Here is a proof coming from a french Math forum :
http://www.les-mathematiques.net/phorum/read.php?12,376956,377123#msg-377123
I translate the solution for the non french readers.
So here comes the solution (credit goes to egoroff) :
For $T(\omega)<\infty$, fix $\mathcal{H}$ as the collection of sets that do not separate $\omega$ and $\omega'$, i.e. sets $A$ s.t. either $\{\omega,\omega'\}\in A$ or $ \in A^c$. Then it is easy to see that $\mathcal{H}$ is a $\sigma$-field.
This was the first step. Next for every $(n+1)$-tuple $t_0<...<t_n\le T(\omega)$ and every Borel sets $A_{t_i}$, the set $(X_{t_i})_{i=0,...,n}\in \Pi_{i=0}^n A_{t_i}$ is in $\mathcal{H}$, by hypothesis over $\omega$ and $\omega'$, so $\mathcal{F}_t\subset \mathcal{H}$ for every $t\le T(\omega)$ as those set generate $\mathcal{F}_t$ .
Now $T(\omega)$ is known and finite we have :
-$S=T\wedge T(\omega)$ is a stopping time and moreover $S\in \mathcal{F}_{T(\omega)}\subset \mathcal{H}$ we have $S(\omega)=S(\omega')$, and so $T(\omega)\le T(\omega')$.
-On the other hand the event $\{T\le T(\omega)\}$ is in $\mathcal{F}_{T(\omega)}$, as $T$ is a stopping time so it is in $\mathcal{H}$, and $\omega\in \{T\le T(\omega)\}$ and so $\omega'$ too, and $T(\omega')\le T(\omega)$.
Finally we have shown that $T(\omega)=T(\omega')$ over $T(\omega)<\infty$ which was the claim to be proved.
Best regards
PS :
I also have a solution of mine based on a variant of Doob's lemma but as it is longer, more technical and far less elegant than this one, I do not post it here.
$(ii) \implies (iii):$ You've assumed that $X_t \to X_\infty$ in $L^1$ as $t \to \infty$. We can then write
$$\bigg |\int_A X_t d \mathbb{P} - \int_A X_\infty d \mathbb{P} \bigg |\leq \int_A |X_t - X_\infty| d \mathbb{P} \leq \|X_t - X_\infty\|_{L^1} \to 0$$
to see that $\int_A X_t d\mathbb{P} \to \int_A X_\infty d\mathbb{P}$.
$(iii) \implies (i):$ To show uniform integrability, we want to show that for $\varepsilon > 0$ there is a $\lambda$ such that
$$\sup_t \int_{\{|X_t| \geq \lambda\}} |X_t| d \mathbb{P} \leq \varepsilon.$$
The authors do this using the standard fact that for $Y \in L^1$, $\int_A |Y| d \mathbb{P} \to 0$ as $\mathbb{P}(A) \to 0$ in the sense that for $\varepsilon > 0$, there is a $\delta>0$ such that $\mathbb{P}(A) < \delta$ implies that $\int_A |Y| d \mathbb{P} < \varepsilon$.
Here is a proof of that fact in your specific case. We have
\begin{align}
\int_{\{|X_t| \geq \lambda\}} |X_\infty| d\mathbb{P} =& \int_{\{|X_t| \geq \lambda\}\cap \{|X|_\infty < K\}} |X_\infty| d\mathbb{P} + \int_{\{|X_t| \geq \lambda\} \cap \{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P}
\\ \leq & K \mathbb{P}(|X_t| \geq \lambda) + \int_{\{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P}.
\end{align}
We can pick $K$ such that $\int_{\{|X|_\infty \geq K\}} |X_\infty| d\mathbb{P} < \frac{\varepsilon}{2}$. For this $K$ we have shown that
$$\sup_t \int_{\{|X_t| \geq \lambda\}} |X_t| d \mathbb{P} \leq K \sup_t \mathbb{P}(|X_t| \geq \lambda) + \frac{\varepsilon}{2}.$$ Since $\mathbb{P}(|X_t| \geq \lambda) \to 0$ uniformly in $t$ as $\lambda \to \infty$, the right hand side of this is less than $\varepsilon$ for $\lambda$ large enough.
Best Answer
There are fewer downcrossings than upcrossings if $X_0 < \alpha$ and at most one more if $X_0 \ge \alpha$, so once you bound upcrossings you've also bound downcrossings.