If $f$ Riemann Integrable then $U(f,D_n)-L(f,D_n) \to 0$

Question

In this question $U(f,D)$ and $L(f,D)$ denote upper and lower Riemann sums respectively with respect to the dissection $D$.

I'm trying to show that if $f:[0,1] \to \mathbb{R}$ is Riemann integrable then $U(f,D_n)-L(f,D_n) \to 0$ where $D_n$ is the dissection $D_n = \{0<\frac{1}{n}<\frac{2}{n}<...<\frac{n-1}{n}<1\}$.

I know that, since $f$ is Riemann Integrable, $\forall \varepsilon >0 \exists \text{ dissection of } [0,1] D \text{ s.t } U(f,D)-L(f,D) < \varepsilon$

What I've tried (don't think it's the right way about this so feel free to skip reading the rest):

Let $m(D)=max\{a_i-a_{i-1}\}$, $M(D)=min\{a_i-a_{i-1}\}$ where $D=\{0=a_0<a_1<...<1\}$.

Note $m(D_n)=\frac{1}{n}$.

Given $\varepsilon >0$, we can find a dissection $D$ with $U(f,D)-L(f,D) < \varepsilon$. Now we can find a $n \in \mathbb{N}$ s.t $\frac{1}{n}<M(D)$, or equivalently, a $D_n$ with $\frac{1}{n}=m(D_n)<M(D)\leq a_i-a_{i-1}$.

So $$\begin{eqnarray} U(f,D_n)&=&\sum_{i=1}^{n} \frac{1}{n} sup\{f(x):\frac{i-1}{n}<x<\frac{i}{n}\} \\ &<&\sum_{i=1}^{n}(a_i-a_{i-1})sup\{f(x):\frac{i-1}{n}<x<\frac{i}{n}\} \end{eqnarray}$$ But this doesn't help with $L(f,D_n)$. I was also thinking as $n$ gets large $\{f(x):\frac{i-1}{n}<x<\frac{i}{n}\}$ would become very bounded so the sup and inf would become similar, but here i'm assuming $f$ is continuous which i can't assume. Finally, i was thinking if i can show $D \subset D_n$ then the result follows immediately, but there's no reason why this should hold.

Answer 1

Using this answer one can prove that for every $\epsilon >0$ there is a corresponding $\delta>0$ such that $U(f, D) - L(f, D) <\epsilon $ whenever the norm of $D$ is less than $\delta $. Now you just need to take $n>1/\delta$ and then $U(f, D_n) - L(f, D_n) <\epsilon $ and thus the desired result is proved.