proof of extending a capacity to a Cartesian product

Let $(X,ℱ)$ be a paved space such that $ℱ$ is closed under finite unions and finite intersections, and $(K,𝒦)$ be a compact paved space. Define $𝒢$ to be the closure under finite unions and finite intersections of the paving $ℱ\times 𝒦$ on $X\times K$. For an $ℱ$-capacity $I$, define

$\tilde{I}:𝒫(X\times K)\to ℝ,$

$\tilde{I}(S)=I({\pi }_X(S)),$

where ${\pi }_X$ is the projection map onto $X$. We show that $\tilde{I}$ is a $𝒢$-capacity and that ${\pi }_X(S)\in {ℱ}_{\delta }$ whenever $S\in {𝒢}_{\delta }$.

Clearly, the property that $\tilde{I}$ is an increasing set function follows from the fact that $I$ satisfies this property. Furthermore, if $S_n\subseteq X\times K$ is an increasing sequence of sets with $S={\bigcup }_n{S}_n$ then ${\pi }_X(S_n)$ is an increasing sequence and

$$\tilde{I}(S)=I({\pi }_X(S))=I\left(\bigcup _n{\pi }_X(S_n)\right)=\underset{n\to \mathrm{\infty }}{lim}I({\pi }_X(S_n))=\underset{n\to \mathrm{\infty }}{lim}\tilde{I}(S_n).$$

To prove that $\tilde{I}$ is a $𝒢$-capacity, it only remains to show that if $S_n$ is a sequence in $𝒢$ decreasing to $S\subseteq X\times K$ then $\tilde{I}(S_n)\to \tilde{I}(S)$. Note that any $S$ in $𝒢$ can be written as $S={\bigcap }_{j=1}^m{\bigcup }_{k=1}^{n_j}{A}_{j,k}\times K_{j,k}$ for sets $A_{j,k}\in ℱ$ and $K_{j,k}\in 𝒦$. The projection onto $X$ is then

$${\pi }_X(S)=\bigcup \{\bigcap _{j=1}^m{A}_{j,k_j}:k_j\le n_j,\bigcap _{j=1}^m{K}_{j,k_j}\ne \mathrm{\varnothing }\}$$

which, as $ℱ$ is closed under finite unions and finite intersections, must be in $ℱ$. Furthermore, for any $x\in X$,

$$S_x\equiv \{y\in K:(x,y)\in S\}=\bigcap _{j=1}^m\bigcup \{K_{j,k}:k\le n_j,x\in A_{j,k}\}.$$

This shows that $S_x$ is in the closure ${𝒦}^{*}$ of $𝒦$ under finite unions and finite intersections. Furthermore, since compact pavings are closed subsets of a compact topological space (http://planetmath.org/CompactPavingsAreClosedSubsetsOfACompactSpace), ${𝒦}^{*}$ is itself a compact paving.

Now let $S_n$ be a decreasing sequence of sets in $𝒢$ and set $S={\bigcap }_n{S}_n$. Then ${\pi }_X(S)\subseteq {\pi }_X(S_n)$ for each $n$, giving ${\pi }_X(S)\subseteq {\bigcap }_n{\pi }_X(S_n)$. To prove the reverse inequality, consider $x\in {\bigcap }_n{\pi }_X(S_n)$. Then, ${(S_n)}_x$ is a nonempty set in ${𝒦}^{*}$ for all $n$. By compactness, $S_x={\bigcap }_n{(S_n)}_x$ must also be nonempty and therefore $x\in {\pi }_X(S)$. This shows that

$$\bigcap _n{\pi }_X(S_n)={\pi }_X(S).$$

Furthermore, as we have shown that ${\pi }_X(S_n)\in ℱ$ and, as $I$ is an $ℱ$-capacity,

$$\tilde{I}(S_n)=I({\pi }_X(S_n))\to I({\pi }_X(S))=\tilde{I}(S).$$

So $\tilde{I}$ is a $𝒢$-capacity.

We finally show that if $S\in {𝒢}_{\delta }$ then ${\pi }_X(S)\in {ℱ}_{\delta }$. By definition, there is a sequence $S_n\in 𝒢$ such that $S={\bigcap }_n{S}_n$. Setting $S_n^{\prime }={\bigcap }_{m\le n}{S}_m$ then, since $𝒢$ is closed under finite unions and finite intersections, $S_n^{\prime }\in 𝒢$. Furthermore, $S_n^{\prime }$ decreases to $S$ so, as shown above, ${\pi }_X(S_n^{\prime })\in ℱ$ and

$${\pi }_X(S)=\bigcap _n{\pi }_X(S_n^{\prime })\in {ℱ}_{\delta }$$

as required.

Title	proof of extending a capacity to a Cartesian product
Canonical name	ProofOfExtendingACapacityToACartesianProduct
Date of creation	2013-03-22 18:47:41
Last modified on	2013-03-22 18:47:41
Owner	gel (22282)
Last modified by	gel (22282)
Numerical id	5
Author	gel (22282)
Entry type	Proof
Classification	msc 28A12
Classification	msc 28A05