Definition of double integral for an specific case Well, this problem was in a test of mine and I couldn't do it. I tried to do using the definition based in supremum and infimum theory. I would like to could I solve this one in both ways, or the simplest one to understand. It is the following statement:
Show, using the definition of double integral, that the function
$$ f : (x, y) \in [-1, 1] \times [-1, 1] \mapsto
\begin{cases}
1; (x, y) \in \{(0, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)\} \\
0; (x, y) \not\in \{(0, 0), (-1, -1), (-1, 1), (1, -1), (1, 1)\}
\end{cases}
$$
is integrable. Besides, evaluate the integral of $f$ in $[-1,1]\times[-1,1]$.
 A: I think this problem is easier than it seems. We have to remember that the integral is just  a sum and the double integral is just a double sum. Also, when defining the integral we observed that the length (similarly area) for the partitions are not fixed.
Hence, we get the same definition if $\Delta x = (b-a)/n$ and $\Delta y = (c-d)/m$ where $n,m$ denote the number of intervals in the partitions of the $x,y$-axis respectively. Hence it follows immediately that,
\begin{align*} \iint_{[-1,1]^2} f \ dA =  \lim_{n,m \to \infty} \sum_{i,j}^{m,n}f(p_{ij}) \ \Delta x \Delta y &= \lim_{n,m \to \infty} \sum_{\textrm{$i,j$ s.t $f \not = 0$}} f(p_{ij}) \Delta x \Delta y  \\ \\ & \leq \lim_{n,m \to \infty} \Delta x \Delta y = 0\end{align*}
A: First we will prove the following easier question
Let $f:[0,1]\rightarrow\mathbb{R}$ be given by $f(1)=0$ and $f(x)=0$ otherwise. Let me prove that $\int f dx =0$
for every partition $P=\{0,x_1,...,x_n=1\}$ of $[0,1]$ we have that $$\sum_{i=1}^n \min_{x\in [x_{i-1},x_i]} f(x) |x_i-x_{i-1}| = 0$$
It follows that the lower integral is always zero.
Now it is left to show that so is the upper integral, that is, we need to show that 
$$\inf_{P=\{x_1,...,x_n\}} \sum_{i=1}^n \max_{x\in[x_{i-1},x_i]}f(x)|x_i-x_{i-1}|=0$$
by the definition of $f$ we have that $\max_{x\in[x_{i-1},x_i]}f(x)=0$ so we only have to deal with the last interval.
Now I will give a sketch, first note that for every partition $P$ of $[0,1]$ you have that $$\sum_{i=1}^n \max_{x\in[x_{i-1},x_i]}f(x)|x_i-x_{i-1}|$$ is non-negative, therefore it is enough to finte a sequence of partitions $P_n$ such that the corresponding sum convergence to $0$ (why?). 
Here is how to construct these partitions:
set $P_n = \{0,1-1/n,1\}$ you have that the sum corresponds to $P_n$ is
$$\max_{x\in[0,1-1/n]} f(x) (1-1/n) + \max_{x\in[1-1/n,1]} f(x) (1/n)=0\cdot(1-1/n)+1\cdot 1/n=1/n\rightarrow 0$$
Thus we show that $f$ is integrable with integral zero.
How do we generalize this to the 2 dimensional case? so the Lower integral is again always zero (because the minimum on every rectangle is $0$) so again we only have to show that the upper integral is zero, how we do that? again you only have to find a partition (now a partition is a set of rectangles) such that the sums corresponding to the partitions convergence to zero. How we choose the partition? we take $P_n$ to be a set of rectangles with edge of lenth $1/n$ containing the "bad points" which in your case are $\{(0,0),(-1,-1),(-1,1),(1,-1),(1,1)\}$ and we don't care about the other rectangles because the maximum is anyway zero.
