Do regions of integration cover *all* possible regions? Now, I'm not real concerned about the integration itself but more about integration regions.
Is it true that all possible regions in two-space are covered by the following setup:
$$\int^a_b \int^{g_1(x)}_{g_2(x)} f(x,y) dydx$$
Any function for g is allowed. Note that I'm only concerned about the bounds, not f itself. If you could say which functions other than the standard elementary calculus set is needed please specify.
My main concern is regions that are not connected. I think any connected single shape is representable this way but about other shapes? For instance two squares not touching?
I know this question is pretty broad but it's more about what integral bounds cover and don't cover, and what is in there means. And at what point you MUST use two integrals.
 A: The short answer is no, there are regions which don't lend themself to this sort of single-integral decomposition. But there is more to say, as usual.
A simple example is the complement of an annulus; that is, the white region(s) in the figure below:
 
You can still put it into a single integral if you make some weird change of coordinates. However, if you're willing to make that dramatic of a change, you can achieve the same effect with less work: simply write any integral of $f$ over some set $D$ as 
$$ \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\Bbb 1_D(x,y) dxdy.$$
where $\Bbb 1_D$ is the indicator function; outputs 1 if $(x,y)\in D$ and 0 otherwise.
As a final caveat, it's worth noting that not every subset $D\subseteq\Bbb R^2$ has a coherent theory of (Lebesgue) integration attached to it; the ones that do are called (Lebesgue) measurable sets. These have quite a technical definition, and pretty much any set that you can think of is measurable. But nonmeasurable sets do exist (assuming the Axiom of Choice) and it's worth keeping them in mind.
