The Unapologetic Mathematician

Mathematics for the interested outsider

Product Measures

We continue as yesterday, considering the two \sigma-finite measure spaces (X,\mathcal{S},\mu) and (Y,\mathcal{T},\nu), and the product measure space (X\times Y,\mathcal{S}\times\mathcal{T}).

Last time we too a measurable set E\subseteq X\times Y and defined the functions f(x)=\nu(E_x) and g(y)=\mu(E^y). We also showed that

\displaystyle\int f\,d\mu=\int g\,d\nu

That is, for every measurable E we can define the real number

\displaystyle\lambda(E)=\int\nu(E_x)\,d\mu=\int\mu(E^y)\,d\nu

I say that this function \lambda is itself a \sigma-finite measure, and that for any measurable rectangle A\times B we have \lambda(A\times B)=\mu(A)\nu(B). Since measurable rectangles generate the \sigma-ring \mathcal{S}\times\mathcal{T}, this latter condition specifies \lambda uniquely.

To see that \lambda is a measure, we must show that it is countably additive. If \{E_n\} is a sequence of disjoint sets then we calculate

\displaystyle\begin{aligned}\lambda\left(\biguplus\limits_{n=1}^\infty E_n\right)&=\int\nu\left(\left(\biguplus\limits_{n=1}^\infty E_n\right)_x\right)\,d\mu\\&=\int\nu\left(\biguplus\limits_{n=1}^\infty(E_n)_x\right)\,d\mu\\&=\int\sum\limits_{n=1}^\infty\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\int\nu\left((E_n)_x\right)\,d\mu\\&=\sum\limits_{n=1}^\infty\lambda(E_n)\end{aligned}

where we have used the monotone convergence theorem to exchange the sum and the integral.

We verify the \sigma-finiteness of \lambda by covering each measurable set E by countably many measurable rectangles with finite-measure sides. Since the sides’ measures are finite, the measure of the rectangle itself is the product of two finite numbers, and is thus finite.

We call the measure \lambda the “product” of the measures \mu and \nu, and we write \lambda=\mu\times\nu. We thus have a \sigma-finite measure space (X\times Y,\mathcal{S}\times\mathcal{T},\mu\times\nu) that we call the “cartesian product” of the spaces (X,\mathcal{S},\mu) and (Y,\mathcal{T},\nu).

July 23, 2010 Posted by | Analysis, Measure Theory | 6 Comments