We find on Petr Mandl's book Analytical treatment of one-dimensional Markov processes page 10 the following definition of conservative markov process
My question concerns the proof of theorem 4 that is made into an exercise (problem 2 pg 18)
The question is to construct a Markov process from the transition probability function $P(t,x;B)$.
The first step is to construct a measure for the case where $\tilde{P}$ is conservative, that is $P(t,x;I) = 1$ for every $t,x$. This is obtained via the Kolmogorov theorem of extension of measures in cartesian products) The familly of compatible measures is constructed using the following scheme $$ \Pi_{x,s,t+s}( A \times B) =\Pi_{x,s,t+s}(X_s \in A, X_{t+s} \in B) = \int P(t,y; B) P(t,x; dy) $$
finite dimensionals are obtained by induction:$s_1,\leq s_2\leq \ldots \leq s_k$
$$ \Pi_{x,s_1,s_2,\ldots,s_k,s_k + t}(A_1 \times A_2 \times \ldots\times A_k \times B) = \Pi_{x,s_1,s_2,\ldots,s_k,s_k + t}(X_{s_1} \in A_1,X_{s_2} \in A_2\ldots, X_{s_k} \in A_k, X_{t+s_k} \in B) = \int_{A_1}\int_{A_2} \ldots \int_{A_k} P(t,y_k; B) P(s_k- s_{k-1}, y_{k-1}; dy_k) \ldots P(s_2- s_1, y_{1}; dy_2) P(s_1, x; dy_1) $$
The compatibility follows from the Chapman-Kolmogorov property (18) in the quoted text.
$$ P(t+s,x ;B) = \int P(s,y;B) P(t,x;dy) $$
Therefore we obtain a family of maesures $P_x$ ($P_x(X_0 = x) = 1$). Define in this case $\zeta = \infty$
Define now $F(t_1,x_1 t_2,x_2,\ldots, t_n, x_n) = P_x( X_{t_1} \leq x_1, X_{t_2} \leq x_2, \ldots, X_{t_n} \leq x_n)$
and $F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m ) = \int_{\infty}^{x_1}\int_{-\infty}^{x_2} \ldots \int_{-\infty}^{x_n} P(t_n - t_{n-1},y_{n-1}; dy_n) \ldots P(t_2- t_1, y_{1}; dy_2) P(t_1-s_m, y_m; dy_1) $
By the property 2 in the list of properties of transition probability functions quoted above we obtain that $(y_1,\ldots y_m) \mapsto F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m )$ is Borel measurable, non decreasing and continuous to the right in $x_1,\ldots ,x_n$ and satisfies
$$\int_{-\infty}^{z_m} F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m ) F(s_m,dy_m) = \int_{-\infty}^{z_1}\int_{\infty}^{x_1}\int_{-\infty}^{x_2} \ldots \int_{-\infty}^{x_n} P(t_n - t_{n-1},y_{n-1}; dy_n) \ldots P(t_2- t_1, y_{1}; dy_2) P(t_1-s_m, y'_m; dy_1)F(s_m,y'_m) = F(s_m,y_m,t_1, x_1,\ldots, t_n, x_n) $$
Moreover,
$$\int_{-\infty}^{z_1} \ldots \int_{-\infty}^{z_m} F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m ) F(s_1, dy_1,\ldots, s_m,dy_m) = \int_{-\infty}^{z_1}\ldots \int_{-\infty}^{z_m}\int_{\infty}^{x_1}\int_{-\infty}^{x_2} \ldots \int_{-\infty}^{x_n} P(t_n - t_{n-1},y_{n-1}; dy_n) \ldots P(t_2- t_1, y_{1}; dy_2) P(t_1-s_m, y_m; dy_1) F(s_1,dy'_1,\ldots ,s_m,dy'_m)= \int_{-\infty}^{z_1}\ldots \int_{-\infty}^{z_m}\int_{\infty}^{x_1}\int_{-\infty}^{x_2} \ldots \int_{-\infty}^{x_n} P(t_n - t_{n-1},y_{n-1}; dy_n) \ldots P(t_2- t_1, y_{1}; dy_2) P(t_1-s_m, y_m; dy_1) P(s_m- s_{m-1}, y'_{m-1}; dy'_m)P(s_2- s_{1}, y'_{1}; dy'_2)F(s_1,dy'_1) = F(s_1,z_1,\ldots,s_m,z_m,t_1, x_1,\ldots, t_n, x_n) $$
We note that $F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m )$ does not depend on $x$ and is the conditional distribution of $X_{t_1},\ldots, X_{t_n}$ given the condition $X_{s_1} = y_1\, ldots, X_{s_m} = y_m$. The fact that $F(t_1, x_1,\ldots, t_n, x_n\mid s_1,y_1,\ldots,s_m,y_m )$ does not depend on $s_1,y_1,\ldots,s_{m-1},y_{m-1}$ is the markovian property of the process.
So far we have used properties (1), (2), (3) of the transition probability function. Now we consider the case when $P$ is not conservative, that is: $$P(t,x, I)< 1\quad \text{ for some } t, x $$
Item $b$ is clear, but when it comes to item (c) this is not so.
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Questions The question is where do we use right continuity in $t$ of transitions functions?
I think that to prove that $F$ has outer measure $0$ I believe it suffices to show that $F^c$ has inner measure $0$ but this follows from the fact that $F^c = \{\exists t, s \mid X_t = \vartheta, X_s \in I\}$. But since $P(s-t, \vartheta, I) = 0$ the probability of this event is zero.
Is the above reasoning correct? Me question is don't we need some sort of right continuity of paths to prove that the denumerable union $$\Cup{t \leq s \in Q_+} \{X_t = \vartheta, X_s \in I\}$$ is equal to $F^c$ or in a sense exaust that set from within (as we are looking for the inner measure of $F^c$)
Second question is are we looking for the inner measure of $F^c$?
Finally, to prove that $\zeta =\inf\{t \mid X_t - \vartheta\}$ satisfies $X_\zeta = \vartheta$ when $\zeta < \infty$ do we use right continuity of paths? or is this a consequence of right continuity of the transition functions?