> 0$, then the claim follows by choosing $N$ large enough and setting $\epsilon = 1/N$ and $E = \set{\tau \ge s+1/N}$. But by right-continuity of $X$, the set $\set{\omega:\tau\brac{\omega} > s}$ is equal to $\set{\omega:\Theta_s(\omega) \text{ is regular}}$ and it remains to show that $\Theta_s$ is regular with strictly positive probability. By Condition \[def:full\_support\] it holds that $\conv(\supp(X_s)) = D$ and we can find $d+1$ convex independent points[^3] $\xi^0, \dotsc, \xi^d$ in $\supp(X_s)$. Recalling the definition of $H$ in , it follows that $H(\xi^0, \dotsc, \xi^d)$ is regular. Since the set of regular matrices is open we find $\delta > 0$ such that even $H(y_0, \dotsc, y_d)$ is regular for all $y_i \in U_{\delta}(\xi_i), i \in \set{0, \dotsc, d}$, where $U_{\delta}(\xi_i)$ is the open ball of radius $\delta$ centered at $\xi_i$. Now, by independence of $X^0, \dots, X^d$, it follows that $$\begin{aligned}
 \PP \left(\Theta_{s} \; \text{is regular}\right) & \ge \PP \Big(X_s^i \in U_\delta(\xi_i) \quad \forall\, i \in \set{0, \dotsc, d}\Big)\\
 &= \prod_{i=0}^d \PP\left(X_s \in U_\delta(\xi_i)\right). 
 \end{aligned}$$ Since for each $i \in \set{0, \dotsc, d}$ the intersection of $U_\delta(\xi_i)$ with the support of $X_s$ is non-empty, all probabilities are strictly positive, and the proof is complete.

Similar to the $\RR^{(d+1) 