I'm reading about Support of Potential measures for Lévy Processes. $\Sigma$ denotes the support of $U(0,\cdot),$ where $U(x,B)=\int_{0}^{\infty}P_{x}(X_{t}\in B)\,\mathrm dt$ is the potential measure of the process $(X_{t})_{t\geq 0},$ and qualify the points in $\Sigma$ as possible.

Then the book says:

  1. $x\in\mathbb{R}^{d}$ is possible if and only if $\forall\epsilon>0,\exists t>0$ such that $P(X_{t}\in B(x,\epsilon))>0$ if and only if $P(T_{B}<\infty)>0$ for every ball $B$ centred at $x.$ Here $T_{B}=\inf\{t>0:X_{t}\in B\}.$
  2. If $x,y\in\Sigma$ then $x+y\in\Sigma.$

I'm trying to prove 1) and 2) but I'm stuck. For 1) I have the second equivalence because of the definition of $T_{B},$ but the other is not clear to me. I think I'm not understanding the meaning of $x\in\Sigma.$

For 2) the book says that consider $x,y\in\Sigma$ and $B(z,\eta)$ where $\eta>0.$ Is enough to apply Markov property at the first passage time into $B(x,\epsilon)$ to see that $P(T_{B(x+y,2\epsilon)}<\infty)>0.$ How the Markov property works here? I don't understand how the author utilizing here.

Any kind of help is thanked in advanced.

  • $\begingroup$ I've edited the formatting slightly to make things easier to read, but I've had to change your list labels to numbers rather than letters. (The site only seems to support numbered lists in questions.) Feel free to roll back my edit if you don't like what I've done. $\endgroup$ – Theoretical Economist Apr 23 '18 at 5:32
  • $\begingroup$ Also, the first bullet point reads a bit awkwardly; you may want to break up the two "if and only if" statements. By "possible", do you mean that a point is in $\Sigma$? $\endgroup$ – Theoretical Economist Apr 23 '18 at 5:33
  • $\begingroup$ Thanks to answer @TheoreticalEconomist. Yes, a point $x\in\Sigma$ is called "possible." $\endgroup$ – Squird37 Apr 23 '18 at 5:43
  • $\begingroup$ Clearly, by definition of support of a measure, if $x \in \Sigma$, then $U(0,B(x,\epsilon))>0$ for all $\epsilon>0$. But as $U(0,B(x,\epsilon))=\int_0^{\infty}P_0(X_t \in B(x,\epsilon))\,d t$, this means that there must exist a $t$ for which $P_0(X_t \in B(x,\epsilon))>0$. $\endgroup$ – Sayantan Apr 23 '18 at 6:04
  • $\begingroup$ I'm a little confused by Q.2. If $\Sigma$ contains an open set and satisfies the property mentioned in the question then $\Sigma = \mathbb{R}^d$. Does that agree with your intuition about the potential measures? $\endgroup$ – Sayantan Apr 23 '18 at 6:07

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