Let $[0,1]^2 = [0,1] \times [0,1]$. Let $f: [0,1]^2\rightarrow$ $\mathbb R$ be defined by setting $f(x,y) = 0$ if $x \not= y$ and $f(x,y) = 1$ if $x = y$. Show that $f$ is integrable over $[0,1]^2$
Attempt at a solution :
Let the partition $P = \{0,\dfrac1n,\dfrac2n,\ldots,1\}\times\{0,\dfrac1n,\dfrac2n,\ldots,1\}$
Then $L(f,P) = 0$ and $$U(f,P) = \sum_{i=0}^{n} M_R f * V(R) = 1 * \sum_{i=0}^n \left[\frac{i}{n} - \frac{i-1}{n}\right]^2 = \frac{n}{n^2} = \frac{1}{n}$$
So $U(f,P) \rightarrow 0$ as $n \rightarrow\infty$
setting $\epsilon = \frac{1}{n}$ we will get $U(f,P)-L(f,P) \le\epsilon$ And hence, $f$ is integrable.
I'm really bad with proofs, I'm trying to use my winter break to sort of catch up and get better, any feedback would be appreciated, thank you.
Your proof works fine. Here's a bit more detail and a slightly different angle.
Partition $[0,1] \times [0,1]$ into squares of side length $1/n$ in the obvious way. Let $S$ be one such square. Then $\sup_{x \in S} f(x) = 1$ if the square intersects the diagonal and $0$ otherwise. The diagonal is covered by the squares with bottom left vertex at $(j/n,j/n)$ for $j=0,\ldots,n-1$, and so there are $n$ squares intersecting the diagonal, each with area $1/n^2$. Then the upper sum is $n \cdot 1/n^2 = 1/n$. Since $n$ was arbitrary we have $\inf U(f,P) \leq \inf_{n \in \mathbb{N}} 1/n = 0$, where the $\inf$ is over all partitions.
Similarly we see $\sup L(f,P) \geq 0$ over all partitions. This shows that the integral exists and is $0$.