Archimedean Sequence of Partitions of a Square

Suppose $f: \vect{I} \to \R^2$, where $\vect{I} = [0,1] \times [0,1]$,

$f(x,y) = \begin{cases} 0 &\text{if } (x,y) \in \vect{I} \land y \ge x\\ 1 &\text{if } (x,y) \in \vect{I} \land y < x \end{cases}$

$P_k := \{0,1/k,\dots,1\}, \vect{P}_k := (P_k,P_k)$ .

It’s said that $\mathcal{U}(f,\vect{P}_k) - \mathcal{L}(f,\vect{P}_k) < 2/k$ .

However, at first, I don’t understand why there are fewer than $2k$ squares which contribute to the difference between $\mathcal{U}(f,\vect{P}_k)$ and $\mathcal{L}(f,\vect{P}_k)$. I wrongly think that only the $k$ diagonal squares make such a contribution. Actually, $\forall\,i = 1,\dots,k - 1, [i / k,(i + 1) / k] \times [(i - 1) / k,i / k]$ also causes the difference between those two Darboux sums because

\begin{aligned} \inf\,\{f(\vect{x}) \mid \vect{x} \in [i / k,(i + 1) / k] \times [(i - 1) / k,i / k]\} &= f(i / k,i / k) = 0 \text{ and}\\ \sup\,\{f(\vect{x}) \mid \vect{x} \in [i / k,(i + 1) / k] \times [(i - 1) / k,i / k]\} &= 1. \end{aligned}