Product Measures
Build a measure on a product, then swap the order of integration — and know when you may not.
The idea
The product measure extends two measures to the product of their spaces, so that a function of two variables can be integrated one variable at a time: fix $x$, integrate in $y$, then integrate the results in $x$.
Let $(X, \mathcal{M}, \mu)$ and $(Y, \mathcal{N}, \nu)$ be measure spaces. The product $\sigma$-algebra $\mathcal{M} \otimes \mathcal{N}$ is the $\sigma$-algebra on $X \times Y$ generated by the measurable rectangles $A \times B$ with $A \in \mathcal{M}$ and $B \in \mathcal{N}$, and the product measure $\mu \times \nu$ is the measure on it satisfying $(\mu \times \nu)(A \times B) = \mu(A)\,\nu(B).$ It is the only such measure provided $\mu$ and $\nu$ are $\sigma$-finite, meaning each space is a countable union of sets of finite measure.
Two theorems justify integrating slice by slice, and they differ only in what they require of $f$.
Theorem (Tonelli's theorem).
If $\mu$ and $\nu$ are $\sigma$-finite and $f \colon X \times Y \to [0, \infty]$ is measurable, then $\iint f \, d(\mu \times \nu) = \int_{X} \left( \int_{Y} f(x, y) \, d\nu(y) \right) d\mu(x) = \int_{Y} \left( \int_{X} f(x, y) \, d\mu(x) \right) d\nu(y),$ including the case where all three are $+\infty$.
Theorem (Fubini's theorem).
If $\mu$ and $\nu$ are $\sigma$-finite and $f$ is integrable, meaning $\iint |f| \, d(\mu \times \nu) < \infty$, then the same three integrals are equal and finite.
Neither hypothesis can be dropped, so the standard order of work is Tonelli applied to $|f|$ first, and then, if the result is finite, Fubini applied to $f$.
Ways to work on it
- Walkthrough. Build a measure on a product space and learn when a double integral splits into iterated ones.
- Practice. Check Tonelli's hypothesis, then compute a double integral one variable at a time.
- Hardest. The classic counterexample: two iterated integrals of the same function that disagree.
Not sure where to start? Take the ten-question placement test.