# Measure on Product Set

Consider a finite sequence of $\sigma$-finite measure spaces $(\Omega_i, \mathcal{F}_i, \mu_i)$. Constructing the product measurable space $$(\Omega_1 \times \cdots \times \Omega_n, \mathcal{F}_1 \otimes \cdots \otimes \mathcal{F}_n),$$ we can define the product measure $\mu = \mu_1 \otimes \cdots \otimes \mu_n$ such that $$\mu(A_1 \times \cdots \times A_n) = \mu_1(A_1) \cdots \mu_n(A_n)$$ whenever $A_i \in \mathcal{F}_i$.

Is it that the product measure only behaves like this when measuring products? E.g., let $n=2$ and consider the "triangle" set $A = \{(x,y): x \in (0,a], y \in (0,x]\} \subset \mathbb{R}^2$ and the Borel $\sigma$-algebra $\mathcal{B}$. Perhaps naively I would put $A_1 = (0,a]$ and $A_2 = (0,x]$, but then I don't think that $A = A_1 \times A_2$. I would then try to measure $\mu(A) = \mu_1(A_1)\mu_2(A_2)$, which I would want to be $\frac{1}{2}a^2$. What would be the proper way to measure non-cartesian product sets?

• The measure of such a set can be calculated using Fubini's theorem (also known as Cavallieris principle in this context). – PhoemueX Jun 30 '15 at 17:36

You're correct that the you can only apply the rule $\mu(A) = \mu_1(A_1) \mu_2(A_2)$ when $A = A_1\times A_2.$ A triangle does not meet this criteria. However, you can create sets of rectangles that tell us what the .
So, for the sake of convenience, let $a=1$ and cut the square $[1,0] \times [1,0]$ up into $n^2$ equal squares in the usual way. [Perhaps first convince yourself that whether a square contains its boundary or not doesn't change its measure.] I claim that each such square has measure $1/n^2$, by the definition of the product measure. Furthermore, there is a collection of $n(n+1)/2$ squares that cover the triangle, and a collection of $n(n-1)/2$ squares that are covered by the triangle.
Therefore, $$\frac{1}{n^2} \frac{n(n-1)}{2} = \frac{1}{2} -\frac{1}{2n} \leq \mu(A) \leq \frac{1}{n^2} \frac{n(n+1)}{2} = \frac{1}{2} +\frac{1}{2n}$$ so we may conclude that $\mu(A) = 1/2$, as this must be true for all $n$.