isn't it? $\endgroup$ Nov 26, 2020 at 13:58
  • $\begingroup$ @FabricePautot here's what I mean: If we are taking the limit of a rational function, say f(x)/g(x), then we can rewrite the single limit as the ratio of the two limits. The limit of ratio of the two equipartitioned Riemann sums results in the ratio of the two integrals only because we it literally is just a certain definition of one integral divided by the definition of another integral. There is no way to "prove" this definition is equivalent to other definitions we may use such as the Fundamental Theorem of Calculus... $\endgroup$
    – Algebraic
    Nov 27, 2020 at 1:01
  • $\begingroup$ @FabricePautot I also want to point out there are some issue with taking the limit dx->0 specifically because this means the length of the intervals will have to decrease as different rates. How might we account for this? One way I want to suggest is by using a procedure that doubles the amount of intervals every step by including the midpoint of the interval. I'm unsure if this will get you where you want to go, but that's my thoughts. Hope this helps. $\endgroup$
    – Algebraic
    Nov 27, 2020 at 1:06

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