Proof of Stopping Times on a Compact Time Interval: Elementary Operations, the Prior Sigma-Algebra, Dyadic Approximation, Sampling, and Hitting Times

Throughout, for a $\sigma$-algebra $\mathcal{G}$ on $\Omega$, a real-valued function $f$ on $\Omega$ is called $\mathcal{G}$-measurable when it is measurable with respect to $\mathcal{G}$ and the Borel $\sigma$-algebra $\mathcal{B}(\mathbb{R})$; by claim 3 of the Borel generator lemma this holds if and only if $\{f>a\}\in\mathcal{G}$ for every real $a$ (the criterion). All set operations used are finite or countable, so they stay inside the $\sigma$-algebras involved ($\sigma$-algebra axioms).

Claim 1. For the constant $c\in[0,T]$ and $t\in[0,T]$, $\{c\le t\}$ is $\Omega$ if $c\le t$ and $\emptyset$ otherwise, both in $\mathcal{F}_t$. For stopping times $\sigma,\tau$: $\{\min(\sigma,\tau)\le t\}=\{\sigma\le t\}\cup\{\tau\le t\}$ and $\{\max(\sigma,\tau)\le t\}=\{\sigma\le t\}\cap\{\tau\le t\}$, both in $\mathcal{F}_t$. Let $\tau$ be a stopping time and $t\in[0,T]$. If $t=0$ then $\{\tau<0\}=\emptyset$. If $t>0$, then

$$\{\tau<t\}=\bigcup_{m\in\mathbb{N},\ t-1/m\ge0}\{\tau\le t-1/m\}:$$

the inclusion $\supseteq$ is clear, and if $\tau(\omega)<t$ there is, by the Archimedean property, a natural number $m$ with $1/m\le t-\tau(\omega)$, so $t-1/m\ge\tau(\omega)\ge0$ and $\omega$ lies in the $m$-th set. Each $\{\tau\le t-1/m\}$ lies in $\mathcal{F}_{t-1/m}\subseteq\mathcal{F}_t$ (the filtration being nondecreasing), so $\{\tau<t\}\in\mathcal{F}_t$ as a countable union. Then $\{\tau=t\}=\{\tau\le t\}\setminus\{\tau<t\}$, $\{\tau\ge t\}=\Omega\setminus\{\tau<t\}$ and $\{\tau>t\}=\Omega\setminus\{\tau\le t\}$ lie in $\mathcal{F}_t$. Finally, for real $a$: $\{\tau>a\}=\Omega$ if $a<0$; $\{\tau>a\}=\emptyset$ if $a\ge T$; and $\{\tau>a\}=\Omega\setminus\{\tau\le a\}$ with $\{\tau\le a\}\in\mathcal{F}_a\subseteq\mathcal{F}_T$ if $0\le a<T$. By the criterion, $\tau$ is $\mathcal{F}_T$-measurable.

Claim 2. $\mathcal{F}_\tau$ is a $\sigma$-algebra contained in $\mathcal{F}_T$. Containment holds by definition. $\Omega\in\mathcal{F}_\tau$ since $\Omega\cap\{\tau\le t\}=\{\tau\le t\}\in\mathcal{F}_t$. If $A\in\mathcal{F}_\tau$ then $\Omega\setminus A\in\mathcal{F}_T$ and $(\Omega\setminus A)\cap\{\tau\le t\}=\{\tau\le t\}\setminus\big(A\cap\{\tau\le t\}\big)\in\mathcal{F}_t$. If $A_1,A_2,\dots\in\mathcal{F}_\tau$ then $\bigcup_jA_j\in\mathcal{F}_T$ and $\big(\bigcup_jA_j\big)\cap\{\tau\le t\}=\bigcup_j\big(A_j\cap\{\tau\le t\}\big)\in\mathcal{F}_t$. Measurability of $\tau$. For real $a$, $\{\tau>a\}\in\mathcal{F}_T$ by claim 1, and for $t\in[0,T]$: $\{\tau>a\}\cap\{\tau\le t\}$ equals $\{\tau\le t\}$ if $a<0$, equals $\emptyset$ if $a\ge t$, and equals $\{\tau\le t\}\setminus\{\tau\le a\}$ with $\{\tau\le a\}\in\mathcal{F}_a\subseteq\mathcal{F}_t$ if $0\le a<t$; in every case it lies in $\mathcal{F}_t$. Hence $\{\tau>a\}\in\mathcal{F}_\tau$ and the criterion applies. Constants. Let $\tau\equiv c$. If $A\in\mathcal{F}_\tau$ then $A=A\cap\{\tau\le c\}\in\mathcal{F}_c$. Conversely if $A\in\mathcal{F}_c$ then $A\in\mathcal{F}_T$, and $A\cap\{\tau\le t\}$ equals $A\in\mathcal{F}_c\subseteq\mathcal{F}_t$ when $t\ge c$ and $\emptyset$ otherwise. Monotonicity. Let $\sigma\le\tau$ pointwise and $A\in\mathcal{F}_\sigma$. Since $\{\tau\le t\}\subseteq\{\sigma\le t\}$, we get $A\cap\{\tau\le t\}=\big(A\cap\{\sigma\le t\}\big)\cap\{\tau\le t\}\in\mathcal{F}_t$, and $A\in\mathcal{F}_T$; so $A\in\mathcal{F}_\tau$. Splitting. Let $s\in[0,T]$ and $A\in\mathcal{F}_s$; put $A'=A\cap\{\tau>s\}$, which lies in $\mathcal{F}_s\subseteq\mathcal{F}_T$ by claim 1. Let $t\in[0,T]$. If $t\ge s$ then $\{\min(s,\tau)\le t\}=\Omega$ and $A'\cap\{\min(s,\tau)\le t\}=A'\in\mathcal{F}_s\subseteq\mathcal{F}_t$. If $t<s$ then $\{\min(s,\tau)\le t\}=\{\tau\le t\}$, which is disjoint from $\{\tau>s\}$, so $A'\cap\{\min(s,\tau)\le t\}=\emptyset\in\mathcal{F}_t$. Hence $A'\in\mathcal{F}_{\min(s,\tau)}$.

Claim 3. $D_n$ is a finite set containing $0$ and $T$, so $\tau_n(\omega)$ is well defined, lies in $D_n$, and satisfies $\tau_n(\omega)\ge\tau(\omega)$. Upper bound. Let $d^-=\max\{d\in D_n:d\le\tau(\omega)\}$, which exists since $0\in D_n$. If $d^-=\tau(\omega)$ then $\tau_n(\omega)=\tau(\omega)$. Otherwise $d^-<\tau(\omega)\le T$, so $d^-=kT2^{-n}$ with $k<2^n$ and $d^-+T2^{-n}=(k+1)T2^{-n}\in D_n$; this point exceeds $\tau(\omega)$ (otherwise it would be a grid point in $(d^-,\tau(\omega)]$, contradicting maximality of $d^-$), hence $\tau_n(\omega)\le d^-+T2^{-n}\le\tau(\omega)+T2^{-n}$. Monotonicity in $n$. $D_n\subseteq D_{n+1}$ because $kT2^{-n}=2kT2^{-(n+1)}$; the minimum over the larger set $\{d\in D_{n+1}:d\ge\tau(\omega)\}$ is at most that over $\{d\in D_n:d\ge\tau(\omega)\}$, so $\tau_{n+1}\le\tau_n$. The two-sided bound $\tau\le\tau_n\le\tau+T2^{-n}$ and $T2^{-n}\to0$ give $\tau_n(\omega)\to\tau(\omega)$ for every $\omega$. Level sets. $\tau_n(\omega)=0$ holds exactly when $0\ge\tau(\omega)$, i.e. $\tau(\omega)=0$; for $k\ge1$, $\tau_n(\omega)=kT2^{-n}$ holds exactly when $kT2^{-n}\ge\tau(\omega)$ and $(k-1)T2^{-n}<\tau(\omega)$. Stopping time. Fix $t\in[0,T]$ and let $d^-(t)=\max\{d\in D_n:d\le t\}$. Since $\tau_n$ takes values in $D_n$, $\tau_n(\omega)\le t$ holds if and only if $\tau_n(\omega)\le d^-(t)$; and this holds if and only if $\tau(\omega)\le d^-(t)$ (if $\tau(\omega)\le d^-(t)$ then $d^-(t)$ is a grid point not below $\tau(\omega)$, so $\tau_n(\omega)\le d^-(t)$; conversely $\tau\le\tau_n$). Thus $\{\tau_n\le t\}=\{\tau\le d^-(t)\}\in\mathcal{F}_{d^-(t)}\subseteq\mathcal{F}_t$. The remaining assertions. $\mathcal{F}_\tau\subseteq\mathcal{F}_{\tau_n}$ follows from the monotonicity part of claim 2, as $\tau\le\tau_n$. If $\sigma\le\tau$ pointwise, then $\{d\in D_n:d\ge\tau(\omega)\}\subseteq\{d\in D_n:d\ge\sigma(\omega)\}$, so $\sigma_n(\omega)\le\tau_n(\omega)$.

Claim 4(i). Fix $t\in[0,T]$ and put $\rho=\min(t,\tau)$, a stopping time by claim 1 with values in $[0,t]$. $\rho$ is $\mathcal{F}_t$-measurable: for real $a$, $\{\rho>a\}$ equals $\emptyset$ if $a\ge t$, equals $\Omega$ if $a<0$, and equals $\{\tau>a\}\in\mathcal{F}_a\subseteq\mathcal{F}_t$ if $0\le a<t$ (claim 1); apply the criterion. Consider the map $\phi:\Omega\to[0,t]\times\Omega$, $\phi(\omega)=(\rho(\omega),\omega)$, and let $\mathcal{C}$ be the family of those sets $E$ in the product $\sigma$-algebra $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$ (with $\mathcal{B}_{[0,t]}$ as in the definition of progressive measurability) for which $\phi^{-1}(E)\in\mathcal{F}_t$. Then $\mathcal{C}$ is a $\sigma$-algebra on $[0,t]\times\Omega$: it contains $[0,t]\times\Omega$ because $\phi^{-1}([0,t]\times\Omega)=\Omega\in\mathcal{F}_t$, and preimages commute with relative complements and countable unions. It contains every measurable rectangle $B\times A$ with $B\in\mathcal{B}_{[0,t]}$ and $A\in\mathcal{F}_t$: indeed $\phi^{-1}(B\times A)=\{\rho\in B\}\cap A$, and $\{\rho\in B\}\in\mathcal{F}_t$, because for $t>0$ one has $B=S\cap[0,t]$ with $S\in\mathcal{B}(\mathbb{R})$ and then $\{\rho\in B\}=\{\rho\in S\}$ by $\rho$ taking values in $[0,t]$, which lies in $\mathcal{F}_t$ by the $\mathcal{F}_t$-measurability of $\rho$; while for $t=0$ the only members of $\mathcal{B}_{[0,0]}$ are $\emptyset$ and $\{0\}$, with preimages $\emptyset$ and $\Omega$. Since the product $\sigma$-algebra is the $\sigma$-algebra generated by the measurable rectangles, minimality gives $\mathcal{C}=\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$; that is, $\phi$ is measurable with respect to $\mathcal{F}_t$ and the product $\sigma$-algebra. Let $\Psi_t$ denote the restriction of $(s,\omega)\mapsto X_s(\omega)$ to $[0,t]\times\Omega$, measurable with respect to $\mathcal{B}_{[0,t]}\otimes\mathcal{F}_t$ and $\mathcal{B}(\mathbb{R})$ by progressive measurability. Then $X^\tau_t(\omega)=X_{\rho(\omega)}(\omega)=\Psi_t(\phi(\omega))$, and for $S\in\mathcal{B}(\mathbb{R})$, $(X^\tau_t)^{-1}(S)=\phi^{-1}\big(\Psi_t^{-1}(S)\big)\in\mathcal{F}_t$. So $X^\tau_t$ is $\mathcal{F}_t$-measurable, and $X^\tau$ is adapted.

Claim 4(ii). Let $a$ be real and $t\in[0,T]$. On $\{\tau\le t\}$ one has $\min(t,\tau)=\tau$, hence $X_\tau=X^\tau_t$ there, and

$$\{X_\tau>a\}\cap\{\tau\le t\}=\{X^\tau_t>a\}\cap\{\tau\le t\}\in\mathcal{F}_t$$

by claim 4(i) and the stopping-time property. Taking $t=T$, where $\{\tau\le T\}=\Omega$, gives $\{X_\tau>a\}\in\mathcal{F}_T$. Hence $\{X_\tau>a\}\in\mathcal{F}_\tau$ for every real $a$, and the criterion, applied with the $\sigma$-algebra $\mathcal{F}_\tau$ of claim 2, shows that $X_\tau$ is $\mathcal{F}_\tau$-measurable, in particular $\mathcal{F}_T$-measurable.

Claim 4(iii). Fix $\omega$ at which the path of $X$ is right-continuous. By claim 3, $(\tau_n(\omega))_n$ is a sequence in $[\tau(\omega),T]$ converging to $\tau(\omega)$, so right-continuity at $s=\tau(\omega)$ gives $X_{\tau_n(\omega)}(\omega)\to X_{\tau(\omega)}(\omega)$, i.e. $X_{\tau_n}(\omega)\to X_\tau(\omega)$. For the path of $X^\tau$, let $t\in[0,T]$ and let $(t_j)$ be a sequence in $[t,T]$ converging to $t$. If $t\ge\tau(\omega)$ then $\min(t_j,\tau(\omega))=\tau(\omega)=\min(t,\tau(\omega))$ for all $j$, and the sequence $X^\tau_{t_j}(\omega)$ is constant equal to $X^\tau_t(\omega)$. If $t<\tau(\omega)$, put $s_j=\min(t_j,\tau(\omega))$; then $s_j\in[t,T]$ and $0\le s_j-t\le t_j-t$, so $s_j\to t$, and right-continuity of the path of $X$ at $t$ gives $X^\tau_{t_j}(\omega)=X_{s_j}(\omega)\to X_t(\omega)=X^\tau_t(\omega)$. Thus the path of $X^\tau$ at $\omega$ is right-continuous. Finally, if $X$ is adapted with every path right-continuous, then $X$ is progressively measurable by claim 2 of the progressive measurability toolkit, so $X^\tau$ is adapted by claim 4(i), every path of $X^\tau$ is right-continuous by what was just shown, and $X^\tau$ is progressively measurable by the same claim 2 of the toolkit.

Claim 5. Fix $\omega\in\Omega$ and write $H=H(\omega)$, $Y_t=Y_t(\omega)$. Continuity of the path at a point $t\in[0,T]$ means, by the definition of continuity for the metric of the real line: for every $\varepsilon>0$ there is $\delta>0$ with $|Y_s-Y_t|<\varepsilon$ for all $s\in[0,T]$ with $|s-t|<\delta$.

(a) The threshold is reached at $\tau$ when $H\neq\emptyset$. Suppose $H\neq\emptyset$; $H$ is bounded below by $0$, so $\tau(\omega)=\inf H$ exists by the approximation lemma for infima, and by its claim 4 there is, for each natural number $m$, some $s_m\in H$ with $s_m<\tau(\omega)+1/m$; also $s_m\ge\tau(\omega)$ as $\tau(\omega)$ is a lower bound of $H$. Hence $s_m\to\tau(\omega)$, and by continuity of the path at $\tau(\omega)$, $Y_{s_m}\to Y_{\tau(\omega)}$. Since $Y_{s_m}\ge c$ for all $m$, we get $Y_{\tau(\omega)}\ge c$: otherwise, with $\varepsilon=c-Y_{\tau(\omega)}>0$, all large $m$ would give $Y_{s_m}<Y_{\tau(\omega)}+\varepsilon=c$. If $\tau(\omega)<T$ then $H\neq\emptyset$ by the definition of $\tau$, so the conclusion holds in that case too.

(b) Below $\tau$ the threshold is not reached. Let $t<\tau(\omega)$. If $H=\emptyset$ then $Y_t<c$ trivially. If $H\neq\emptyset$ then $\tau(\omega)$ is a lower bound of $H$, so $t\notin H$, i.e. $Y_t<c$.

(c) $\tau$ is a stopping time. For $t=T$, $\{\tau\le T\}=\Omega\in\mathcal{F}_T$. For $t=0$: by (a) and the definition of $\tau$ (note $T>0$), $\tau(\omega)=0$ holds if and only if $0\in H(\omega)$, i.e. $\{\tau\le0\}=\{Y_0\ge c\}=\Omega\setminus\{Y_0<c\}$, which lies in $\mathcal{F}_0$ because $Y_0$ is $\mathcal{F}_0$-measurable. Now fix $t\in(0,T)$ and let $E_t$ be the set of $\omega$ for which there is $s\in[0,t]$ with $Y_s(\omega)\ge c$. Then $\{\tau\le t\}=E_t$: if $\tau(\omega)\le t<T$ then $H(\omega)\neq\emptyset$ (for $H(\omega)=\emptyset$ would give $\tau(\omega)=T>t$) and $Y_{\tau(\omega)}(\omega)\ge c$ by (a), with $\tau(\omega)\in[0,t]$; conversely, if $s\in[0,t]$ and $Y_s(\omega)\ge c$, then $s\in H(\omega)$ and $\tau(\omega)=\inf H(\omega)\le s\le t$. We show

$$E_t=\bigcap_{j\in\mathbb{N}}\ \bigcup_{m\in\mathbb{N}}\ \bigcup_{k=0}^{m}\Big\{Y_{kt/m}>c-\tfrac1j\Big\}.$$

For $\subseteq$: let $\omega\in E_t$ with $Y_s(\omega)\ge c$, $s\in[0,t]$, and let $j\in\mathbb{N}$. By continuity of the path at $s$ (with $\varepsilon=1/j$) there is $\delta>0$ with $|Y_r(\omega)-Y_s(\omega)|<1/j$ whenever $r\in[0,T]$ and $|r-s|<\delta$. Choose $m\in\mathbb{N}$ with $t/m<\delta$ (Archimedean property) and let $k$ be the largest integer with $kt/m\le s$; then $0\le k\le m$ (as $0\le s\le t$) and $0\le s-kt/m<t/m<\delta$, so $Y_{kt/m}(\omega)>Y_s(\omega)-1/j\ge c-1/j$. For $\supseteq$: let $\omega$ belong to the right side. The path $s\mapsto Y_s(\omega)$ restricted to $[0,t]$ is continuous on $[0,t]$ (the $\varepsilon$--$\delta$ condition at each point of $[0,t]$, quantified over $s\in[0,T]$, holds a fortiori when quantified over $s\in[0,t]$), so by the extreme value theorem it attains a maximum value at some $s^*\in[0,t]$. For every $j$ there is a grid point $q_j\in[0,t]$ with $Y_{q_j}(\omega)>c-1/j$, hence $Y_{s^*}(\omega)\ge Y_{q_j}(\omega)>c-1/j$ for every $j$, and therefore $Y_{s^*}(\omega)\ge c$ (if $Y_{s^*}(\omega)<c$, the Archimedean property would give $j$ with $1/j<c-Y_{s^*}(\omega)$, a contradiction); so $\omega\in E_t$. Finally, for each grid point $q=kt/m\in[0,t]$ the set $\{Y_q>c-1/j\}$ lies in $\mathcal{F}_q\subseteq\mathcal{F}_t$, because $Y$ is adapted; the displayed expression is a countable intersection of countable unions of finite unions of such sets, hence lies in $\mathcal{F}_t$. Thus $\{\tau\le t\}\in\mathcal{F}_t$ for every $t\in[0,T]$, and $\tau$ is a stopping time. $\blacksquare$