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