Multidimensional Integration 4 | Fubini's Theorem in Action [dark version]

Fubini’s theorem becomes usable even when the integration region isn’t a rectangle—by extending the function to a larger Cartesian product and setting it to zero outside the original domain. That move preserves the original integral while allowing the order of integration to be swapped, which can make calculations dramatically simpler.

The setup starts with a measurable function f on a Cartesian product A×B in R^{n+m}. Under standard conditions—either f is nonnegative or f is integrable in the absolute-value sense—the n+m dimensional integral can be computed as an iterated integral over A and B. Crucially, the theorem guarantees that the order of these iterated integrals does not matter.

The practical problem is that many real regions aren’t rectangles. A function might be defined on a curved set U in R^2 rather than on a product of intervals. In that case, Fubini’s theorem can’t be applied directly because the domain isn’t of the form A×B. The workaround is to embed U into a rectangle A×B, then define an extended function f~ on the rectangle: f~(x,y)=f(x,y) when (x,y) lies in U, and f~(x,y)=0 everywhere else. Since f~ is zero outside U, integrating f~ over the rectangle gives exactly the same value as integrating f over U. Once this extension is in place, Fubini’s theorem applies, and the iterated integrals can be chosen in whichever order makes the “zero regions” easiest to handle.

An example in R^2 makes the method concrete. The region U is bounded by y=3−x^2, with x ranging from 0 to 1 and y running up to the parabola; the region is not a Cartesian product because the upper boundary depends on x. The function being integrated is constant and equal to 1 on U, so the integral represents the area of U (interpretable as the volume under the surface z=1 above the (x,y)-plane).

After extending the constant function to the surrounding rectangle by setting it to zero outside U, the iterated integral can be written with y as the inner variable: for each fixed x, the integrand is 1 only when y lies between 1 and 3−x^2, and it is effectively zero otherwise. That observation collapses the inner integral to “upper limit minus lower limit,” yielding 3−x^2−1=2−x^2. The remaining outer integral over x from 0 to 1 is then straightforward, producing the final area value 5/3.

The takeaway is operational: when the region is curved, extend the function by zero to a rectangle, then use Fubini to reduce a multi-dimensional integral to a sequence of one-dimensional integrals—often with the best order chosen to exploit where the integrand vanishes.