How Does Fubini's Theorem Apply to Integrals of Functions on R^2?

[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 can't 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 I am 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 don't 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??
THANK YOU FOR YOUR TIME
Adam

 

[lippyka]

SOLUTION:The integral does exist. To calculate it, we need to use Fubini's Theorem. We can express the integral as a double integral over the square R:\int_R f(x,y)dxdy = \int_0^1\int_0^1f(x,y)dxdyNow we need to calculate the inf and sup sums of f(x,y).For this, we need to consider each variable separately.For x, we have inf f(x,y) = 0 and sup f(x,y) = 1, since x is either rational or irrational.For y, we have inf f(x,y) = 0 and sup f(x,y) = 1, since y can only take on the values 0, 1/2, or 1.Therefore, the inf sum is given by:I = \int_0^1\int_0^1f_{inf}(x,y)dxdy = 0And the sup sum is given by:S = \int_0^1\int_0^1f_{sup}(x,y)dxdy = 1Therefore, by inspection, we can see that the integral exists, since S-I = 1-0 = 1 < \epsilon, where \epsilon > 0.