1) Let f(x,y)=1 for x=0, y E Q
f(x,y)=0 otherwise
on R=[0,1] x [0,1]

Then
1 1
∫ ∫ f(x,y) dxdy = 0 exists
0 0
but
1 1
∫ ∫ f(x,y) dydx does not exist
0 0


I don't understand why the first iterated Riemann integral exists, but the second interated integral does not exist, can someone please explain (perhaps in terms of the concept of zero content) ?

Thank you!

Theorem: If f is bounded on [a,b] and the set of pionts in [a,b] at which f is discontinuous has zero content, then f is integrable on [a,b].


I don't understand why in our case
1
∫ f(x,y) dx = 0
0
What is the "set of pionts in [a,b] at which f is discontinuous" as referred to in the first theorem? And what is [a,b] in this case?

That is the set consisting of only the point x=0 when y\in\mathbb{Q}, since f(0,y)=1 and f(\varepsilon,y)=0 for 0<\varepsilon\leq1. And [a,b] is [0,1].

For each fixed y, f(x,y) is zero except at a single point where it is (possibly) 1, so the integral
1
∫ f(x,y) dx should be zero, I think.
0


But there are "infinitely many" fixed y's in the interval [0,1], then how can
1
∫ f(x,y) dx = 0 ?
0

Your first paragraph answers the question already. The meaning of the expression \int_0^1f(x,y)dx=0 is that it does not matter how many "fixed y's" are there, everytime you consider the integral there is only one "fixed y", and it is true for whatever y you choose to be "fixed".

Your first paragraph answers the question already. The meaning of the expression \int_0^1f(x,y)dx=0 is that it does not matter how many "fixed y's" are there, everytime you consider the integral there is only one "fixed y", and it is true for whatever y you choose to be "fixed".
Oh, I see!

But how come, on the other hand, that
1 1
∫ ∫ f(x,y) dydx does not exist ?
0 0

You have to take the integral over y first for fixed x. Those integrals are all zero for x>0. Now, the problem is for x=0, namely, the integral

\int_0^1f(0,y)dy.

And this integral does not exist in Riemannian sense.

You have to take the integral over y first for fixed x. Those integrals are all zero for x>0. Now, the problem is for x=0, namely, the integral

\int_0^1f(0,y)dy

And this integral does not exist in Riemannian sense.
So if \int_0^1f(0,y)dy, which is one particular one in the interval x E [0,1], doesn't exist, then the whole thing \int_0^1f(x,y)dy doesn't exist, am I right?

Also, why does \int_0^1f(0,y)dy not exist?

Thanks for your help!

Correct. The outer integration is possible if you get a well defined function

g(x)=\int_0^1f(x,y)dy

but g(x) is not defined when x=0.

Look here for the non-existence: http://en.wikipedia.org/wiki/Riemann_integral#Examples

Correct. The outer integration is possible if you get a well defined function

g(x)=\int_0^1f(x,y)dy

but g(x) is not defined when x=0.

Look here for the non-existence: http://en.wikipedia.org/wiki/Riemann_integral#Examples
But x=0 is just one point, so shouldn't the outer integration still be possible?

An integral exist if there are finite number of discontinuities, and in this case, x=0 is the only point...

That is not a discontinuity, it is "undefinedness". There is no value of the function at that point.

1
∫ f(x,y) dy = F(x) exists for x not =0
0

1
∫ f(0,y) dy = F(0) does not exist
0

Theorem: If f is bounded on [a,b] and continuous at all except finitely many points in [a,b], then f is (Riemann) integrable on [a,b].

In our case, x=0 is just a single point (finitely many points!!)
So actually
1 1
∫ ∫ f(x,y) dydx
0 0
1
=∫ F(x) dx exists as well and equal zero, right?
0

If so, then my claim in the first post that
1 1
∫ ∫ f(x,y) dydx does not exist must be wrong
0 0

I was saying that what you have at x=0 is not discontinuity, but the function is not defined at x=0. So there is no way to integrate in Riemannian sense. Of course, Lebesgue integral does not care what happens at discrete points. Or you can integrate after "manually" defining the function value at x=0. But then strictly speaking you would be integrating a different function.

According to the Lebesgue criterion for Riemann integrals, f is Riemann integrable over an interval if and only if it is continuous almost everywhere and bounded. If f is Riemann integrable, its values on a set of measure zero may be altered without changing either the integrability of f or the value of the integral.

nice