Define the General Subdivision $S$ of a rectangle $R$ in $\Bbb R^2$ as a collection $E_1,...,E_k$ of Jordan regions such that none of them has interior points in common, and:
$$R \subset \bigcup_{i=1}^k E_i$$
The norm of $S$ is defined as:
$$d(S)=\max \left[\operatorname{diam}(E_i:1<i<k)\right]$$
Where $diam$ denotes denotes the diameter of each rectangle. Also, if $f$ is a continuous in $R$ and for each $i$, $(x_i, y_i)$ is in $E_i$ then the sum
$$S(f,S,(x_i,y_i))= \sum_{i=i}^k f(x_i,y_i) \operatorname{vol}(E_i) $$
is named the Riemann sum of $f$ in S. I need to show that for every $\epsilon >0$ there is a $\delta >0$ such that if $d(S)<\delta$ then:
$$\left| \iint_R f \ dA -S(f,S,(x_i,y_i)) \right| < \epsilon $$
The definition of integral I've been studying is given by the superior and inferior sums:
$$\iint_R f \ dA= \sup \{L(f,P) \mid \text {P is a partition of } \mathbb{R}\}= \inf \{ U(f,P) \mid \text {P is a partition of } \mathbb{R} \}$$
I tried using the Cauchy Criterion but i couldn't prove what I need. Any help will be great, thanks in advance.
We assume $f$ is integrable, hence, bounded on the compact rectangle $R \subset \Bbb R^2$.
If $I$ is the integral, then for any $\epsilon > 0 $, there is a partition $P_\epsilon$ such that $U(f,P_\epsilon) < I + \epsilon/4$. (The integral is the infimum of upper sums).
A partition can be characterized as the collection of vertices $(x_i,y_j)$ that form non-overlapping sub-rectangles $[x_{i-1},x_i] \times [y_{j-1},y_j]$ covering R:
$$P_{\epsilon} = \{(x_i,y_j): 0 \leq i \leq M, 0 \leq j \leq N, R = \cup_{i = 1}^M \cup_{j = 1}^N[x_{i-1},x_i] \times [y_{j-1},y_j]\}.$$
Let
$$\delta = \frac{\epsilon}{4MNDd(R)}$$
where $D = \sup\{|f(x)-f(y)|:x,y \in R \}$ denotes the maximum oscillation of $f$, $MN$ is the number of sub-rectangles in the partition $P_\epsilon$, and $d(R)$ is the diameter of $R$.
Now let $P$ be any partition with $d(P)<\delta$.
Form the common refinement $Q$ of $P$ and $P_{\epsilon}.$ Each sub-rectangle of $P$ and $P_{\epsilon}$ is a finite union of sub-rectangles of $Q$. The upper and lower sums of a partition bracket the sums for a refined partition:
$$L(f,P_{\epsilon}) \leqslant L(f,Q) \leqslant U(f,Q) \leqslant U(f,P_{\epsilon}), \\ L(f,P) \leqslant L(f,Q) \leqslant U(f,Q) \leqslant U(f,P). $$
Each sub-rectangle $R_{ij}$ of $P$ is of the form
$$R_{ij} = [x_{i-1}^P,x_i^P] \times [y_{j-1}^P,y_j^P]$$
and can be decomposed as a union of sub-rectangles $R_{ijkl} \subset R_{ij}$ of $Q$:
$$R_{ij} = \bigcup_{k=1}^{m_{ij}} \bigcup_{l=1}^{n_{ij}} R_{ijkl}.$$
The upper sums $U(f,P)$ and $U(f,Q)$ are
$$U(f,P) = \sum_{i,j} \sup_{x \in R_ij} f(x) vol(R_ij), \\ U(f,Q) = \sum_{i,j}\sum_{k,l} \sup_{x \in R_{ijkl}} f(x) vol(R_{ijkl}).$$
As $R_{ijkl} \subset R_{ij}$, we have
$$\sup_{x \in R_{ij}}f(x)-\sup_{x \in R_{ijkl}}f(x) \leqslant D.$$
Hence,
$$U(f,P) - U(f,Q) \leqslant \sum_{i,j}\sum_{k,l} [\sup_{x \in R_{ij}}f(x)-\sup_{x \in R_{ijkl}}f(x) ]vol(R_{ijkl}) \leqslant D\sum_{i,j}\sum_{k,l} vol(R_{ijkl}). $$
Each sub-rectangle $R_{ijkl}$ of $Q$ is of the form
$$R_{ijkl} = [x_{k-1}^{(i)},x_k^{(i)}] \times [y_{l-1}^{(j)},y_l^{(j)}], \\ vol(R_{ijkl}) = (x_{k}^{(i)}-x_{k-1}^{(i)})(y_{l}^{(j)}-y_{l-1}^{(j)}). $$
We formed the common refinement $Q$ from $P$ and $P_{\epsilon}$ (with $MN$ sub-rectangles). The partition $Q$ has no more than $MN$ more partition points than $P,$ in either dimension.
Hence,
$$U(f,P) - U(f,Q) \leqslant D\sum_{i,j}\sum_{k,l} vol(R_{ijkl}) \\ = D\sum_{i}\sum_{j} \sum_{k} \sum_{l }(x_{k}^{(i)}-x_{k-1}^{(i)}) (y_{l}^{(j)}-y_{l-1}^{(j)}) \\ =D\sum_{i}\sum_{k} (x_{k}^{(i)}-x_{k-1}^{(i)}) \sum_{j} \sum_{l } (y_{l}^{(j)}-y_{l-1}^{(j)}) \\ \leqslant D\sum_{i}\sum_{k} (x_{k}^{(i)}-x_{k-1}^{(i)}) d(R) \\ \leqslant D(MN \delta) d(R).$$
Substituting for $\delta$ we find
$$U(f,P) - U(f,Q) \leqslant MND d(R)\delta= \epsilon/4.$$
It follows that
$$U(f,P) \leqslant U(f,Q) + \epsilon/4 \\ < U(f,P_\epsilon)+\epsilon/4 \\ < I + \epsilon/2.$$
By a similar argument, you can show $L(f,P)> I - \epsilon/2$.
If $S(f,P)$ is any Riemann sum with respect to the partition P, then
$$I - \epsilon/2 < L(f,P) \leqslant S(f,P) \leqslant U(f,P) < I + \epsilon/2,$$
and
$$|S(f,P) - I| < \epsilon.$$