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I have a question about this paper BH, which is one of the basic literature on reflecting Brownian motions.

For any bounded open subset $D$, the semigroup $\{T_t\}_{t>0}$ on $L^{2}(D,dx)$ generated by Neumann laplacian admits an integral kernel. If $D$ is bounded Lipschitz domain, $\{T_t\}_{t>0}$ has the integral kernel $p_{t}(x,y)$ which is continuous on $(0,\infty) \times \bar{D} \times \bar{D}$.

The authors of this paper constructed reflecting Brownian motion on $\bar{D}$ in the following procedure:

  1. They proved $H^{1}(D)$ (1st order Sobolev space with Neumann boundary condition) is a regular Dirichlet form on $L^{2}(\bar{D},dx)$. Since, $H^{1}(D)$ is regular on $L^{2}(\bar{D},dx)$, there exists a Hunt process $X=(X_t,P_x)$ on $\bar{D}$ associated with $H^{1}(D)$.
  2. However, $X=(X_t,P_x)$ does not necessarily have transition density. So they introduced new probability measures $\{Q_x\}_{x \in \bar{D}}$. Each $Q_{x}$ is defined by \begin{equation*} Q_{x}(A \circ \theta_{t})=\int_{D}p_{t}(x,y)P_{y}(A)\,dy,\quad x\in\bar{D},\ t>0,\ A \in\mathcal{F}. \end{equation*} Here, $\theta_t$ is the shift operator of $X=(X_t,P_x)$ and $\mathcal{F}$ is the filtration generated by $X$.

My question

They claimed $Q_{x}$ uniquely determines a probability measure $Q_x$ on $\mathcal{F}$. However, I don't know why... For example, how do they define $Q_{x}(\{X_0=x\})$ ?

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  • $\begingroup$ I did not look into the linked paper, but this seems to be a standard trick to get rid of the "properly exceptional set". Think of a 2-D BM which is stopped at the origin. With probability 1 it never hits the origin, so the semigroup of this process is the same as that of the usual BM, except that $T_t f(0) = f(0)$ for the former one, so technically it does not have transition densities. Under $Q_x$, however, the stopped Brownian motion becomes the usual one. $\endgroup$ Jan 31, 2018 at 10:17
  • $\begingroup$ (continued) In any case, the expression for $Q_x$ that you gave, together with the obvious condition $Q_x(\{X_0 = x\}) = 1$, defines the measure on all cylinder sets, so indeed it does determine the probability measure uniquely (as long as it defines it at all, but this is a different question; for Hunt processes it does). $\endgroup$ Jan 31, 2018 at 10:23
  • $\begingroup$ In other words, are you saying that $(X_t ,Q_x)$ will become Hunt process under condition $Q_{x}(\{X_0=x\})=1$? $\endgroup$
    – sharpe
    Jan 31, 2018 at 12:04
  • $\begingroup$ Yes, I think so. (The condition $Q_x(\{X_0=x\})$ should actually follow automatically if all paths are càdlàg, because $Q_x(\{|X_t-x|<r\})=\int_{B(x,r)}p_t(x,y)dy\to 1$ as $t\to 0$, so $Q_x(\{X_{0+}=x\})=1$). $\endgroup$ Jan 31, 2018 at 13:40
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    $\begingroup$ No, I only used the definition of $Q_x$ and the properties of $p_t(x,y)$: $Q_x(\{|X_t - x|<r\})=\int_D p_t(x,y)P_y(\{|X_0 - x|<r\})dy=\int_{B(x,r)}p_t(x,y)dy$. $\endgroup$ Jan 31, 2018 at 15:01

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