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Using Riemann integration, you can find a multiple integral by switching the order of integration. Using Lebesgue integration, Fubini's theorem allows one to do a similar thing.

Are the arguments for Fubini's theorem (e.g switching the order of measures) analogous to those for Riemann integration? Or put another way, does Fubini's theorem extend the Riemann logic by arguing that under Lebesgue integration, discontinuities or other irregularities under Riemann integration have measure zero?

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In Lebesgue's theory, there is no interest in discontinuities. Fubini's theorem is proved by pure measure-theoretic ideas. Here you can read a proof: tau.ac.il/~tsirel/Courses/Analysis3/lect9.pdf –  Siminore Sep 14 '12 at 13:35
@Siminore: Your link treats the Riemann version (which I don't think you intended). –  t.b. Sep 14 '12 at 13:43
@t.b. Sorry, here is a better link: math.bu.edu/people/prakashb/Math/… –  Siminore Sep 14 '12 at 14:19

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In this context one should distinguish between a multiple integral and an iterated integral. In an iterated integral, there's such a thing as order of integration. One integrates to find the function $g(y) = \int_a^b f(x,y)\,dx$, and then one integrates to find $\int_c^d g(y)\,dy$. That's iterated integration. Multiple integration, on the other hand, is defined in terms of Lebesgue measure on higher-dimensional spaces rather than by iteration of integrals defined on $1$-dimensional space. Fubini's theorem says the multiple integral is equal to the iterated integral if the multiple integral of the absolute value of the function being integrated is finite. If the latter is not finite, it sometimes happens that both the positive part of the function (i.e. $f_+ = f(x)$ if $f(x)\ge 0$ and $f_+(x)=0$ if $f(x)<0$) and the negative part diverge to infinity. In that case, the multiple integral does not exist, but sometimes the iterated integrals still exist, and sometimes integrating in different orders gives different numbers as the value of the iterated integral. Thus $$ \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dx\,dy \ne \int_0^1\int_0^1 \frac{x^2-y^2}{(x^2+y^2)^2}\,dy\,dx. $$ (Just first-year calculus; so work it out.)

It's not true that "under Lebesgue integration, discontinuities or other irregularities under Riemann integration have measure zero". On the contrary, that describes Riemann integration: a function of a real variable is Riemann-integrable precisely if the set of points in its domain where it is discontinuous has measure $0$. Some functions that are nowhere continuous are Lebesgue integrable.

Nor is the reasoning by which results are proved the same. One of the advantages of Lebesgue's theory is that it's easier to prove convergence theorems ("dominated" and "monotone") because the methods of proof are different. The desire for such theorems came originally from the study of trigonometric series, including Fourier series.

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