The Unapologetic Mathematician

Mathematics for the interested outsider

Measures on Product Spaces

After considering the product of measurable spaces, let’s consider what happens if our spaces are actually equipped with measures. That is, let (X,\mathcal{S},\mu) and (Y,\mathcal{T},\nu) be \sigma-finite measure spaces, and consider the product space (X\times Y,\mathcal{S}\times\mathcal{T}).

Now, if E is any measurable subset of X\times Y, then we can define two functions — f:X\to\mathbb{R} and g:Y\to\mathbb{R} — by the formulæ f(x)=\nu(E_x) and g(y)=\nu(E^y), where E_x and E^y are the X-section determined by x and Y-section determined by y, respectively. I say that both f and g are non-negative measurable functions, and that

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

where we take the first integral over X and the second over Y.

Let \mathcal{E} be the collection of subsets of X\times Y for which the assertions we’ve made are true. It’s straightforward to see that \mathcal{E} is closed under countable disjoint unions. Indeed, if \{E_n\} is a sequence of disjoint sets for which the assertions hold — call the functions \{f_n\} and \{g_n\} respectively — then if E is the disjoint union of the E_n we can calculate

\displaystyle\begin{aligned}f(x)&=\nu(E_x)\\&=\nu\left(\left(\bigcup\limits_{n=1}^\infty E_n\right)_x\right)\\&=\nu\left(\bigcup\limits_{n=1}^\infty(E_n)_x\right)\\&=\sum\limits_{n=1}^\infty\nu\left((E_n)_x\right)\\&=\sum\limits_{n=1}^\infty f_n(x)\end{aligned}

Since each of the f_n are measurable, so is f, and g is similarly measurable. The equality of the integrals should be clear.

Now, since \mu and \nu are \sigma-finite, every measurable subset of X\times Y can be covered by a countable disjoint union of measurable rectangles, each side of each of which has finite measure. Thus we just have to verify that the result holds for such measurable rectangles with finite-measure sides, and it will hold for all measurable subsets of X\times Y, and that \mathcal{E} is a monotone class. That \mathcal{E} is a monotone class follows from the dominated convergence theorem and the monotone convergence theorem, and so we have only to show that the assertions hold for measurable rectangles with finite-measure sides.

Okay, so let E=A\times B be a measurable rectangle with \mu(A)<\infty and \nu(B)<\infty. We can simplify our functions to write

\displaystyle\begin{aligned}f(x)&=\nu(B)\chi_A\\g(y)&=\mu(A)\chi_B\end{aligned}

These are clearly measurable, and we find that

\displaystyle\int f\,d\mu=\mu(A)\nu(B)=\int g\,d\nu

so the assertions hold for measurable rectangles with finite-measure sides, and thus for all measurable subsets of X\times Y.

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July 22, 2010 - Posted by | Analysis, Measure Theory

1 Comment »

  1. [...] We continue as yesterday, considering the two -finite measure spaces and , and the product measure space [...]

    Pingback by Product Measures « The Unapologetic Mathematician | July 23, 2010 | Reply


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