Borel right process

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Let E be a locally compact separable metric space. We will denote by \mathcal E the Borel subsets of E. Let Ω 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 Xt the coordinate map at t; for each \omega \in \Omega , X_t(\omega) \in E is the value of ω 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_tf(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 that the mapping given by t \rightarrow 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) \rightarrow 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^*\right)^{\lambda\otimes \mu}-measurable for all finite measures λ on \mathcal B([0,\infty)) and μ 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. Now, this shows that for any bounded universally measurable function f on E, the mapping t\rightarrow 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^x\left[ \int_0^\infty e^{-\alpha t} f(X_t) dt \right].

The followings are the defining properties of Borel right processes:

  • Hypothesis Droite 1: For each probability measure μ 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 μ and transition semigroup \{ P_t : t \in [0,\infty) \}.
  • Hypothesis Droite 2: Let f be α-excessive for the resolvent on (E, \mathcal E^*). Then, for each probability measure μ on (E,\mathcal E), a mapping given by t \rightarrow f(X_t) is Pμ almost surely right continuous on [0,\infty).