Optional stopping under uniform integrability

Fix a stopping time $\rho$ equal to either $\sigma$ or $\tau$. For $r\ge n$, F2 applied to $\rho \wedge n\le r$ gives $M_{\rho \wedge n}=\mathbb{E}[M_r\mid {\mathcal{F}}_{\rho \wedge n}]\mathrm{.}$ Let $r\to \mathrm{\infty }$ in $L^1$ using F1 and conditional contraction to obtain $M_{\rho \wedge n}=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\rho \wedge n}]\mathrm{.}$

The sigma-algebras ${\mathcal{F}}_{\rho \wedge n}$ increase with $n$. Thus the sequence in step 1.1 is a closed, hence uniformly integrable, martingale by F1. Since $\rho <\mathrm{\infty }$ almost surely, F3 gives $M_{\rho \wedge n}\to M_{\rho }$ almost surely; F1's UI convergence implication upgrades this to $L^1$, proving $M_{\rho }\in L^1$.

For $A\in {\mathcal{F}}_{\rho }$, the event $A\cap \{\rho \le n\}$ belongs to ${\mathcal{F}}_{\rho \wedge n}$ (check the defining finite-level intersections). Apply the conditional identity in step 1.1 on this event, then let $n\to \mathrm{\infty }$. The left side converges by step 2.1; the right side converges by dominated convergence for $M_{\mathrm{\infty }}{1}_{A\cap \{\rho \le n\}}$. Hence ${\int }_A{M}_{\rho }dP={\int }_A{M}_{\mathrm{\infty }}dP,$ so $M_{\rho }=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\rho }]$.

Since $\sigma \le \tau$ pointwise, F6 gives ${\mathcal{F}}_{\sigma }\subseteq {\mathcal{F}}_{\tau }$. Apply the tower property in F4 to the two representations from step 3.1: $\mathbb{E}[M_{\tau }\mid {\mathcal{F}}_{\sigma }]=\mathbb{E}[\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\tau }]\mid {\mathcal{F}}_{\sigma }]=\mathbb{E}[M_{\mathrm{\infty }}\mid {\mathcal{F}}_{\sigma }]=M_{\sigma }\mathrm{.}$ Taking expectations finishes. AC is the background hypothesis recorded in F5; the conditional-expectation identities come from F1 and F4.