where we recall that $\mathcal{F}_{v}^{P}=\sigma(\beta_{s},\,0\leq s\leq v)\vee\mathcal{N}_{P}$, $v\geq 0$. Consequently, $$\mathbf{E}\left[g\left(\beta_{u}\right)|\mathcal{F}_{t}^{\beta}\right]=\lim_{n\rightarrow\infty}\mathbf{E}\left[g\left(\beta_{u}\right)|\mathcal{F}_{t_{n}}^{P}\right]\,, \; \mathbf{P}\textrm{-a.s.}$$ From Theorem \[thm:Let–and\] and the definition of $G_{t_n,u}$ by we know that, as $t_{n}<u$, $\mathbf{P}$-a.s. $$\begin{aligned}
\nonumber\lefteqn{\mathbf{E}\left[g\left(\beta_{u}\right)|\mathcal{F}_{t_{n}}^{P}\right]=\mathbf{E}\left[g\left(\beta_{u}\right)|\beta_{t_{n}}\right]}\\
&=&g\left(0\right)\mathbb{I}_{\left\{ \tau\leq t_{n}\right\} }+g\left(0\right)\int_{\left(t_{n},u\right]}\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)\,\mathbb{I}_{\left\{ t_{n}<\tau\right\} }\label{eq:equiv1}\\
\nonumber&&+\int_{\left(u,+\infty\right)}G_{t_n,u}\left(r,\beta_{t_n}\right)\,\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)\,\mathbb{I}_{\left\{ t_{n}<\tau\right\} }\,.\end{aligned}$$ We want to prove that $$\lim_{n\rightarrow\infty}\mathbf{E}\left[g\left(\beta_{u}\right)|\mathcal{F}_{t_{n}}^{P}\right]=\mathbf{E}\left[g\left(\beta_{u}\right)|\beta_{t}\right],\;\mathbf{P}\textrm{-a.s.}$$ Using (\[eq:equiv1\]) and Theorem \[thm:Let–and\] we see that this latter relation holds true if the following two identities are satisfied, $\mathbf{P}$-a.s. on $\{t<\tau\}$: $$\begin{aligned}
\lim_{n\rightarrow\infty}\int_{\left(t_{n},u\right]}\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)&= \int_{\left(t,u\right]}\phi_{t}\left(r,\beta_{t}\right)dF\left(r\right)\,,\label{eq:star2}\\
\lim_{n\rightarrow\infty}\int\limits_{\left(u,+\infty\right)}\!\!\!G_{t_n,u}\left(r,\beta_{t_{n}}\right)\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)&=\!\!\! \int\limits_{\left(u,+\infty\right)}\!\!\!G_{t,u}\left(r,\beta_{t}\right)\phi_{t}\left(r,\beta_{t}\right)dF\left(r\right)\,.\label{eq:star3}\end{aligned}$$ Relation (\[eq:star2\]) can be derived as follows.

*Proof of (\[eq:star2\]).* Recalling that by (\[eq:densitybeta2\]) $$\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)=\frac{\varphi_{t_{n}}\left(r,\beta_{t_{n}}\right)}{{\displaystyle \int_{\left(t_{n},+\infty\right)}}\varphi_{t_{n}}\left(v,\beta_{t_{n}}\right)dF(v)},\quad t_n<r\,,$$ the integral on the left-hand side of (\[eq:star2\]) can be rewritten as $$\int_{\left(t_{n},u\right]}\phi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)=\frac{{\displaystyle \int_{\left(t_{n},u\right]}}\varphi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF(r)}{{\displaystyle \int_{\left(t_{n},+\infty\right)}}\varphi_{t_{n}}\left(v,\beta_{t_{n}}\right)dF(v)}\,.$$ The plan is to apply Lebesgue’s bounded convergence theorem to the numerator and the denominator of the above expression. To this end we have to prove $\mathbf{P}$-a.s. pointwise convergence and uniform boundedness of the integrand $\varphi_{t_n}(\cdot, \beta_{t_n})$ $\mathbf{P}$-a.s. We begin by focusing our attention on the function $(t,r,x)\mapsto\varphi_{t}(r,x)$ defined by , which is continuous on $(0,+\infty)\times[0,+\infty)\times\mathbb{R}\backslash\{ 0\}$. Setting $\varphi_{t}(+\infty,x):=p(t,x,0)$ for every $t>0$ and $x\in\mathbb{R},$ we see that the resulting function $(t,r,x)\mapsto\varphi_{t}(r,x)$, now defined on $(0,+\infty)\times[0,+\infty]\times\mathbb{R}\backslash\{ 0\}$, is continuous, too. Hence $\lim_{n\rightarrow \infty}\varphi_{t_{n}}(r,\beta_{t_{n}})=\varphi_{t}(r,\beta_{t})$, $\mathbf{P}$-a.s. on $\tau>t$, providing pointwise convergence. For this we note that $\beta_{t}\neq0$ $\mathbf{P}$-a.s. 