Consider: $\int \int_A 1 \ dx dy, \quad$ where $A = [3,3]\times\mathbb{R}^2$.
I know this is probably an extremely basic question, but I've forgotten everything from my multivariable calculus class many years ago, and I'm having trouble with some conceptual questions regarding the above problem.
First question: How does one evaluate this integral? Intuitively, I feel like it should be equal to 0 since you're basically taking the area of a line. And indeed:
$\int^\infty_{-\infty} \int^3_3 1 \ dxdy = \int^\infty_{-\infty} 0 dy = 0$.
But when I switched the order of integration, I obtained:
$\int^3_3 \int^\infty_{-\infty} 1 \ dy dx, \quad$ which does not converge.
Here I thought that this can't be right, and I know that in the one variable case, an infinite integral can be expressed as a limit. So I wrote the above again as:
$\lim_{a \rightarrow \infty} \lim_{b \rightarrow -\infty}\int^a_{b} \int^3_3 1 \ dxdy = \lim_{a \rightarrow \infty}\lim_{b \rightarrow -\infty} \int^3_3 \int^a_{b} 1 \ dydx = 0$.
But I'm not entirely convinced whether I can use this one-variable concept of limits in the multi-variable case, and furthermore that I can pull this limit out of the integral. What is the best and most rigorous way to evaluate this (or show that it does not exist)?
Second question: In a more general sense, I understand Fubini's theorem to imply that if
$\int \int_A f(x,y) \ dx dy, \quad$ exists, then assuming that you have the correct bounds for integration,
$\int \int f(x,y) \ dx \ dy = \int \int f(x,y) \ dy \ dx$
If my understanding of Fubini's theorem is true (and if not please correct me), does the converse hold? i.e. if:
$\int \int f(x,y) \ dx \ dy = \int \int f(x,y) \ dy \ dx$
then does
$\int \int_A f(x,y) \ dx dy, \quad$ exists?
I hope those questions made sense, and thank you in advance!
Note: I haven't yet learned Lebesgue integration, nor any measure theory, so if it possible to answer the question using only Riemann integrals, and no measure theory, that would be great!