- #1

Davidllerenav

- 424

- 14

- Homework 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{i-1}{n},\frac{j-1}{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{i-1}{n},\frac{j-1}{n})##, right?

What should I do next?

Last edited: