Proving monotone function of two variables is integrable 
Let $f:[0,1]^2\rightarrow \mathbb R$ be a monotone function of two variables, that is, $x\leq x'$ and $y\leq y' \implies f(x,y)\leq f(x',y').$ Prove that $f$ is Riemann integrable.

I want to "copy" and generalize the argument for the one dimensional case. Well, what I tried so far was to consider the partition $P=\{P_{ij}: i,j = 1,\cdots, N\}, N\in \mathbb N, $ given by $P_{ij} = (\frac{i-1}{N},\frac{i}{N})\times (\frac{j-1}{N},\frac{j}{N})$. This is a partition of the square by small squares of area $1/N^2$. As in the one dimensional case, I want the sum $R(f,P)-L(f,P)$ to telescope and be something like: $\frac{f(1,1)-f(0,0)}{N}$ where $f(1,1)$ and $f(0,0)$, as we may notice, is the maximum and minimum of the function on the square. But, it doesn't seem that this sum will be telescoping, because for each small square, its maximum and minimum of the function is reached at the opposed diagonal vertices (on the right) and will always remain diagonal vertices which will not "kill each other". 
Is this the right approach? Any hint on how to prove this?
 A: The key idea is that you have $N^2$ squares each of area $\dfrac{1}{N^2}$, but after telescoping the sum "as much as possible", there are only $2N-1$ summands left, each summand being of the form $f(p) - f(q)$; which is bounded by $f(1,1) - f(0,0)$. Let $S$ denote an arbitrary subrectangle determined by the partition $P$ you have constructed. Then,
\begin{align}
U(f,P) - L(f,P) &= \sum_{S \in P} \left( M_S(f) - m_S(f) \right) \cdot \text{area}(S) \\
&=  \dfrac{1}{N^2} \sum_{S \in P} \left( M_S(f) - m_S(f) \right) \\
&=  \dfrac{1}{N^2}  \left( \text{sum of $(2N-1)$ terms of the form $f(p) - f(q)$} \right) \\
& \leq \dfrac{1}{N^2} \sum_{i=1}^{2N-1} f(1,1) - f(0,0) \\
&= \dfrac{2N-1}{N^2} \cdot \left( f(1,1) - f(0,0) \right)
\end{align}
As $N \to \infty$, the RHS $\to 0$; hence you're done.
Now, I know I didn't index my terms properly etc, because I think this is one of those problems where a picture makes it a hundred times clearer. Consider the figure below:



Here, I chose $N=5$, so there are $N^2 = 25$ rectangles. The purple diagonal lines indicate how the telescoping works. The red dots are to be taken with a $+$ sign, while the blue dots are to be taken with a $-$ sign. The $2N-1$ I got above is because there are $2N-1 = 9$ red dots. So for this partition, just to make things explicit, we have
\begin{align}
25 \cdot \left( U(f,P) - L(f,P) \right) &= 
\left[ f \left( \dfrac{1}{5},1  \right)  - f \left(0, \dfrac{4}{5} \right)\right] +
\left[ f \left( \dfrac{2}{5},1  \right)  - f \left(0, \dfrac{3}{5} \right)\right] \\
&+ \left[ f \left( \dfrac{3}{5},1  \right)  - f \left(0, \dfrac{2}{5} \right)\right]
+ \left[ f \left( \dfrac{4}{5},1  \right)  - f \left(0, \dfrac{1}{5} \right)\right] \\\\
&+ \left[ f \left( 1,1  \right)  - f \left(0, 0 \right)\right] \\\\ 
&+ \left[ f \left( 1, \dfrac{4}{5}  \right)  - f \left(\dfrac{1}{5}, 0 \right)\right] 
+ \left[ f \left( 1, \dfrac{3}{5}  \right)  - f \left(\dfrac{2}{5}, 0 \right)\right] \\
&+ \left[ f \left( 1, \dfrac{2}{5}  \right)  - f \left(\dfrac{3}{5}, 0 \right)\right] 
+ \left[ f \left( 1, \dfrac{1}{5}  \right)  - f \left(\dfrac{4}{5}, 0 \right)\right] \\\\
& \leq 9 \cdot \left[ f \left( 1,1  \right)  - f \left(0, 0 \right)\right]
\end{align}
Hence,
\begin{align}
U(f,P) - L(f,P) &\leq \dfrac{9}{25} \cdot \left[ f \left( 1,1  \right)  - f \left(0, 0 \right)\right] \\\\
&= \dfrac{2(5) -1}{5^2} \left[ f \left( 1,1  \right)  - f \left(0, 0 \right)\right]
\end{align}

I think it's a notational nightmare to precisely explain where the $2N-1$ comes from, but pictorially it's a trivial counting exercise. So, as a recap, the crux of the proof was that the number of summands grows like $N$, while the area grows like $N^2$, so that their ratio  goes to $0$ for large $N$. I'm sure this proof can be extended to $\mathbb{R^d}$ as well, where now, the number of summands grows like $N^{d-1}$, while the volume of each cube grows like $N^d$.
