Limit of the ratio of two non-Riemann sums.

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,

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