How to prove that if $S(S,f,\{p_i\})=\sum f(p_j)A(D_j)$ then $|\iint_{R}f-S|(S,f,\{p_i\})|<\varepsilon$ Let $$s(S,f,\{p_i\})=\sum_{j=1}^m f(p_j)A(D_j)$$ where $p_j\in D_j$ and $S$ is a general division of a close rectangle $R$, i.e., a finite number of domains $D_j$ each having area, and together covers $R$ and such that no pair have common interior points. The norm $d(S)$ is the largest of the diameters of the set $D_j$.  How can I prove that if $f$ is continuous on $R$, then for $\varepsilon>0$ there is $\delta>0$ such that $$\left| \iint_R f-s(S,f,\{p_j\}) \right| $$ whenever $d(S)<\delta$ ??
I saw this problem from Buck's Advanced Calculus, and I'm wondering how can I prove it.
Attempt: When one integrates over domains $D_j$ which are not rectangles, then we choose a rectangle $R_j$ which contains $D_j$ and defined $F=f$ on $D_j$ and $0$ outside $D_j$, so that $$\iint_{D_j}f=\iint_RF $$ The associated Riemann sum, is given by $$ S(N,f,\{p_{ij}\}) = \sum_{i,j}f(p_{i,j})A(R_{ij}) $$ where $A(R_{ij})$ denotes the area of rectangles and $N$ is a grid. One can think that if the difference of this two sums are equal, then we are done, I mean making small the difference
$$\left| \sum_{i,j}f(p_{ij})A(R_{ij})-\sum_{j}f(p_j)A(D_j)\right|<\varepsilon$$
The uniform continuity of $f$ can make $|f(p_j)-f(p_{i,j})|$ small but how to deal with $|A(R_{ij})-A(D_j))|$ and also is there a way of operating with only one sum in $\sum_{i,j}f(p_{i,j})A(R_{ij})$??
Here there is the same question: but in a different complicated approach (uses lower and upper bounds, I'm wondering if using the uniform continuity of $f$ could be done easier) Riemann Integrability in $\Bbb R^2$
 A: We can show that under the given hypotheses that
$$\int_R f = \int_{\cup_j D_j} f = \sum_j \int_{D_j} f$$
We will use the fact that if $f_1$ and $f_2$ are Riemann integrable on $Q$ then $\int_Q f_1 \pm f_2 = \int_Qf_1 \pm \int_Q f_2$.
First assume that $f$ is both continuous and nonnegative and consider two sets $D_1$ and $D_2$.  Let $f_{D_j}(x) = f(x)$ for $x \in D_j$ and $f_{D_j}(x) = 0$ for $x \in D_j^c$.
We have
$$f_{D_1\cup D_2}(x) = \max(f_{D_1}(x), f_{D_2}(x))= \frac{1}{2}\left(f_{D_1}(x)+f_{D_2}(x)+|f_{D_1}(x)-f_{D_1}(x)| \right),\\ \quad f_{D_1\cap D_2}(x) = \min(f_{D_1}(x), f_{D_2}(x))= \frac{1}{2}\left(f_{D_1}(x)+f_{D_2}(x)-|f_{D_1}(x)-f_{D_1}(x)| \right),$$
and, both $f_{D_1\cup D_2}$ and $f_{D_1\cap D_2}$ are Riemann integrable.
Since, $f_{D_1\cup D_2}  = f_{D_1}+ f_{D_2} - f_{D_1\cap D_2}$, it follows that if $m(D_1\cap D_2) = 0$, then
$$\int_{D_1\cup D_2} f = \int_{D_1} f +\int_{D_2} f - \int_{D_1\cap D_2} f= \int_{D_1} f +\int_{D_2} f, $$
By induction this result generalizes to
$$\int_{\cup_j D_j} f = \sum_j \int_{D_j} f,$$
We can easily prove this for the general case where $f$ is not nonnegative or nonpositive, by applying  the previous result to the positive and negative parts, $f_+ = \max (f,0)$ and $f_- = \max (-f,0)$.
