Use the Integrability Criterion to show that the function $f: I \to \Bbb R$ is integrable. Question:
For the generalized rectangle $I= [0,1]\times [0,1]$ in the plane $\Bbb R^2$
$$f(x,y)=\begin{cases} 5 & if\ \  (x,y)\ is\ in\ I\ and\ x> 1/2 \\ 1 & if\ (x,y)\ is\ in\ I\ \ and\ x\le 1/2 \end{cases}$$
Use the $\cal{Integrability\ Criterion}$ to show that the function $f: I \to \Bbb R$ is integrable. 

$\cal{Integrability\ Criterion:}$ 
Bounded function $f: I \to \Bbb R$ is integrable $\iff$ for each $\epsilon >0$ there is a partition $P$ of $I$ such that $$U(f, P)-L(f,P)<\epsilon$$

Solution trial:
Let $\epsilon >0$ 
For $k\in \Bbb N$, let $P_k$ be the partition of the interval $[0,1]$ of equal length $1/k$. 

After then, how should I solve the question? 
 A: Let $P_k$ be the partition of $[0,1]\times[0,1]$ consisting of the following three parts: $\left[0,\dfrac12-\dfrac1k\right]\times\left[0,1\right],\left[\dfrac12-\dfrac1k,\dfrac12+\dfrac1k\right]\times\left[0,1\right],\left[\dfrac12+\dfrac1k,1\right]\times\left[0,1\right]$.
Then $U(f,P_k)=\left(\dfrac{1}{2}-\dfrac1k\right)\cdot1+\left(\dfrac{2}{k}\right)\cdot5+\left(\dfrac{1}{2}-\dfrac1k\right)\cdot5$ and 
$L(f,P_k)=\left(\dfrac{1}{2}-\dfrac1k\right)\cdot1+\left(\dfrac{2}{k}\right)\cdot1+\left(\dfrac{1}{2}-\dfrac1k\right)\cdot5$.

Therefore $U(f,P_k)-L(f,P_k)=\dfrac2k\cdot4=\dfrac8k\xrightarrow[k\to\infty]{}0$ and $f$ is integrable.
A: I would use the partition $P_\epsilon$ with only three parts:
$$
[0,\frac12-\frac\epsilon{20}]\times[0,1],\qquad [\frac12-\frac\epsilon{20},\frac12+\frac\epsilon{20}]\times[0,1]\quad\text{and}\quad [\frac12+\frac\epsilon{20},1]\times[0,1].
$$
In other words along the $x$-axes use split points at $1/2\pm\epsilon/20$, and along the $y$-direction you don't need any split points at all as the function does not depend on $y$.
Calculate the upper and lower sums related to these partitions, and see what comes out.
