# Analysis + Fubini's Theorem

1. Mar 1, 2006

### Pearce_09

hello,
I am having trouble with this problem involving Fubini's Theorem. I have done a question already similar to this ( i will post it as well ), but this question is a bit different, which is causing the problem.

(question that i have completed)

Let f be the function on Rdefined by
$$f(x)=\left{\begin{array}{cc}1,&\mbox{ if } x is rational \\0, & \mbox{ if } x is irrational \end{array}\right$$
show that $$\int f(x)dx$$ does not exist for any a<b
(a and b are the endpoints for the integral)

sollution:
consider inf sums and sup sums.

Inf sum: I = $$\sum_\alpha (inf f) \Delta x$$

sup sum: S = $$\sum_\alpha (sup f) \Delta x$$

inf-sum $$\leq \int f \leq$$ sup-sum

therefor by inspection
inf f = 0 and sup f = 1
therefor
I = $$\int f_{inf} = \int 0dx = 0$$
S = $$\int f_{sup} = \int 1dx = 1$$

so S-I cannot be made < $$\epsilon$$
__________
( now the problem i cant seem to figure out )

Let f be the function on $$R^2$$ defined by

$$f(x)=\left{\begin{array}{cc}1,&\mbox{ if } x is rational/ and y =0, 1/2, or 1\\0, & \mbox otherwise\end{array}\right$$

and R the square R = {(x,y) $$are in R^2: 0 \leq x \leq 1, 0\leq y \leq 1}$$ determine if the integral exists.

show that $$\int_R f(x)dxdy$$
______
that is the question... now it is similar to the one i have done but y is involved.. now im confused.. because does it make that much of a difference.
i know i have to consider the inf and sup sums again. but what i dont know is the values of the inf and sup sums. Is it 0,1 again. any help is amazing and greatly appreciated.
(in this problem instead of integrating once i would do it twice.. now would that give me 2 inf and 2 sup sums.. or somthing different?)

also if there is any confusion ill try to clear it up??