Borel right process

Let $E$ be a locally compact separable metric space.We will denote by $mathcal\; E$ the Borel subsets of $E$.Let $Omega$ be the space of right continuous maps from $[0,infty)$ to $E$ that have left limits in $E$,and for each $t\; in\; [0,infty)$, denote by $X\_t$ the coordinate map at $t$; foreach $omega\; in\; Omega$, $X\_t(omega)\; in\; E$ is the value of $omega$ at $t$. We denote the universal completion of $mathcal\; E$ by $mathcal\; E^*$.For each $tin\; [0,infty)$, let

: $mathcal\; F\_t\; =\; sigmaleft\{\; X\_s^\{-1\}(B)\; :\; sin\; [0,t]\; ,\; B\; in\; mathcal\; E\; ight\},$

: $mathcal\; F\_t^*\; =\; sigmaleft\{\; X\_s^\{-1\}(B)\; :\; sin\; [0,t]\; ,\; B\; in\; mathcal\; E^*\; ight\},$

and then, let

: $mathcal\; F\_infty\; =\; sigmaleft\{\; X\_s^\{-1\}(B)\; :\; sin\; [0,infty),\; B\; in\; mathcal\; E\; ight\},$

: $mathcal\; F\_infty^*\; =\; sigmaleft\{\; X\_s^\{-1\}(B)\; :\; sin\; [0,infty),\; B\; in\; mathcal\; E^*\; ight\}.$

For each Borel measurable function $f$ on $E$, define, for each $x\; in\; E$,

: $U^alpha\; f(x)\; =\; mathbf\; E^xleft\; [\; int\_0^infty\; e^\{-alpha\; t\}\; f(X\_t),\; dt\; ight]\; .$

Since $P\_tf(x)\; =\; mathbf\; E^xleft\; [f(X\_t)\; ight]$ and the mapping given by $t\; ightarrow\; X\_t$ is right continuous, we see that for any uniformly continuous function $f$, we have that the mapping given by $t\; ightarrow\; P\_tf(x)$ is right continuous. Therefore, together with the monotone class theorem, one can show that for any universally measurable function $f$, the mapping given by $(t,x)\; ightarrow\; P\_tf(x)$, is jointly measurable, that is, $mathcal\; B(\; [0,infty))otimes\; mathcal\; E^*$ measurable, and subsequently, the mapping is also $left(mathcal\; B(\; [0,infty))otimes\; mathcal\; E^*\; ight)^\{lambdaotimes\; mu\}$-measurable for all finite measures $lambda$ on $mathcal\; B(\; [0,infty))$ and $mu$ on $mathcal\; E^*$.Here, $left(mathcal\; B(\; [0,infty))otimes\; mathcal\; E^*\; ight)^\{lambdaotimes\; mu\}$ is the completion of$mathcal\; B(\; [0,infty))otimes\; mathcal\; E^*$ with respectto the product measure $lambda\; otimes\; mu$. Now, this shows that for any bounded universally measurable function $f$ on $E$,the mapping $t\; ightarrow\; P\_tf(x)$ is Lebeague measurable, and hence, for each $alpha\; in\; [0,infty)$, one can define

$U^alpha\; f(x)\; =\; int\_0^infty\; e^\{-alpha\; t\}P\_tf(x)\; dt.$

There is enough joint measurability to check that $\{U^alpha\; :\; alpha\; in\; (0,infty)\}$ is a Markov resolvent on $(E,mathcal\; E^*)$,which uniquely associated with the Markovian semigroup $\{\; P\_t\; :\; t\; in\; [0,infty)\; \}$. Consequently, one may apply Fubini's theorem to see that

$U^alpha\; f(x)\; =\; mathbf\; E^xleft\; [\; int\_0^infty\; e^\{-alpha\; t\}\; f(X\_t)\; dt\; ight]\; .$

The followings are the defining properties of Borel right processes:

Hypothesis Droite 1:

For each probability measure $mu$ on $(E,\; mathcal\; E)$, there existsa probability measure $mathbf\; P^mu$ on $(Omega,\; mathcal\; F^*)$ such that$(X\_t,\; mathcal\; F\_t^*,\; P^mu)$ is a Markov process with initial measure $mu$and transition semigroup $\{\; P\_t\; :\; t\; in\; [0,infty)\; \}$.

Hypothesis Droite 2:

Let $f$ be $alpha$-excessive for the resolvent on $(E,\; mathcal\; E^*)$.Then, for each probability measure $mu$ on $(E,mathcal\; E)$, a mappinggiven by $t\; ightarrow\; f(X\_t)$ is $P^mu$ almost surely right continuous on $[0,infty)$.