I'm reading about Potential measures of Lévy processes. There is a theorem which proof is not totally clear to me. The teorem is the next:

Suposse that $H=\mathbb{R}^{d},$ where $H$ is the spanned of the support of potential measure $U(x,\cdot)$ of the Lévy process $(X_{t})_{t\geq 0}$ Then the process is recurrent if and only if for every Borel set $K$ with positive Lebesgue measure, $U(x,B)=\infty$ for $a.e.\space x\in\mathbb{R}^{d}.$

To get context, $T_{B}=\inf\{t\geq 0:X_{t}\in B\}$ and $\mathcal{R}=\{x\in\mathbb{R}^{d}:P(\exists s\geq t, X_{s}\in x+B)=1 \forall t\geq 0\}$

The proof is the next:

To start with, we prove that $U(x,B)=\infty$ for every $x\in\mathbb{R}^{d}$ and non-empty open ball $B.$

With no loss of generality, we may suppose that $B$ is centred at the origin. Let $B^{'}$ be another open ball centred at the origin, with radius less than half of that of $B^{'}.$ Then probability $P(T_{x+B^{'}}<\infty)=1$ since $x\in\mathcal{R}.$ Denote by $\overline{B^{'}}$ the closure of $B^{'}$ and recall that $\overline{B^{'}}-\overline{B^{'}}\subseteq B.$ As a consecuence, we have that $B^{'}\subset x-y+B$ for every $y\in x+\overline{B^{'}},$ and then $U(0,x-y+B)\geq U(0,B^{'}).$

$\textbf{Here comes my first doubt:}$ We now deduce from the Markov property applied to the first entrance time into $x+B^{'}$ that: \begin{eqnarray*} U(0,x+B)&\geq&\int_{\mathbb{R}^{d}}P(X_{T_{x+B^{'}}}\in dy)U(0,x-y+B)\\ &\geq&P(X_{T_{x+B^{'}}}<\infty)\inf_{y\in x+\overline{B^{'}}}U(0,x-y+B)\\ &\geq&P(X_{T_{x+B^{'}}}<\infty)U(0,B^{'}). \end{eqnarray*}

I don't understand how the Markov property is used. The chain of inequalities follows because of properties of Potential measure.

Since $U(0,B^{'})=\infty,$ this shows that $U(0,x+B)=U(-x,B)=\infty.$

Next, we suppose that $C$ and $K$ are both compact sets with positive Lebesgue measure. Fubini's theorem yields $$\int_{C}U(x,K)dx=\int_{\mathbb{R}^{d}}U(0,dy)\int_{C}1_{K}(x+y)dx.$$

$\textbf{Second doubt:}$ why the function $y\rightarrow\int_{C}1_{K}(x+y)dx$ is continuous and not identically zero? What is the reason to prove this?

Then $\int_{C}U(x,K)dx=\infty,$ and since this holds for all compact sets $C$ with positive lebesgue measure, $U(x,K)=\infty$ for almost every $x$ (by inner regularity of Lebesgue measure). This extends immediatly to the general case when $K$ is a Borel set with positive Lebesgue measure.

$\textbf{third doubt:}$ Why the above is follow by inner regularity of Lebesgue measure and how thos extends to general case?

Any kind of help is thanked in advanced.


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