Product Measures and Fubini - Main Theorem

TheoremFubini's Theorem

Let $(X, \mathcal{F}, \mu)$ and $(Y, \mathcal{G}, \nu)$ be $\sigma$ -finite measure spaces. If $f \in L^1(X \times Y, \mu \otimes \nu)$ (i.e., $f$ is integrable with respect to the product measure), then:

For almost every $x \in X$ , the function $y \mapsto f(x, y)$ is in $L^1(Y, \nu)$
For almost every $y \in Y$ , the function $x \mapsto f(x, y)$ is in $L^1(X, \mu)$
The functions $x \mapsto \int_Y f(x, y) \, d\nu(y)$ and $y \mapsto \int_X f(x, y) \, d\mu(x)$ are integrable
$\int_{X \times Y} f \, d(\mu \otimes \nu) = \int_X \left(\int_Y f(x, y) \, d\nu(y)\right) d\mu(x) = \int_Y \left(\int_X f(x, y) \, d\mu(x)\right) d\nu(y)$

Fubini's Theorem is the fundamental result for evaluating multiple integrals via iterated integration. Unlike Tonelli, it requires integrability but applies to general (not necessarily non-negative) functions.

ExampleComputing Areas and Volumes

To find the volume under $f(x, y) = x^2 y$ over $[0, 1] \times [0, 1]$ :

Since $|f(x, y)| \leq 1$ on the square, $f \in L^1([0, 1]^2)$ . By Fubini: $\int_0^1 \int_0^1 x^2 y \, dy \, dx = \int_0^1 x^2 \left(\int_0^1 y \, dy\right) dx = \int_0^1 x^2 \cdot \frac{1}{2} \, dx = \frac{1}{6}$

We could also integrate in the opposite order: $\int_0^1 \int_0^1 x^2 y \, dx \, dy = \int_0^1 y \left(\int_0^1 x^2 \, dx\right) dy = \int_0^1 y \cdot \frac{1}{3} \, dy = \frac{1}{6}$

Remark

Important aspects of Fubini's Theorem:

Integrability is essential: The hypothesis $f \in L^1(\mu \otimes \nu)$ cannot be omitted. Without it, the iterated integrals may not exist, or may exist but give different values (as in the counterexample with $\frac{xy}{(x^2+y^2)^2}$ ).
Checking integrability: To verify $f \in L^1(\mu \otimes \nu)$ , often use Tonelli's Theorem on $|f|$ : $\int_{X \times Y} |f| \, d(\mu \otimes \nu) = \int_X \left(\int_Y |f(x, y)| \, d\nu(y)\right) d\mu(x) < \infty$
Null sets: The "almost every" qualifications are necessary. For instance, if $X = Y = [0, 1]$ and $f(x, y) = \chi_{\{x = y\}}$ , then $f(x, \cdot) = 0$ for a.e. $x$ , but $f$ is not identically zero.
Connection to Riemann integration: For continuous functions on rectangles in $\mathbb{R}^n$ , Fubini reduces to the classical fact that double Riemann integrals can be computed as iterated integrals.

Fubini's Theorem is indispensable in analysis, probability theory (for computing expectations of products), and physics (for evaluating multiple integrals in coordinate systems).