Proof of Optional Stopping for Bounded Right-Continuous Square-Integrable Martingales on a Compact Time Interval

Throughout, the stopping-time toolkit is cited by claim number, $D_n=\{kT2^{-n}:k\in\{0,\dots,2^n\}\}$ is the $n$-th dyadic grid, and $\sigma_n,\tau_n$ are the dyadic approximations from above of $\sigma,\tau$ (toolkit, claim 3). A real-valued function on $\Omega$ is called $\mathcal{G}$-measurable, for a $\sigma$-algebra $\mathcal{G}\subseteq\mathcal{F}$, when it is measurable with respect to $\mathcal{G}$ and the Borel $\sigma$-algebra of the real line. Every random variable $U$ with $|U|\le C$ everywhere is square-integrable ($\mathbb{E}[U^2]\le C^2$ by monotonicity of the integral), hence integrable; this is used without further comment for all the bounded random variables below.

Step 0: two forms of right-continuity. Let $\omega\in\Omega$ be a point at which the path $t\mapsto M_t(\omega)$ is right-continuous at every $t\in[0,T)$ in the sense of the hypothesis (in particular, any $\omega\in\Omega_0$). The path is then right-continuous in the sequential sense of the toolkit: let $t\in[0,T]$ and let $(s_j)$ be a sequence in $[t,T]$ converging to $t$. If $t=T$ then $s_j=T$ for all $j$ and there is nothing to prove. If $t<T$, given $\varepsilon>0$ choose $\eta>0$ from the hypothesis; for all large $j$ we have $t\le s_j\le\min(t+\eta,T)$, hence $|M_{s_j}(\omega)-M_t(\omega)|\le\varepsilon$. Conversely, if a path is right-continuous in the sequential sense at every point of $[0,T]$, it is right-continuous at every $t\in[0,T)$ in the sense of the hypothesis: otherwise there would be $t\in[0,T)$, $\varepsilon>0$ and, for each $j\in\mathbb{N}$, a point $s_j\in[t,\min(t+1/j,T)]$ with $|M_{s_j}(\omega)-M_t(\omega)|>\varepsilon$, contradicting the sequential form along $(s_j)$.

Step 1: measurability and convergence of the dyadic samples (claim (a)). Since the pair $\sigma\le\tau$ in the statement is arbitrary, whatever is proved below for $\tau$ (respectively for the pair $\sigma\le\tau$) holds for every stopping time $\rho$ (respectively every pair $\rho\le\rho'$) of $(\mathcal{F}_t)_{t\in[0,T]}$; claims (a) and (b) are used in that generality in Step 4. Fix $n$. At every $\omega$ exactly one of the sets $\{\tau_n=d\}$, $d\in D_n$, contains $\omega$, so pointwise

Each $\{\tau_n=d\}$ lies in $\mathcal{F}_d\subseteq\mathcal{F}_T$ (toolkit, claim 1) and each $M_d$ is $\mathcal{F}_d$-measurable, so $M_{\tau_n}$ is an $\mathcal{F}_T$-measurable random variable by claims 1--3 of the arithmetic lemma for measurable functions, with $|M_{\tau_n}|\le K$. For $\omega\in\Omega_0$, $(\tau_n(\omega))_n$ is a sequence in $[\tau(\omega),T]$ converging to $\tau(\omega)$ (toolkit, claim 3), so Step 0 gives $M_{\tau_n}(\omega)\to M_\tau(\omega)$. Hence $M_{\tau_n}\mathbf{1}_{\Omega_0}\to M_\tau\mathbf{1}_{\Omega_0}$ at every point of $\Omega$ (both sides vanish off $\Omega_0$), and the limit is a random variable on $(\Omega,\mathcal{F},P)$ by claim 5 of the arithmetic lemma ($\mathbf{1}_{\Omega_0}$ being $\mathcal{F}$-measurable as $\Omega_0\in\mathcal{F}$), bounded by $K$. The same applies to $\sigma$. If every path of $M$ is right-continuous at every $t\in[0,T)$, then by Step 0 every path is right-continuous in the sequential sense, $M$ is adapted, so $M$ is progressively measurable by claim 2 of the progressive measurability toolkit, and $M_\tau$ is $\mathcal{F}_\tau$-measurable by claim 4(ii) of the stopping-time toolkit. This proves (a).

Step 2: optional stopping on the dyadic grid. Fix $n$ and list $D_n$ as $t_0<t_1<\dots<t_{2^n}$. By toolkit claim 3, $\sigma_n\le\tau_n$ pointwise. For every $\omega$, with $\sigma_n(\omega)=t_i$ and $\tau_n(\omega)=t_{i'}$, $i\le i'$, the indicator of $\{\sigma_n\le t_k\}\cap\{\tau_n>t_k\}$ at $\omega$ equals $1$ exactly for $i\le k\le i'-1$, so the telescoping identity

holds pointwise (an empty sum when $i=i'$). Let $D\in\mathcal{F}_\sigma$. Since $\sigma\le\sigma_n$, toolkit claim 3 gives $D\in\mathcal{F}_{\sigma_n}$, so $D\cap\{\sigma_n\le t_k\}\in\mathcal{F}_{t_k}$ by the definition of the prior $\sigma$-algebra; and $\{\tau_n>t_k\}\in\mathcal{F}_{t_k}$ by toolkit claim 1. Hence $A_k=D\cap\{\sigma_n\le t_k\}\cap\{\tau_n>t_k\}\in\mathcal{F}_{t_k}$, and the martingale property in averaged form of the martingale definition, for the times $t_k\le t_{k+1}$, gives $\mathbb{E}[M_{t_{k+1}}\mathbf{1}_{A_k}]=\mathbb{E}[M_{t_k}\mathbf{1}_{A_k}]$. Multiplying the telescoping identity by $\mathbf{1}_D$, taking expectations (all terms are bounded random variables) and using linearity of the integral,

Step 3: passage to the limit (claim (b)). Put $g_n=(M_{\tau_n}-M_{\sigma_n})\mathbf{1}_D$ and $f_n=g_n\mathbf{1}_{\Omega_0}$. Both are random variables bounded by $2K$ everywhere and agree on $\Omega_0$, so $\mathbb{E}[f_n]=\mathbb{E}[g_n]=0$ by claim 2 of the lemma on almost sure equality of bounded random variables and Step 2. By Step 1, $f_n\to f=(M_\tau-M_\sigma)\mathbf{1}_{\Omega_0}\mathbf{1}_D$ at every point of $\Omega$, and $|f_n|\le2K$ with the constant $2K$ integrable on the probability space. The dominated convergence theorem yields $\mathbb{E}[f]=\lim_n\mathbb{E}[f_n]=0$, which by linearity is the identity of (b). For the particular case take $\sigma$ to be the constant $0$ (a stopping time with $\sigma\le\tau$, toolkit claim 1) and $D=\Omega$, which lies in $\mathcal{F}_\sigma=\mathcal{F}_0$ by toolkit claim 2: then $\mathbb{E}[M_\tau\mathbf{1}_{\Omega_0}]=\mathbb{E}[M_0\mathbf{1}_{\Omega_0}]=\mathbb{E}[M_0]$, the last equality again by claim 2 of the almost-sure-equality lemma. Finally, suppose every path of $M$ is right-continuous at every $t\in[0,T)$. By Step 1, $M_\sigma$ is $\mathcal{F}_\sigma$-measurable and $M_\tau$ is a random variable, both bounded by $K$, hence square-integrable; and for $D\in\mathcal{F}_\sigma$,

the outer equalities by the almost-sure-equality lemma and the middle one by (b). These are conditions (i)--(iii) of the definition of conditional expectation, so $M_\sigma$ is a conditional expectation of $M_\tau$ given $\mathcal{F}_\sigma$.

Step 4: the stopped process (claim (c)). Claims (a) and (b) are used here for arbitrary stopping times, as justified at the start of Step 1. Fix $0\le s\le t\le T$ and $D\in\mathcal{F}_s$. Put $\sigma'=\min(s,\tau)$ and $\tau'=\min(t,\tau)$, stopping times by toolkit claim 1 with $\sigma'\le\tau'$ pointwise, so that $M^\tau_s=M_{\sigma'}$ and $M^\tau_t=M_{\tau'}$. By Step 1 applied to $\sigma'$ and $\tau'$, both $M_{\sigma'}\mathbf{1}_{\Omega_0}$ and $M_{\tau'}\mathbf{1}_{\Omega_0}$ are random variables bounded by $K$. Split $D$ into $E_1=D\cap\{\tau\le s\}$ and $E_2=D\cap\{\tau>s\}$; both lie in $\mathcal{F}_s$ ($\{\tau\le s\}$ by the definition of a stopping time, $\{\tau>s\}$ by toolkit claim 1), so $\mathbf{1}_{E_1}$ and $\mathbf{1}_{E_2}$ are random variables and $\mathbf{1}_D=\mathbf{1}_{E_1}+\mathbf{1}_{E_2}$. On $E_1$ we have $\min(t,\tau)=\tau=\min(s,\tau)$, so $M_{\tau'}\mathbf{1}_{\Omega_0}\mathbf{1}_{E_1}=M_{\sigma'}\mathbf{1}_{\Omega_0}\mathbf{1}_{E_1}$ pointwise. By the splitting assertion of toolkit claim 2, $E_2\in\mathcal{F}_{\sigma'}$, so claim (b) applied to the stopping times $\sigma'\le\tau'$ and the event $E_2$ gives $\mathbb{E}[M_{\tau'}\mathbf{1}_{\Omega_0}\mathbf{1}_{E_2}]=\mathbb{E}[M_{\sigma'}\mathbf{1}_{\Omega_0}\mathbf{1}_{E_2}]$. Adding the two contributions by linearity of the integral yields the identity of (c).

Suppose finally that every path of $M$ is right-continuous at every $t\in[0,T)$. As in Step 1, $M$ is progressively measurable, so $M^\tau$ is adapted by toolkit claim 4(i); $|M^\tau_t(\omega)|=|M_{\min(t,\tau(\omega))}(\omega)|\le K$ for all $t,\omega$, so each $M^\tau_t$ is square-integrable; and for $s\le t$ and $D\in\mathcal{F}_s$ the identity of (c) together with claim 2 of the almost-sure-equality lemma (applied on each side to remove the factor $\mathbf{1}_{\Omega_0}$) gives $\mathbb{E}[M^\tau_t\mathbf{1}_D]=\mathbb{E}[M^\tau_s\mathbf{1}_D]$, the martingale property in averaged form. Thus $M^\tau$ is a square-integrable martingale with respect to $(\mathcal{F}_t)_{t\in[0,T]}$, with time index restricted to $[0,T]$. By Step 0 every path of $M$ is right-continuous in the sequential sense, so by toolkit claim 4(iii) every path of $M^\tau$ is right-continuous in the sequential sense, hence, by the converse direction of Step 0, right-continuous at every $t\in[0,T)$ in the sense of the hypothesis. $\blacksquare$