Show that the double integral $\iint_R f(x,y) dx dy $ does not exist

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A function is defined on the rectangle R = [0,1;0,1] as follows:

$f(x)=\begin{cases} \frac{1}{2}, & \texttt{ when y is rational }\\ x, &\texttt{ when y is irrational} \end{cases}$

Show that the double integral $\iint_R f(x,y) dx dy $ does not exist:

This is a solved problem in my textbook.

Rectangular partition is made of size 1x1 with each subrectangle size $\frac{1}{2n}X\frac{1}{2n}$

The rectangle is divided into half as shown in figure. I did not understand how did he calculate infimum and supremum values of f in each subrectangle(as shown in box in 2nd image ).

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Every rectangle $R_{ij} =[x_{i-1},x_i] \times [y_{j-1},y_j]$ contains points where $y$ is irrational $(f(x,y) = x)$ and points where $y$ is rational $(f(x,y) = 1/2)$.

If $(x,y) \in R_{ij}$ we have $x \leqslant 1/2$ for $i \leqslant n$, and $x \geqslant 1/2$ for $i > n$.

Hence,

$$\sup_{R_{ij}} \, f(x,y) = \begin{cases}\max(1/2,x_i) = 1/2, & i \leqslant n \\ \max(1/2,x_i) = x_i, & i > n \end{cases}$$

$$\inf_{R_{ij}} \, f(x,y) = \begin{cases}\min(1/2,x_{i-1}) = x_{i-1}, & i \leqslant n \\ \min(1/2,x_{i-1}) = 1/2, & i > n \end{cases}$$

There appears to be a typographical error in the lower sum printed in the book.