Extension of the stochastic integral in the book "PDE and Martingale Methods in Option Pricing"

Question

In chapter 4.4 the authors of PDE and Martingale Methods in Option Pricing do the following: They take a filtered proabability space $(\Omega,\mathcal A,(\mathcal F_t)_{t\in[0,\:T]},\operatorname P)$ and an $\mathcal F$-progressively measurable $u:\Omega\times[0,T]\to\mathbb R$ with $\operatorname P\left[u\in\mathcal L^2\left(\left.\lambda^1\right|_{[0,\:T]}\right)\right]=1$ (where $\lambda^1$ denotes the Lebesgue measure on $\mathbb R$) and define $$A_t(\omega):=\begin{cases}\displaystyle\int_0^tu(\omega)^2\:{\rm d}\lambda^1&\text{, if }u(\omega)\in\mathcal L^2\left(\left.\lambda^1\right|_{[0,\:T]}\right)\\0&\text{, otherwise}\end{cases}\;\;\;\text{for }(\omega,t)\in\Omega\times[0,T]\;.$$ We can show that $A$ is an $\mathcal F$-adapted continuous stochastic process on $(\Omega,\mathcal A,\operatorname P)$. Now, they define $$\tau_n:=\inf\left\{t\in[0,T]:A_t\ge n\right\}\wedge T$$ for $n\in\mathbb N$ and state that $$\operatorname P\left[\tau_n\xrightarrow{n\to\infty}T\right]=1\tag 1$$ and $$\bigcup_{n\in\mathbb N}F_n=\Omega\setminus N\tag 2$$ for some $\operatorname P$-null set $N$ and $$F_n:=\left\{\tau_n=T\right\}=\left\{A_T\le n\right\}\;.\tag 3$$

I've got vaarious problems with $(1)$, $(2)$ and $(3)$:

1. It's clear that each $\tau_n$ is an $\mathcal F$-stopping time with $\tau_n\le\tau_{n+1}$. However, isn't it obvious that $$\tau_n\xrightarrow{n\to\infty}T\tag{3'}$$ holds surely? Why do they state that $(3')$ only holds with probability $1$?
2. Isn't the equality in $(3)$ wrong? It's clear that $t\mapsto A_t$ is increasing. Let $$\tau_n':=\inf\left\{t\in[0,T]:A_t>n\right\}\;.$$ Then, $$\left\{A_T\le n\right\}\subseteq\left\{\tau_n'=T\right\}\;,$$ but the other inclusion doesn't seem to hold. So, what's really correct?
3. What is $N$ exactly? I guess $N=\left\{\Phi\not\in\mathcal L^2\left(\left.\lambda^1\right|_{[0,\:T]}\right)\right\}$, but that doesn't make sense (for the same reason why $(1)$ doesn't make sense). On the one hand, the author set $A=0$ on this $N$. On the other hand, he acts as he hadn't done this.

Answer 1

This is an answer to 2.

Let $\omega\in\left\{\tau_n=T\right\}$. Either $A_T(\omega)\ge n$ or $A_T(\omega)<n$. In the second case, we're done. In the first case, $$A_t(\omega)<n\;\;\;\text{for all }t\in[0,T)\tag 4$$ (by definition of $\tau_n(\omega)=T$) and since $A$ is continuous, we obtain $$A_T(\omega)=\lim_{t\to T-}A_t(\omega)\le n\tag 5\;.$$ Thus, $$A_T(\omega)=n\;.\tag 6$$ So, we can conclude that $$\left\{\tau_n=T\right\}=\left\{A_T=n\right\}\cup\left\{A_T<n\right\}=\left\{A_T\le n\right\}\;.\tag 7$$

Answer 2

1. You are correct.

2. $\cup_n F_n=\{A_T<\infty\}$, but the hypotheses do not imply that $\operatorname P[A_T<\infty]=1$. Example: $T=1$ and $u(\omega,s)=(1-s)^{-1/2}$ for all $\omega$ and $0\le s<1$. Then $A_t(\omega)<\infty$ for all $\omega$ and $t\in[0,1)$, $\tau_n(\omega)=1-e^{-n}$, and $\cup_nF_n =\emptyset$. (Or is there a typo in your statement of the matter, so that we should be assuming $s\mapsto u(\omega,s)\in \mathcal L^2(\lambda|_{[0,T]})$ for a.e. $\omega$?)