Let $I=[0,1]\times[0,1]$ and $f:I \rightarrow \mathbb{R}$, $f(x) =\begin{cases} 1&\text{if } (x,y)=(\frac{1}{2},\frac{1}{2})\\ 0&\text{if }(x,y) \neq (\frac{1}{2},\frac{1}{2})\end{cases}$
Prove that f is Riemann integrable in $I$ and its integral is $0$.
What I want to do is to prove that for any $\epsilon \gt 0$ there exists a partition $P$ such that $U(f,P)-L(f,P) \lt \epsilon$. If $P$ is a partition of $I$ there exists only one $J \in P$ such that $(\frac{1}{2},\frac{1}{2}) \in P$. My idea is:
Let $\epsilon \gt 0$. If we take a partition $P=\{P_1, P_2\}$ of $[0,1]\times[0,1]$ such that P divides $I$ in intervals of length $\sqrt{\frac{\epsilon}{2}}$ then $sup \{ f(x): x \in J \} \cdot V(J)=1 \cdot \sqrt{\frac{\epsilon}{2}} \cdot \sqrt{\frac{\epsilon}{2}} = \frac{\epsilon}{2}$ and $inf \{ f(x): x \in J \}V(J)=0$. In any other $J' \in P$ we have that $(\frac{1}{2},\frac{1}{2})\notin J'$ so $sup \{ f(x): x \in J' \} \cdot V(J')=inf \{ f(x): x \in J' \} \cdot V(J')=0$ and then $U(f,P)-L(f,P) = \frac{\epsilon}{2} \lt \epsilon$.
The problem is, how can I explicitly give that partition? And how to be sure that it is indeed a partition of $I$?
Hint:
Pick $n \in \mathbb{N}$ such that $\frac1n < \sqrt{\frac\epsilon2}$. Define two partitions $0 = x_0 < x_1 < \ldots < x_n = 1$, $0 = y_0 < y_1 < \ldots < y_n = 1$ of $[0,1]$ as $$x_i = y_i = \frac{i}{n}, \quad\forall i = 0, \ldots, n$$
Now $[x_{i-1}, x_i] \times [y_{j-1}, y_j]$ for $i,j = 0, \ldots, n$ will be the desired partition of $[0,1]^2$.