# extending a capacity to a Cartesian product

A capacity on a set $X$ can be extended to a set function on a Cartesian product $X\times K$ simply by projecting any subset onto $X$, and then applying the original capacity.

###### Theorem.

Suppose that $(X,\mathcal{F})$ is a paved space such that $\mathcal{F}$ is closed under finite unions and finite intersections  , and that $(K,\mathcal{K})$ is 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 $X\times K$.

If $I$ is an $\mathcal{F}$-capacity and $\pi_{X}\colon X\times K\to X$ is the projection map, we can form the composition   $\displaystyle I\circ\pi_{X}\colon\mathcal{P}(X\times K)\to\mathbb{R},$ $\displaystyle I\circ\pi_{X}(S)=I(\pi_{X}(S)).$

Then $\pi_{X}(S)\in\mathcal{F}_{\delta}$ for any $S\in\mathcal{G}_{\delta}$, and $I\circ\pi_{X}$ is a $\mathcal{G}$-capacity.

This result justifies looking at capacities when considering projections from the Cartesian product $X\times K$ onto $X$. We see that the property of being a capacity is preserved under composing with such projections. However, additivity of set functions is not preserved, so the corresponding result would not be true if “capacity” was replaced by “measure  ” or “outer measure   ”.

Recall that if $S\subseteq X\times K$ is $(\mathcal{G},I\circ\pi_{X})$-capacitable then, for any $\epsilon>0$, there is an $A\in\mathcal{G}_{\delta}$ such that $A\subseteq S$ and $I\circ\pi_{X}(A)>I\circ\pi_{X}(S)-\epsilon$. However, $\pi_{X}(A)\subseteq\pi_{X}(S)$ and, by the above theorem, $\pi_{X}(A)\in\mathcal{F}_{\delta}$. This has the following consequence.

###### Lemma.

Let $S\subseteq X\times K$ be $(\mathcal{G},I\circ\pi_{X})$-capacitable. Then, $\pi_{X}(S)$ is $(\mathcal{F},I)$-capacitable.

Title extending a capacity to a Cartesian product ExtendingACapacityToACartesianProduct 2013-03-22 18:47:38 2013-03-22 18:47:38 gel (22282) gel (22282) 6 gel (22282) Theorem msc 28A12 msc 28A05