Can you help me to understand a result I found in Krylov's book "Introduction to stochastic calculus". First, I will introduce some notations:
$w_t,t\ge 0$ denotes a Wiener process.
$\mathcal{B}(R)$ denotes a Borel sigma-field on real line
$\tau(\omega)=\inf\{t: w_t\not\in (a,b)\}$ is the first time Wiener process exits the interval $(a,b)$.
$B_s=w_{\tau+s}-w_{\tau}$
$$\mathcal{F}_{\le \tau}^w:=\sigma\{ \{\omega: w_{s\wedge\tau}\in B\}, s\ge 0, B\in \mathcal{B}(R)\},$$ $$ \mathcal{F}_{\ge \tau}^w:=\sigma\{ \{\omega: w_{\tau+s}-w_\tau\in B\}, s\ge 0, B\in \mathcal{B}(R)\}, $$
It is known that Wiener process has a strong Markov property, that is
- Let $\tau$ be almost surely finite $\mathcal{F}_t$-measurable stopping time then sigma-fields $ \mathcal{F}_{\le \tau}^w$ and $ \mathcal{F}_{\ge \tau}^w$ are independent.*
Theorem 1. Let $u$ be twice continuosly differentable function defined on $[-\infty, \infty]$ and $u, u', u''$ are bounded functions. Then for every $\lambda> 0$ $$ u(0)=E\int_0^{\tau}e^{-\lambda t}\left(\lambda u(w_t)-\frac{1}{2}u''(w_t)\right)dt $$ where $w_t, t\ge 0$ is a Wiener process.
I understood its proof. And based on theorem 1, the author obtains
Theorem 2. Let $-\infty<a<0<b<+\infty$, $u$ be twice continuosly differentable function defined on $[a,b]$ and $u, u', u''$ are bounded functions. Then for every $\lambda\ge 0$ $$ u(0)=E\int_0^{\tau}e^{-\lambda t}\left(\lambda u(w_t)-\frac{1}{2}u''(w_t)\right)dt+Ee^{-\lambda\tau}u(w_\tau) $$ where $w_t, t\ge 0$ is a Wiener process.
Proof
From theorem 1 we know that (denoting $f=\lambda u-\frac 12 u''$) $$ u(0)=E\int_0^{\tau}e^{-\lambda t}f(w_t)dt=E\int_0^{\tau}\ldots+E\int_\tau^{\infty}\ldots= $$ $$ =E\int_0^{\tau}e^{-\lambda t}f(w_t)+E\int_0^{\infty}e^{-\lambda t}e^{-\lambda \tau}f(w_{\tau+t})dt $$ $$ =E\int_0^{\tau}e^{-\lambda t}f(w_t)+\int_0^{\infty}e^{-\lambda t}Ee^{-\lambda \tau}f(w_{\tau}+B_t)dt=:I+J. $$
It is easy to prove that the stopping time $\tau$ is $\mathcal{F}_{\le \tau}^w$-measurable. Evidently, $w_{\tau\wedge t}$ is $\mathcal{F}_{\le \tau}^w$-measurable as well, so is $w_{\tau\wedge t}1_{\tau<\infty}$. As $\tau$ is almost surely finite, $w_{\tau\wedge t}1_{\tau<\infty}\to w_{\tau}$ as $t\to \infty$, therefore $w_{\tau}$ is $\mathcal{F}_{\le \tau}^w$-measurable too.
Question 1. Howthe author futher splits $$Ee^{-\lambda \tau}f(w_{\tau}+B_t)=Ee^{-\lambda \tau}Ef(w_{\tau}+B_t)?$$
We can split it in case of independence, but I dont see it here.
Because if this is valid, then $$ J=\int_0^{\infty}e^{-\lambda t}Ee^{-\lambda \tau}Ef(w_{\tau}+B_t)dt=Ee^{-\lambda \tau}v(w_\tau) $$ where $$ v(y):=E\int_0^{\infty}e^{-\lambda t}f(y+B_t)dt $$ Applying theorem 1 to this expectation, we obtain that $v(y)=u(y)$. So the theorem is proved for $\lambda >0$ (the theorem 1 had a limitation that $\lambda >0$).
Question 2 Author says that we can pass to the limit when $\lambda\to 0$ using dominated convergence theorem provided that $E\tau<\infty$. I dont see why it is necessary to have the finiteness of $E\tau$ to apply dominated convergence.
I would be really, really thankful for your help.