Reverse martingale convergence

Fix $N$. Read $X_N,X_{N-1},\dots ,X_0$ in that order with filtration ${\mathcal{G}}_N\subseteq \cdots \subseteq {\mathcal{G}}_0$; F1 makes this a finite ordinary martingale. An upcrossing of $X_0,\dots ,X_N$ becomes a downcrossing of the reversed list, hence an upcrossing of its negative through $[-b,-a]$. F3 bounds its expectation by endpoint positive parts, uniformly in $N$, because F2 gives uniform $L^1$ bounds. The same argument directly bounds downcrossings.

For every rational $a<b$, the total upcrossing and downcrossing counts are finite almost surely. Intersecting these countably many full-measure events, the usual rational-interval argument gives a finite or extended-real limit. Uniform integrability bounds the positive and negative tails uniformly, so Fatou excludes both infinite values. Denote the finite almost-sure limit by $X_{\mathrm{\infty }}$.

The almost-sure convergence from step 2.1 gives convergence in probability by F4. Using the uniformly integrable family from F2, F4 then upgrades it to $X_n\to X_{\mathrm{\infty }}$ in $L^1$.

Fix $r$. For all $n\ge r$, $X_n$ is ${\mathcal{G}}_n$-measurable and ${\mathcal{G}}_n\subseteq {\mathcal{G}}_r$, so F5 makes $X_{\mathrm{\infty }}$ measurable for ${\mathcal{G}}_r$. This holds for every $r$, hence $X_{\mathrm{\infty }}$ is ${\mathcal{G}}_{\mathrm{\infty }}$-measurable. If $A\in {\mathcal{G}}_{\mathrm{\infty }}$, then $A\in {\mathcal{G}}_n$ for every $n$ and F1 gives ${\int }_A{X}_{n}dP={\int }_A{X}_{0}dP\mathrm{.}$ The $L^1$ limit passes through the left integral, so F5 identifies $X_{\mathrm{\infty }}=\mathbb{E}[X_0\mid {\mathcal{G}}_{\mathrm{\infty }}]$. AC is used exactly as recorded in F6.