Lebesgue’s bounded convergence theorem and the properties of $G_{t_n,u}(r,\beta_{t_n}(\omega))$ and $\varphi_{t_{n}}(r,\beta_{t_{n}}(\omega))$ (and hence $\phi_{t_{n}}(r,\beta_{t_{n}}(\omega))$) of $\mathbf{P}$-a.s. boundedness and pointwise convergence verified above.

The first part of the proof of the theorem is now finished. In the second part of the proof we consider the case $t=0$ which is divided into two steps. In the first step we assume that there exists $\varepsilon>0$ such that $\mathbf{P}(\tau>\varepsilon)=1$ and in the second step we will drop this condition.

Let us assume that there exists $\varepsilon>0$ such that $\mathbf{P}(\tau>\varepsilon)=1$. Let $(t_{n})_{n\in\mathbb{N}}$ be a decreasing sequence of strictly positive real numbers converging to $0$: $0<...<t_{n+1}<t_{n},\: t_{n}\downarrow0$ as $n\rightarrow\infty$. Without loss of generality we assume $t_{n}<\varepsilon$ for all $n\in\mathbb{N}$. Then (\[eq:densitybeta2\]) can be rewritten as follows: $$\label{Phi-rewritten}
\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)=\frac{\frac{1}{\sqrt{2\pi t_{n}}}\sqrt{\frac{r}{r-t_{n}}}\exp\left[-\frac{\beta_{t_{n}}^{2}r}{2t_{n}\left(r-t_{n}\right)}\right]}{{\displaystyle \int_{\left(\varepsilon,+\infty\right)}}\frac{1}{\sqrt{2\pi t_{n}}}\sqrt{\frac{s}{s-t_{n}}}\exp\left[-\frac{\beta_{t_{n}}^{2}s}{2t_{n}\left(s-t_{n}\right)}\right]dF\left(s\right)}\,\mathbb{I}_{\left(t_n,+\infty\right)}\left(r\right)\,.$$ We have the following auxiliary result:

\[lem:On-the-set-2\] Suppose that $\mathbf{P}(\tau>\varepsilon)=1$. Then the function $r\mapsto\phi_{t_{n}}(r,\beta_{t_{n}})$ is $\mathbf{P}_\tau$-a.s. uniformly bounded by some constant $K=K(\varepsilon,\omega)<+\infty$ and, for all $r>0$, $$\lim_{n\rightarrow\infty}\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)=1\,,\;\mathbf{P}\textrm{-a.s.}$$

See Appendix \[sec:Proofs-of-Section\].

Now we turn to the proof of the Markov property of $\beta$ with respect to $\mathbb{F}^\beta$ at $t=0$ under the additional assumption that $\mathbf{P}(\tau>\varepsilon)=1$. Since $\mathbf{E}\left[g(\beta_u)|\mathcal{F}^\beta_0\right]= \lim_{n\rightarrow\infty} \mathbf{E}\left[g(\beta_u)|\mathcal{F}^P_{t_n}\right]$, as in the first part, it is sufficient to verify that $$\lim_{n\rightarrow\infty}\mathbf{E}\left[g\left(\beta_{u}\right)|\mathcal{F}^P_{t_{n}}\right]=\mathbf{E}\left[g\left(\beta_{u}\right)|\beta_{0}\right],\;\mathbf{P}\textrm{-a.s.}\label{eq:equiv 2}$$ Using the formula of 