Pondering about the independence of dimension of Monte Carlo Integration, I came up with the following explanation:
An integral over a square is not harder, thus has the same rate of convergence, than an integral over an interval, because computing an integral really just means "we have so and so many 2s, so and so many 1.5s, etc..." (Lebesgue!). In other words, there is no geometry/topology involved.
Of course this is also the reasoning that is done when considering integrals as expectations of random variables. However, I would like to continue my argument slightly differently as follows:
Theorem: There is a measurable bijection $\theta\colon (0,1)^2\to (0,1)$ (both with Lebesgue algebra) which pushes Lebesgue's measure forward to Lebesgue's measure.
If this Theorem was true, then $\int_{(0,1)^2}fdx=\int_{(0,1)}f\circ\theta^{-1}dx$, and we would see that integrating an arbitrary (i.e., no a-priori regularity knowledge to lose) function on $(0,1)^2$ is really just as hard as integrating an arbitrary function on $(0,1)$, no matter how we do it at the end.
I thought to remember that a lot of spaces actually fulfilled the statement of the Theorem and that those spaces had a name, but I can't remember.
If I understood your question correctly, you are talking about standard probability spaces. You may also be interested in checking out Borel isomorphism (that does not necessarily preserve measures though).