# Borel right process

In the mathematical theory of probability, a Borel right process, named after Émile Borel, is a particular kind of continuous-time random process.
Let $E$ be a locally compact, separable, metric space. We 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 $t$; for each $\omega \in \Omega$, $X_{t}(\omega )\in E$ is the value of $\omega$ at $t$. We denote the universal completion of ${\mathcal {E}}$ by ${\mathcal {E}}^{*}$. For each $t\in [0,\infty )$, let
    ${\mathcal {F}}_{t}=\sigma \left\{X_{s}^{-1}(B):s\in [0,t],B\in {\mathcal {E}}\right\},$
    ${\mathcal {F}}_{t}^{*}=\sigma \left\{X_{s}^{-1}(B):s\in [0,t],B\in {\mathcal {E}}^{*}\right\},$
and then, let
    ${\mathcal {F}}_{\infty }=\sigma \left\{X_{s}^{-1}(B):s\in [0,\infty ),B\in {\mathcal {E}}\right\},$
    ${\mathcal {F}}_{\infty }^{*}=\sigma \left\{X_{s}^{-1}(B):s\in [0,\infty ),B\in {\mathcal {E}}^{*}\right\}.$
For each Borel measurable function $f$ on $E$, define, for each $x\in E$,
    $U^{\alpha }f(x)=\mathbf {E} ^{x}\left[\int _{0}^{\infty }e^{-\alpha t}f(X_{t})\,dt\right].$
Since $P_{t}f(x)=\mathbf {E} ^{x}\left[f(X_{t})\right]$ and the mapping given by $t\rightarrow X_{t}$ is right continuous, we see that
for any uniformly continuous function $f$, we have the mapping given by $t\rightarrow P_{t}f(x)$ is right continuous.
Therefore, together with the monotone class theorem, for any universally measurable function $f$, the mapping given by $(t,x)\rightarrow P_{t}f(x)$, is jointly measurable, that is, ${\mathcal {B}}([0,\infty ))\otimes {\mathcal {E}}^{*}$ measurable, and subsequently, the mapping is also $\left({\mathcal {B}}([0,\infty ))\otimes {\mathcal {E}}^{*}\right)^{\lambda \otimes \mu }$-measurable for all finite measures $\lambda$ on ${\mathcal {B}}([0,\infty ))$ and $\mu$ on ${\mathcal {E}}^{*}$. Here, $\left({\mathcal {B}}([0,\infty ))\otimes {\mathcal {E}}^{*}\right)^{\lambda \otimes \mu }$ is the completion of ${\mathcal {B}}([0,\infty ))\otimes {\mathcal {E}}^{*}$ with respect to the product measure $\lambda \otimes \mu$. Thus, for any bounded universally measurable function $f$ on $E$, the mapping $t\rightarrow P_{t}f(x)$ is Lebeague measurable, and hence, for each $\alpha \in [0,\infty )$, one can define
    $U^{\alpha }f(x)=\int _{0}^{\infty }e^{-\alpha t}P_{t}f(x)dt.$
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
    $U^{\alpha }f(x)=\mathbf {E} ^{x}\left[\int _{0}^{\infty }e^{-\alpha t}f(X_{t})dt\right].$
The following are the defining properties of Borel right processes:
    For each probability measure $\mu$ on $(E,{\mathcal {E}})$, there exists a 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 $\{P_{t}:t\in [0,\infty )\}$.
    Let $f$ be $\alpha$-excessive for the resolvent on $(E,{\mathcal {E}}^{*})$. Then, for each probability measure $\mu$ on $(E,{\mathcal {E}})$, a mapping given by $t\rightarrow f(X_{t})$ is $P^{\mu }$ almost surely right continuous on $[0,\infty )$.