Lecture 8: Product measures

24/10/24

Reference: Lecture notes 4, 7.1

The Carathéodory theorem allows us to extend the definition of the Lebesgue measure on $\R$ to $\R^n$ for any $n$. Namely, take $\mathcal{S}_n \subset P(\R^n)$ to be the set of half-closed $n$-dimensional boxes

$I = [a_1,b_1) \times \cdots \times [a_n,b_n)$

and define $\rho_n \colon \mathcal{S}_n \to [0,\infty]$ by

$\rho_n(I) = \prod_{k=1}^n (b_n-a_n)$.

Then $\mathcal{S}_n$ is a semiring and $\rho_n$ is a premeasure. Therefore, there exists a $\sigma$-algebra $\Sigma_n$ containing $\mathcal{S}_n$ and a complete measure $\lambda_n \colon \Sigma_n \to [0,\infty]$ such that $\lambda_n(I)=\rho_n(I)$ for every box $I$. Moreover, this measure is unique on $\Sigma_n$. We call it the $n$-dimensional Lebesgue measure and $\Sigma_n$ the set of Lebesgue measurable subsets. This is all analogous to the case $n=1$.

However, this construction does not tell us how $(\R^n,\Sigma_n,\lambda_n)$ relates to $(\R,\Sigma_1,\lambda_1)$. For example, if $A,B\subset \R$ are Lebesgue measurable, is their product $A\times B \subset \R^2$ Lebesgue measurable? Is it true that $\lambda_2(A\times B) = \lambda_1(A)\lambda_1(B)$? These are some basic properties we would want because in analysis, geometry, and other applications it is important to be able to reduce a problem from higher dimensions to lower. Recall, for example, how you can compute $3$-dimensional integrals by slicing the domain by $2$-dimensional planes, or how in geometry one can study shapes by considering their projections to lower dimensions. Another interesting example is tomography which aims to reconstruct a function of $2$ variables from all of its integrals along $1$-dimensional lines.

More generally, let $(X,\mathcal{A},\mu)$ and $(X',\mathcal{A}',\mu')$ be two measure spaces. The product

$\mathcal{A}\times\mathcal{A}' = \{ A\times A' \subset X\times X' \ | \ A \in \mathcal{A}, \ A' \in \mathcal{A}' \}$

is not, in general, a $\sigma$-algebra and the function

$A \times A' \mapsto \mu(A)\mu'(A')$

is not a measure. To correct the first issue, we introduce the following.

Definition. The product of $\sigma$-algebras $\mathcal{A}$ and $\mathcal{A}'$ is the $\sigma$-algebra generated by $\mathcal{A}\times\mathcal{A}'$ in $X\times X'$. We denote it by $\mathcal{A}\otimes\mathcal{A}'.$ ∎