on the set $\{t<\tau\}$. Now we fix $\omega\in\Omega$ such that $t<\tau(\omega)$ and $\beta_{t}(\omega)\neq0$. Then the set $\{ t_{n}:\, n\in\mathbb{N}\} \times(t,+\infty]\times\{ \beta_{t_{n}}(\omega):\, n\in\mathbb{N}\} $ is obviously contained in a compact subset of $(0,+\infty)\times[0,+\infty]\times\mathbb{R}\backslash\{0\}$ (depending on $\omega$). This implies that $\varphi_{t_{n}}(r,\beta_{t_{n}}(\omega))$ is bounded (by a constant depending on $\omega$). Using Lebesgue’s bounded convergence theorem we can conclude $$\lim_{n\rightarrow\infty}\int_{\left(t_{n},u\right]}\varphi_{t_{n}}\left(r,\beta_{t_{n}}\left(\omega\right)\right)dF\left(r\right)=\int_{\left(t,u\right]}\varphi_{t}\left(r,\beta_{t}\left(\omega\right)\right)dF\left(r\right).$$ Consequently, we have proven that $$\lim_{n\rightarrow\infty}\int_{\left(t_{n},u\right]}\varphi_{t_{n}}\left(r,\beta_{t_{n}}\right)dF\left(r\right)=\int_{\left(t,u\right]}\varphi_{t}\left(r,\beta_{t}\right)dF\left(r\right)\,\textrm{on }\left\{ t<\tau\right\} \;\mathbf{P}\textrm{-a.s.}$$ This is also valid for intervals $(t_{n},+\infty)$and $(t,+\infty)$ (instead of $(t_{n},u]$ and $(t,u]$). Relation (\[eq:star2\]) follows immediately.

Let us conclude the first part of the proof by showing that equality (\[eq:star3\]) is indeed true.

*Proof of (\[eq:star3\]).  * Recall that $t<t_{n}<u$, $n\in\mathbb{N}$. We start by noting that the function $$y\mapsto p\left(\frac{r-u}{r-t_{n}}\left(u-t_{n}\right),\, y,\,\frac{r-u}{r-t_{n}}\beta_{t_{n}}\right)$$ is a density on $\mathbb{R}$ for all $n$. Denoting by $N(\mu,\sigma^{2})$ the normal distribution with expectation $\mu$ and variance $\sigma^{2}$, it follows that the probability measures $N(\frac{r-u}{r-t_{n}}\beta_{t_{n}}(\omega),\frac{r-u}{r-t_{n}}(u-t_{n}))$ converge weakly to $N(\frac{r-u}{r-t}\beta_{t}(\omega),\frac{r-u}{r-t}(u-t))$. Since the function $g$ is bounded by $M$, we have that $$\begin{aligned}
|G_{t_n,u}(r,\beta_{t_n}(\omega))|&=\left|\int_{\mathbb{R}}g\left(y\right)p\left(\frac{r-u}{r-t_{n}}\left(u-t_{n}\right),\, y,\,\frac{r-u}{r-t_{n}}\beta_{t_{n}\left(\omega\right)}\right)dy\right|\nonumber\\
&\leq M<+\infty\,.\label{eq:HJ1}\end{aligned}$$ Furthermore, $$\begin{aligned}
\lefteqn{\lim_{n\rightarrow\infty} G_{t_n,u}\left(r,\beta_{t_n}(\omega)\right)\nonumber}\\
&=\lim_{n\rightarrow\infty}\int_{\mathbb{R}}g\left(y\right)p\left(\frac{r-u}{r-t_{n}}\left(u-t_{n}\right),\, y,\,\frac{r-u}{r-t_{n}}\beta_{t_{n}}\left(\omega\right)\right)dy\nonumber\\
&=\int_{\mathbb{R}}g\left(y\right)p\left(\frac{r-u}{r-t}\left(u-t\right),\,y,\,\frac{r-u}{r-t}\beta_{t}\left(\omega\right)\right)dy
=G_{t,u}\left(r,\beta_{t}(\omega)\right)\label{eq:HJ2}\end{aligned}$$ immediately follows from the assumption that $g$ is bounded and continuous combined with the weak convergence of the Gaussian measures stated above. Now (\[eq:star3\]) can be derived using 