I am reading the proof of theorem 3.8 in "Brownian Motion and Stochastic Calculus" from Karatzas and Shreve.
Following is something I don't understand.
Here $U_{[0,n](\alpha,\beta,X(\omega))}$ denotes the number of upcrossings in the interval $[0,n]$ between levels $\alpha$ and $\beta$ by the path $X(\omega)$. I am not sure how $A^{(n)}$ contains the points for which $\liminf$ is stricly less than $\limsup$. Can anyone clarify ? Why do we have upcrossings and not necessarily downcrossings ?
Definition of $U$ :
Let $F$ be a finite subset of $I$, then define
$U_F(\alpha,\beta,X(\omega))$ to be number of upcrossings between levels $\alpha$ and $\beta$ in the sequence $\{X_f(\omega) :f \in F\}$.
Then $U_I(\alpha,\beta,X(\omega)) = \sup \{U_F(\alpha,\beta,X(\omega): F\text{ is a finite subset of } I\}$.
Best Answer
If $\omega$ is in the set $$ \Big\{\omega\in\Omega;\underline{\lim}\limits_{s\uparrow t}X_s(\omega)<\overline{\lim}\limits_{s\uparrow t}X_s(\omega)\,,\text{ for some }t\in[0,n]\Big\} $$ then there exist two rationals $\alpha<\beta$ such that for some $t\in[0,n]\,,$ $$ \underline{\lim}\limits_{s\uparrow t}X_s(\omega)<\alpha<\beta<\overline{\lim}\limits_{s\uparrow t}X_s(\omega)\,. $$ By the very definition of $\liminf$ and $\limsup$ this means that the path $s\mapsto X_s(\omega)$ must be infinitely often below $\alpha$ and infinitely often above $\beta$. In other words, the number of upcrossings of $[\alpha,\beta]$ must be infinite. That is $\omega$ is in $A^{(n)}_{\alpha,\beta}$.
Regarding your question about downcrossings: Please repeat in detail the proof of K&S Thm. 3.8 using downcrossings to find that out.
Also in the future: don't post pictures. Type everything in MathJax. Keep defintions self contained. Non one runs around with their copy of K&S all the time.