Quadratic variation of the Brownian motion and Itō's lemma Let


*

*$(\Omega,\mathcal A,\operatorname P)$ be a probability space

*$B=(B_t)_{t\ge 0}$ be a Brownian motion on $(\Omega,\mathcal A,\operatorname P)$

*$\mathcal P$ be a sequence of countable subsets $$\mathcal P^n:=\left\{\cdots<t_i^{(n)}<t_{i+1}^{(n-1)}<\cdots\right\}$$ of $[0,\infty)$ such that


*

*$0\in\mathcal P^n\subseteq\mathcal P^{n+1}$

*$\sup\mathcal P^n=\infty$

*$\displaystyle|\mathcal P^n|:=\sup_{t\in\mathcal P^n}\min_{s\in\mathcal P^n:s\ne t}|s-t|\stackrel{n\to\infty}{\to}0$


*$\mathcal P_T^n:=\mathcal P^n\cap [0,T)$

*$t':=t_{i+1}^{(n)}\wedge T$, if $t=t_i^{(n)}$


How can we show, that $$\sum_{t\in\mathcal P_T^n}f(B_t)(B_{t'}-B_t)^2\stackrel{n\to\infty}{\to}\int_0^Tf(B_s)\;ds$$ in probability?

I've found a proof (Theorem 7.12), but I absolutely don't understand what they are doing. They state, that "[if $\tau$ is] the first exit time from a compact interval [...] it suffices to prove the statement for [the] Brownian motion stopped at $\tau$".
What do they mean by "first exit from a compact interval $K$? Maybe $$\tau:=\inf\left\{t\ge 0:B_t\not\in K\right\}\;?$$
Why is that sufficient?
 A: Define $\tau_L = \inf\{t \geq 0, |B_t|>L\}$. Note that $\tau_L \uparrow \infty$ and therefore $\Bbb{P}(\tau_L<n) \xrightarrow[L \to \infty]{} 0$ by the bounded convergence theorem.
That is sufficient because once you prove 
$$ \lim \sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge T - t^{(n)}_{j}\wedge T) \xrightarrow[n\to \infty]{\text{prob}}\sum\int_0^{t\wedge T} f(B_s)\, ds$$
You will know that
$$  \Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1}  - t^{(n)}_{j}) -\sum\int_0^{t} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) \leq \\\Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge \tau_L - t^{(n)}_{j}\wedge \tau_L) -\sum\int_0^{t\wedge \tau_L} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) + \Bbb{P}(t\geq \tau_L)$$
$\Bbb{P}(t \geq \tau_L) \xrightarrow[L\to \infty]{} 0$
Choose $L $ such that $\Bbb{P}(t \geq \tau_L) <\epsilon/2$ and then choose $n$ such that
$$\Bbb{P}\bigg(\bigg\vert\lim\sum_{j=1}^n f(B_{t^{(n)}_j})(t^{(n)}_{j+1} \wedge \tau_L - t^{(n)}_{j}\wedge \tau_L) -\sum\int_0^{t\wedge \tau_L} f(B_s)\, ds \bigg\vert \geq \epsilon\bigg) <\epsilon/2.$$
