Let the continuous function $\ell:\mathbb R \times(0,\infty)\to[0,\infty)$ be a Lévy-type kernel, such that $$ \sup_{x}\int_0^\infty \min\{1,y\}\ell(x, y)\,dy<\infty, $$ and suppose that $\mathcal Lg(x):=\int_0^\infty (g(x-y)-g(x))\ell(x,y)\,dy$ generates a decreasing Lévy-type process $s\mapsto L^{x}_s$, where $x\in\mathbb R$ denotes the starting point. Define the first exit time from $(0,\infty)$ as $\tau_0(x):=\inf\{s>0: L_s^{x}\le 0 \}$. Questions:

(i) Is this enough to prove that $\mathbf P[L_{\tau_0(x)}^{x} =0]=0$ ?

(ii) If (i) is not true, does it become true if we additionally assume that $L^{x}_s$ allows a density for every $s,x>0$ ?

iii) If (i) is not true, is (i) true for the Lévy case $\ell(x,y)=\ell(y)$ ? (I know $\mathbf P[L_{\tau_0(x)}^{x} =0]=0$ is true if $\ell(y) dy=\ell(dy)$ is an infinite measure)


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