Limit of the ratio of two non-Riemann sums.

Question

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?

Answer 1

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,

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

Answer 2

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...