I read "Probability with Martingales" by David Williams. I have trouble to understand the proof of 12.14.
Let $M$ be $\mathcal{L}^2$-martingale with $M_0 = 0$. Of courese, $M^2$ has a doob decomposition: $M^2 = M^2_0 + N + A$. Here, $A$ is an increasing previsible process with $A_0 = 0$.
"Increasing process" means that \begin{equation} \mathbf{P}(A_n \leq A_{n+1}, \forall n) = 1 \end{equation} This means that $\mathbf{P}(\{\omega \in \Omega \mid A_n(\omega) \leq A_{n+1}(\omega), \forall n\}) = 1$.
My question is following:
"Is the process $\displaystyle{\left(\frac{1}{1+A}\right)}$ bounded?"
In other words, does the formula below holds?:
\begin{equation} \left(\frac{1}{1+A_n(\omega)}\right) \leq \left(\frac{1}{1+A_0(\omega)}\right) = 1 \quad (\forall \omega, \forall n) \end{equation}
or I could say my question as:
"Is $A$ non-negative?"
I don't think it holds because $A$ is no more than $\textbf{almost surely}$ increasing. There should be a null set such that $\{\omega \in \Omega \mid A_n(\omega) > A_{n+1}(\omega), \exists n\}$. Therefore, if I take $\omega \in \{\omega \in \Omega \mid A_n(\omega) > A_{n+1}(\omega), \exists n\}$, it may be happen, for example, that \begin{equation} \left(\frac{1}{1+A_1(\omega)}\right) > \left(\frac{1}{1+A_0(\omega)}\right) = 1 \end{equation} this contradicts to $\displaystyle{\left(\frac{1}{1+A}\right)}$ is bounded. So I cannot deduce "bounded".
Thank you.
Formally you are right. What Williams means is that the process is bounded almost surely. This is enough to prove the theorem anyway. (he proves that a limit exists almost surely, so it's not a problem to ignore a set of probability zero here)