Why does the first iterated integral exist but the second one does not?

[kingwinner]

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 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!

 

[rudinreader]

The dirichlet function d(y) = 1 y in Q, d(y) = 0 y not in Q is not Riemann integrable.
So for fixed x = 0, f(x,y) = d(y) is not integrable with respect to y.

But $$\int_0^1 f(x,y) dx = 0$$, (regardless if f(0,y) = 1 or 0) so you can integrate the first expression.

 

[kingwinner]

I haven't learned about the dirichlet function, so I don't quite get your points. Are there any other ways of explaining? Anyone want to give it another try?

Thanks!

 

[EnumaElish]

Please do not cross post.

 

[kingwinner]

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, am I 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 am not too sure about this, could someone please give an affirmative answer?