We know that if $f:A=[a,b]\to \mathbb{R}$ is continuous then $f$ is Riemann integrable. I'm trying to generalize this to when $A\subset \mathbb{R}^n$, where $A$ is a rectangle and we have a partition $P=\{R_1,\dots,R_k\}$ of $A$
To prove this I'm trying to prove a certain proposition. Given $R=I_1\times\dots \times I_n\subset \mathbb{R}^n$ where $I_i=[b_i-a_i]$ for $1\leq i\leq n$, if $v(R)$ is the volume of $R$ and we have that $ v(R)=(b_1-a_1)(b_2-a_2)\cdots(b_n-a_n)< \delta/n$ then $\lVert x-y\rVert <\delta$ for any $ x,y\in R\subset\mathbb{R}^n$.
(Note: I'm not entirely sure if this is true for $\delta/n$ but its a guess and will change it accordingly.)
If I can prove the above, then the my proof would be done. My proof so far is:
Since $A$ is compact and $f$ is continuous then $f$ is uniformly continuous therefore given $\epsilon>0$ there exists a $\delta$ such that if $\lVert x-y \rVert <\delta$ then $\lvert f(x)-f(y)\rvert <\epsilon/v(A)$.
Now let $P$ be partition of $A$ such that $P=\{R1,\dots, R_k\}$ and $v(R_i)<\delta/n$. We have that $U(f,P)-L(f,P)= \sum_i^k \sup_{R_i}f*v(R_i)-\sum_i^k \inf_{R_i}f*v(R_i)$
$=\sum_i^k (\sup_{R_i}-\inf_{R_i}f)v(R_i)$
$=\sum_i^k[ f(x_i)-f(y_i)]v(R_i)$, for some $x_i,y_i \in R_i$ since $R_i$ is compact.
And then if my above proposition is true I can replace $f(x_i)-f(y_i)$ with $\epsilon/v(A)$ and I would end up with $U(f,P)−L(f,P)<\epsilon$ therefore per the Cauchy criterion it is integrable.
In proving my proposition above, so far I only have that for $x,y\in R$
$\lVert x-y\rVert=\sqrt{(x_1-y_1)^2+\cdots+(x_n-y_n)^2}\leq \sqrt{(b_1-a_1)^2+\cdots+(b_n-a_n)^2}$
I'm not sure how to proceed from here though; any help would appreciated.