How can I show that double integral exists? For $Q= [0,1] \times [0,1]$
$$f(x,y) = \begin{cases} 1, \text{ if $x = y$} \\ 0 ,\text{ if $x \neq y$} \end{cases}$$
prove that double integral exists and equals to $ 0$
So I tried to prove this by showing that function is bounded 
$0<f<1$ so there is sup and inf.
If I can show sup and inf are the same, then I can say double integral exists.
But how can I show that they are the same? 
 A: Let $1>\delta>0$ and define a sub-region $Q_{\delta}$ as the region in $Q$ for which $\max(0,x-\delta)<y<\min(x+\delta,1)$.  Notice that 
$$0\le\int \int_{Q_{\delta}} f(x,y)dxdy\le1 \int \int_{Q_{\delta}} dxdy$$ 
$$=1\left(1-\left(1-\delta^2\right)\right)$$
$$=\delta (2-\delta)$$
$$<2\delta$$
Thus, the integral can be made smaller than any pre-chosen $\epsilon>0$ by choosing $\delta<\frac12 \epsilon$.  This should lead immediately to the conclusion that the double integral is zero in the sense of Lebesgue.
If this is a Riemann integral, one partitions the rectangle in a set of sub-rectangles.  Note that for any such partition, one can show that the area of the set of rectangles along the diagonal (on which $f=1$) goes to zero as the norm of the net of this partition goes to zero.  In fact, these rectangles along the diagonal can be contained in a region $Q_{\delta}$ as before.
A: Consider partitioning the unit square $[0,1] \times [0,1]$ into a grid of squares of size $1/n \times 1/n$. Call this partition $\mathcal{P}_n$. Let's calculate the upper sum $U(\mathcal{P}_n,f)$ and the lower sum $L(\mathcal{P}_n,f)$. If we can show that:
$$\lim_{n \rightarrow \infty} U(\mathcal{P}_n,f) = \lim_{n \rightarrow \infty} L(\mathcal{P}_n,f) = 0$$
Then the function is integrable and its integral is zero.
First, $L(\mathcal{P}_n,f) = 0$, since the function $f$ takes the value $0$ at least once on every square.
Second, $U(\mathcal{P}_n,f) = \frac{k}{n^2}$, where $k$ is the number of squares on which $f(x,y) = 1$ for some point in that square. This only happens for squares that intersect the diagonal. Those are the squares on the diagonal (there are $n$ of these), the squares directly above the diagonal ($n-1$ of these) and directly below (also $n-1$). So $k = n + n-1 + n-1 = 3n-2$. And so:
$$\lim_{n \rightarrow \infty} U(\mathcal{P}_n,f) = \lim_{n \rightarrow \infty} \frac{3n - 2}{n^2} = 0$$
