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 5
 Problem Statement

Define ##f:R\subset\mathbb{R}^2\rightarrow\mathbb{R}## by
##f(x,y)=
\begin{cases}
2y & if \ x \ irrational \\
1 & if \ x\ rational
\end{cases}
##
Prove that ##f## is not integrable over the square ##R=[0,1]\times[0,1]##.
Hint: prove that for any partition ##P## of ##R##, ##L(f,P)<3/4## and ##U(f,P)>5/4##.
 Relevant Equations

##L=\sum_{i,j}fm_{ij}A_{ij}##
##U=\sum_{i,j}fM_{ij}A_{ij}##
I'm confused with how Riemann sums work on double integrals. I know that ##L=\sum_{i,j}fm_{ij}A_{ij}## and ##U=\sum_{i,j}fM_{ij}A_{ij}## where ##m_{ij}## is the greatest lower bound and ##M_{ij}## is the least uper bound and ##A_{ij}## is the area of each partition.
##A_{ij}=\frac{1}{n^2}## for every partition. If I get it right, ##M_{ij}## must be where ##x_i## and ##y_i## are the maximum value on the square, so it must be at the the right top corner, while ##m_{ij}## must be at the bottom left corner. So ##M_{ij}=(\frac{i}{n},\frac{j}{n})## and ##m_{ij}=(\frac{i1}{n},\frac{j1}{n})##, right?
What should I do next?
##A_{ij}=\frac{1}{n^2}## for every partition. If I get it right, ##M_{ij}## must be where ##x_i## and ##y_i## are the maximum value on the square, so it must be at the the right top corner, while ##m_{ij}## must be at the bottom left corner. So ##M_{ij}=(\frac{i}{n},\frac{j}{n})## and ##m_{ij}=(\frac{i1}{n},\frac{j1}{n})##, right?
What should I do next?
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