i have a question on the proof of Theorem 2.6 in the book "Stochastic Calculus and Financial Applications" by J. Michael Steele. Here a segment:
I don't understand why $A_{ab}$ is a subset of the given set above. Can someone please give a proof? Thanks!
There is a typo in the proof; the inclusion $$A_{a,b} \subseteq \left\{\omega; \sup_{k \geq m} |M_k(\omega)-M_m(\omega)| \geq \epsilon \right\}$$ hold for all $\epsilon<(b-a)/2$ (and not for $\epsilon>(b-a)/2$).
Let $\omega \in A_{a,b}$, i.e.
$$\liminf_{n \to \infty} M_n(\omega) \leq a < b \leq \limsup_{n \to \infty} M_n(\omega).$$
By the definition of liminf, there exists for any $\varrho>0$ and $m \in \mathbb{N}$ a number $k \geq m$ such that
$$M_k(\omega) \leq a+\varrho.$$
Similarly, by the definition of limsup, there exists $\ell \geq m$ such that
$$M_{\ell}(\omega) \geq b-\varrho.$$
Consequently,
$$|M_{\ell}(\omega)-M_k(\omega)| \geq M_{\ell}(\omega)-M_k(\omega) \geq b-a-2 \varrho.$$
This implies
$$\begin{align*} b-a-2\varrho &\leq |M_{\ell}(\omega)-M_k(\omega)| \\ &\leq |M_{\ell}(\omega)-M_m(\omega)| + |M_m(\omega)-M_k(\omega)| \\ &\leq 2 \sup_{j \geq m} |M_j(\omega)-M_m(\omega)|,\end{align*}$$
i.e.
$$\sup_{j \geq m} |M_j(\omega)-M_m(\omega)| \geq \frac{b-a}{2} - \varrho.$$
Choosing $\varrho>0$ sufficiently small, we find that
$$\sup_{j \geq m} |M_j(\omega)-M_m(\omega)| \geq \epsilon$$
for any $\epsilon \in (0,(b-a)/2)$.