proof of extending a capacity to a Cartesian product

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

 $\displaystyle\tilde{I}\colon\mathcal{P}(X\times K)\to\mathbb{R},$ $\displaystyle\tilde{I}(S)=I(\pi_{X}(S)),$

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

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

 $\tilde{I}(S)=I(\pi_{X}(S))=I\left(\bigcup_{n}\pi_{X}(S_{n})\right)=\lim_{n% \rightarrow\infty}I(\pi_{X}(S_{n}))=\lim_{n\rightarrow\infty}\tilde{I}(S_{n}).$

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

 $\pi_{X}(S)=\bigcup\left\{\bigcap_{j=1}^{m}A_{j,k_{j}}\colon k_{j}\leq n_{j},\ % \bigcap_{j=1}^{m}K_{j,k_{j}}\not=\emptyset\right\}$

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

 $S_{x}\equiv\left\{y\in K\colon(x,y)\in S\right\}=\bigcap_{j=1}^{m}\bigcup\left% \{K_{j,k}\colon k\leq n_{j},x\in A_{j,k}\right\}.$

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

Now let $S_{n}$ be a decreasing sequence of sets in $\mathcal{G}$ 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 $\mathcal{K}^{*}$ 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 $\pi_{X}(S_{n})\in\mathcal{F}$ and, as $I$ is an $\mathcal{F}$-capacity,

 $\tilde{I}(S_{n})=I(\pi_{X}(S_{n}))\rightarrow I(\pi_{X}(S))=\tilde{I}(S).$

So $\tilde{I}$ is a $\mathcal{G}$-capacity.

We finally show that if $S\in\mathcal{G}_{\delta}$ then $\pi_{X}(S)\in\mathcal{F}_{\delta}$. By definition, there is a sequence $S_{n}\in\mathcal{G}$ such that $S=\bigcap_{n}S_{n}$. Setting $S^{\prime}_{n}=\bigcap_{m\leq n}S_{m}$ then, since $\mathcal{G}$ is closed under finite unions and finite intersections, $S^{\prime}_{n}\in\mathcal{G}$. Furthermore, $S^{\prime}_{n}$ decreases to $S$ so, as shown above, $\pi_{X}(S^{\prime}_{n})\in\mathcal{F}$ and

 $\pi_{X}(S)=\bigcap_{n}\pi_{X}(S^{\prime}_{n})\in\mathcal{F}_{\delta}$

as required.

Title proof of extending a capacity to a Cartesian product ProofOfExtendingACapacityToACartesianProduct 2013-03-22 18:47:41 2013-03-22 18:47:41 gel (22282) gel (22282) 5 gel (22282) Proof msc 28A12 msc 28A05