Proof of Anchored Good-Set Clocks on a Subinterval and the Block Escape Bound

Throughout, "the extended lemma" is the extended good-set stopping-time lemma, "the progressive lemma" is the progressive measurability lemma for the realized control, "the causality lemma" is the causality and adaptedness lemma, and "the toolkit" is the integral toolkit on a compact interval. All notation is that of the statement. We record three facts used repeatedly. (F1) For $t_0\le t\le T$ and every $\omega$: $0\le\mathcal{E}_t(\omega)-\mathcal{E}_{t_0}(\omega)\le4R^2(t-t_0)$ and paths of $\mathcal{E}$ are nondecreasing and continuous (claim 5 of the progressive lemma), and $0\le\mathcal{O}_t(\omega)-\mathcal{O}_{t_0}(\omega)\le t-t_0$ with nondecreasing continuous paths (claim 1 of the extended lemma); moreover $\delta^2(\mathcal{O}_t-\mathcal{O}_{t_0})\le\mathcal{E}_t-\mathcal{E}_{t_0}$: at every $(s,\omega)$ one has $\delta^2I^{\mathrm{out}}_s(\omega)\le|\hat{\alpha}(s,\omega)-A_s|^2$ (immediate from the definition of $I^{\mathrm{out}}$ in the preamble of the extended lemma: on $\{I^{\mathrm{out}}_s=1\}$ the right side exceeds $\delta^2$, and otherwise the left side is $0$), so by the splitting fact (F3) below and monotonicity of the integral, $\delta^2(\mathcal{O}_t-\mathcal{O}_{t_0})=\delta^2\int_{[0,T]}\mathbf{1}_{(t_0,t]}I^{\mathrm{out}}_s\,ds\le\int_{[0,T]}\mathbf{1}_{(t_0,t]}\bigl|\hat{\alpha}(s,\cdot)-A_s\bigr|^2ds=\mathcal{E}_t-\mathcal{E}_{t_0}$. (F2) $Y_t$ is $\mathcal{G}_t$-measurable with continuous paths and $0\le Y_t\le K_Y$ (claim 4 of the causality lemma). (F3) (Splitting.) For a bounded measurable $h$ on $[0,T]$ and $0\le t_0\le t\le T$: $\int_{[0,t]}h\,ds-\int_{[0,t_0]}h\,ds=\int_{[0,T]}\mathbf{1}_{(t_0,t]}h\,ds$. Indeed, by claim 2 of the toolkit (zero extension) the two left integrals equal $\int_{[0,T]}\mathbf{1}_{[0,t]}h$ and $\int_{[0,T]}\mathbf{1}_{[0,t_0]}h$, and $\mathbf{1}_{[0,t]}-\mathbf{1}_{[0,t_0]}=\mathbf{1}_{(t_0,t]}$ pointwise, so the claim follows from linearity of the integral. In particular $\mathcal{E}_t(\omega)-\mathcal{E}_{t_0}(\omega)=\int_{[0,T]}\mathbf{1}_{(t_0,t]}(s)|\hat{\alpha}(s,\omega)-A_s|^2ds$, and likewise for $\mathcal{O}$.

Claim 1. All three sets are contained in $[t_0,T]$, so each of $\sigma_Y,\sigma_{\mathcal{E}},\sigma_{\mathrm{out}}$ is $[t_0,T]$-valued ($t_0$ is a lower bound, and the default value is $T$); hence so is $\sigma^*$.

The Lipschitz clocks. Let $(V_t)_{t\in[t_0,T]}$ stand for either $(\mathcal{E}_t-\mathcal{E}_{t_0})_{t\in[t_0,T]}$ or $(\mathcal{O}_t-\mathcal{O}_{t_0})_{t\in[t_0,T]}$, let $c>0$ stand for $c_{\mathcal{E}}$ or $\theta_{\mathrm{out}}$ accordingly, and let $\sigma$ be the associated clock. By (F1), $V_{t_0}=0$, every path of $V$ is nondecreasing with $0\le V_t-V_r\le L(t-r)$ for $t_0\le r\le t\le T$ (with $L=4R^2$ or $L=1$), and $V_t=\mathcal{E}_t-\mathcal{E}_{t_0}$ resp. $\mathcal{O}_t-\mathcal{O}_{t_0}$ is $\mathcal{G}_t$-measurable for $t\ge t_0$ ($\mathcal{E}_t$ and $\mathcal{E}_{t_0}$ are $\mathcal{G}_t$-measurable by claim 5 of the progressive lemma and $\mathcal{G}_{t_0}\subseteq\mathcal{G}_t$; likewise for $\mathcal{O}$ by claim 1 of the extended lemma). The greatest-lower-bound argument of claim 2 of the extended lemma applies unchanged to $V$ on $[t_0,T]$: if the hitting set is nonempty then every element $t$ of it satisfies $t\ge\sigma(\omega)$ and $c\le V_t(\omega)\le V_{\sigma(\omega)}(\omega)+L\,(t-\sigma(\omega))$, so, when $L>0$, $\sigma(\omega)+\bigl(c-V_{\sigma(\omega)}(\omega)\bigr)/L$ is a lower bound of the hitting set, whence $V_{\sigma(\omega)}(\omega)\ge c$ (when $L=0$ the path of $V$ is constant $0<c$ and the hitting set is empty). From this, for $q\in[t_0,T)$: if $V_q(\omega)\ge c$ then $q$ belongs to the hitting set and $\sigma(\omega)\le q$; conversely if $\sigma(\omega)\le q<T$ then the hitting set is nonempty, so $V_{\sigma(\omega)}(\omega)\ge c$ as just shown, and the nondecreasing path gives $V_q(\omega)\ge c$. Hence $\{\sigma\le q\}=\{V_q\ge c\}\in\mathcal{G}_q$; for $q\in[0,t_0)$: $\{\sigma\le q\}=\emptyset\in\mathcal{G}_q$; and $\{\sigma\le T\}=\Omega\in\mathcal{G}_T$. So $\sigma$ is a stopping time of $(\mathcal{G}_t)_{t\in[0,T]}$, hence of $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$ since $\mathcal{G}_q\subseteq\mathcal{F}^{\mathrm{sys}}_q$ for every $q$ (claim 1 of the progressive lemma). This also proves the two displayed identities of claim 1 and, for later use, the hitting values of claim 3 for $\sigma_{\mathcal{E}}$ and $\sigma_{\mathrm{out}}$ (if $\sigma(\omega)<T$ the hitting set is nonempty, so $V_{\sigma(\omega)}(\omega)\ge c$).

The deviation clock. Let $q\in[t_0,T)$ and put $D_q=\bigl([t_0,q]\cap\mathbb{Q}\bigr)\cup\{t_0,q\}$, a countable set: subsets of the countable set of rationals are countable and adjoining finitely many points preserves countability (countability of the rationals and the basic countability lemma). We claim

First equality: if there is such a $t$, then $t\in H^{t_0}_Y(\omega)$ and $\sigma_Y(\omega)\le t\le q$; conversely if $\sigma_Y(\omega)\le q<T$ then $H^{t_0}_Y(\omega)\neq\emptyset$, and $Y_{\sigma_Y(\omega)}(\omega)\ge\varepsilon_1$ — shown as follows, which also proves claim 3 for $\sigma_Y$: every $s\in[t_0,\sigma_Y(\omega))$ has $s\notin H^{t_0}_Y(\omega)$, so $Y_s(\omega)<\varepsilon_1$; if one had $Y_{\sigma_Y(\omega)}(\omega)<\varepsilon_1$ then, by continuity of the path of $Y$ at $\sigma_Y(\omega)$ (F2) and this pre-bound, there would be $\rho>0$ with $Y_s(\omega)<\varepsilon_1$ for every $s\in[t_0,\min(T,\sigma_Y(\omega)+\rho)]$, forcing every element of $H^{t_0}_Y(\omega)$ to exceed $\sigma_Y(\omega)+\rho$ when $\sigma_Y(\omega)+\rho\le T$ (contradicting that $\sigma_Y(\omega)$ is the greatest lower bound) and forcing $H^{t_0}_Y(\omega)=\emptyset$ otherwise (contradicting nonemptiness); so $Y_{\sigma_Y(\omega)}(\omega)\ge\varepsilon_1$, and $t=\sigma_Y(\omega)\in[t_0,q]$ is the required witness. Second equality: if $Y_t(\omega)\ge\varepsilon_1$ for some $t\in[t_0,q]$ then, the restriction of the path to the compact interval $[t_0,q]$ being continuous (claim 1 of restriction stability), its supremum over $[t_0,q]$ is attained (the extreme value theorem) and is at least $\varepsilon_1$; for every $\eta>0$ there is, by continuity at the maximiser and density of $D_q$ in $[t_0,q]$ (density of the rationals, the endpoints belonging to $D_q$), a point $r\in D_q$ with $Y_r(\omega)>\sup_{[t_0,q]}Y(\omega)-\eta$, so $\sup_{D_q}Y(\omega)=\sup_{[t_0,q]}Y(\omega)\ge\varepsilon_1$. Conversely, if $\sup_{D_q}Y(\omega)\ge\varepsilon_1$ then, the supremum over $[t_0,q]$ being attained and at least as large, there is $t\in[t_0,q]$ with $Y_t(\omega)\ge\varepsilon_1$. Now each $Y_t$ with $t\in D_q$ is $\mathcal{G}_q$-measurable (F2 and $\mathcal{G}_t\subseteq\mathcal{G}_q$), so the countable supremum is $\mathcal{G}_q$-measurable by claim 1 of the measurable limits toolkit, applied to an enumeration of $D_q$ (repeating terms if necessary) with the uniform bound $K_Y$ of (F2); hence $\{\sigma_Y\le q\}\in\mathcal{G}_q$. For $q<t_0$, $\{\sigma_Y\le q\}=\emptyset$, and $\{\sigma_Y\le T\}=\Omega$. So $\sigma_Y$ is a stopping time of both filtrations, as above.

Finally $\sigma^*$ is a stopping time of both filtrations by claim 1 of the stopping-time toolkit, applied twice, and $\{t<\sigma^*\}=\Omega\setminus\{\sigma^*\le t\}\in\mathcal{G}_t$.

Claim 2. Pre-bounds: let $t\in[t_0,T]$ with $t<\sigma^*(\omega)$. If $Y_t(\omega)\ge\varepsilon_1$ then $t\in H^{t_0}_Y(\omega)$ and $\sigma_Y(\omega)\le t$, a contradiction; the other two are identical. Stopped bounds: put $u=\min(t,\sigma^*(\omega))\in[t_0,T]$ (note $\sigma^*\ge t_0$ and $t\ge t_0$). For the Lipschitz clocks: if $u<\sigma_{\mathcal{E}}(\omega)$ then $\mathcal{E}_u-\mathcal{E}_{t_0}<c_{\mathcal{E}}$ by the pre-bound (valid with $\sigma_{\mathcal{E}}$ in place of $\sigma^*$ by the same one-line argument); if $u=\sigma_{\mathcal{E}}(\omega)$, the argument of claim 2 of the extended lemma for the stopped bound applies to the nondecreasing $L$-Lipschitz path $V$ started at $V_{t_0}=0$: either $u=t_0$ and $V_u=0<c_{\mathcal{E}}$, or $u>t_0$ and assuming $V_u>c_{\mathcal{E}}$ with $\eta=(V_u-c_{\mathcal{E}})/2>0$ one has $\eta<V_u\le L(u-t_0)$, hence (when $L>0$) $s=u-\eta/L\in(t_0,u)$ satisfies $V_s<c_{\mathcal{E}}$ and $V_u\le V_s+\eta<c_{\mathcal{E}}+\eta=V_u-\eta$, a contradiction (when $L=0$, $V\equiv0$); so $V_u\le c_{\mathcal{E}}$. The same argument with $L=1$ gives $\mathcal{O}_u-\mathcal{O}_{t_0}\le\theta_{\mathrm{out}}$. For $Y$: assume $Y_{t_0}(\omega)<\varepsilon_1$; if $u<\sigma_Y(\omega)$ then $Y_u(\omega)<\varepsilon_1$ (pre-bound); if $u=\sigma_Y(\omega)=t_0$ then $Y_u(\omega)=Y_{t_0}(\omega)<\varepsilon_1$; if $u=\sigma_Y(\omega)>t_0$ then $Y_s(\omega)<\varepsilon_1$ for all $s\in[t_0,u)$ and, were $Y_u(\omega)>\varepsilon_1$, continuity at $u$ would produce $s\in[t_0,u)$ with $Y_s(\omega)>\varepsilon_1$; so $Y_u(\omega)\le\varepsilon_1$. The increment comparison $\delta^2(\mathcal{O}_t-\mathcal{O}_{t_0})\le\mathcal{E}_t-\mathcal{E}_{t_0}$ is (F1).

Claim 3. Proved within claim 1.

Claim 4. For every $\omega$ the path of the realized flow satisfies, by claim 2 of the flow stability lemma and claim 1 of the existence and uniqueness theorem applied to $x_0$ and the admissible representative $s\mapsto\hat{\alpha}(s,\omega)$ (claim 3 of the realized-control lemma) — the flow taking values in $\Delta^l$ by those same claims — together with claim 6 of the affine-rate lemma ($\hat{b}=b$ on $\Delta^l$):

Fix $\omega$ and write $v^\gamma_s=b^\gamma(\Phi_s(\omega),\hat{\alpha}(s,\omega))-b^\gamma(S^*_s,A_s)$, a bounded measurable function of $s$: bounded by $4\sqrt{l}(l-1)B$ by claim 4 of the affine-rate lemma, and measurable by measurability of continuous functions of measurable maps — the path components of $\hat{\alpha}(\cdot,\omega)$ are measurable (claim 3 of the realized-control lemma), $\Phi(\omega)$ and $S^*$ have continuous hence measurable components, and $b$ is sequentially continuous on $\Delta^l\times\mathcal{A}$, since $|b(\Sigma,\alpha)-b(\Sigma',\alpha')|\le\Lambda_b|\Sigma-\Sigma'|+K_2|\alpha-\alpha'|$ by claim 4 of the affine-rate lemma and the triangle inequality. By (F3), for $t\in[t_0,T]$ and each $\gamma$,

Introduce the $\ell^1$-deviation $\overline{Y}^{(1)}_s(\omega)=\sum_{\gamma=1}^l|\Phi^\gamma_s(\omega)-S^{*\gamma}_s|$; its path is continuous, hence measurable in $s$ (claim 4 of the measurable limits toolkit), and bounded by $l$, each coordinate of $\Phi_s$ and of $S^*_s$ lying in $[0,1]$. The elementary inequalities $\bigl(\sum_\gamma|x_\gamma|\bigr)^2\ge\sum_\gamma x_\gamma^2$ and $\bigl(\sum_\gamma|x_\gamma|\bigr)^2\le l\sum_\gamma x_\gamma^2$ for reals $x_1,\dots,x_l$ (the latter from summing $2|x_\gamma||x_\delta|\le x_\gamma^2+x_\delta^2$ over all pairs), applied to $x_\gamma=\Phi^\gamma_s(\omega)-S^{*\gamma}_s$, give $Y_s\le\overline{Y}^{(1)}_s\le\sqrt{l}\,Y_s$, the nonnegative square root being nondecreasing. By the $\ell^1$-$\ell^2$ comparison just recorded (applied to the components of $v_s$), the triangle inequality for the Euclidean norm, and claim 4 of the affine-rate lemma, at every $s$,

where $K_2'=2\sqrt{l}\,(l-1)K_1=K_2$ is the control-Lipschitz constant of that claim. Hence, using $Y_s\le\overline{Y}^{(1)}_s$, the scalar bound $|\int h\,ds|\le\int|h|\,ds$ for each component (from $\pm h\le|h|$ and linearity and monotonicity of the integral), and summing over $\gamma$: for $t\in[t_0,T]$,

Fix $t^*\in[t_0,T]$ and define $w:[0,T]\to[0,\infty)$ by $w_s=\overline{Y}^{(1)}_{t_0}$ for $s<t_0$ and $w_s=\overline{Y}^{(1)}_s$ for $s\ge t_0$; then $w=\overline{Y}^{(1)}_{t_0}\mathbf{1}_{[0,t_0)}+\overline{Y}^{(1)}\mathbf{1}_{[t_0,T]}$ is bounded by $l$ and measurable, as a sum of products of measurable functions with indicators of Borel sets (measurability of continuous functions of measurable maps). Put $a=\overline{Y}^{(1)}_{t_0}+\sqrt{l}K_2J^{t_0}_{t^*}(\omega)$. For $s<t_0$, $w_s\le a$; for $s\in[t_0,t^*]$, the display above and $J^{t_0}_s\le J^{t_0}_{t^*}$ (monotonicity of the integral in the indicator), together with $\int\mathbf{1}_{(t_0,s]}\overline{Y}^{(1)}\le\int_{[0,s]}w$ ($w\ge0$ and $w=\overline{Y}^{(1)}$ on $(t_0,s]$, using claim 2 of the toolkit), give $w_s\le a+\sqrt{l}\Lambda_b\int_{[0,s]}w_r\,dr$. Hence, for every $s\in[0,t^*]$, $w_s\le a+\sqrt{l}\Lambda_b\int_{[0,s]}w$. If $t^*>0$, apply Gronwall's lemma for bounded measurable functions with horizon $t^*$ (in place of its $T$) to the restriction of $w$ to $[0,t^*]$, which is bounded and measurable with respect to the trace Borel $\sigma$-algebra on $[0,t^*]$; this yields $w_{t^*}\le a\,e^{\sqrt{l}\Lambda_bt^*}\le a\,e^{\sqrt{l}\Lambda_bT}$. If $t^*=0$ (so $t_0=0$), then $w_{t^*}=\overline{Y}^{(1)}_{t_0}\le a\le a\,e^{\sqrt{l}\Lambda_bT}$ directly, the exponential being at least $1$. Therefore

Finally, by claim 4 of the toolkit applied on the interval $[t_0,t^*]$ (with $g=1$) to the restriction of $s\mapsto|\hat{\alpha}(s,\omega)-A_s|$, and by (F3),

(the case $t^*=t_0$ being trivial with $J=0$). Here the identifications between the integrals of (F3), which carry the indicator $\mathbf{1}_{(t_0,t^*]}$, and the integrals over the closed interval $[t_0,t^*]$ used by claim 4 of the toolkit hold because the integrands differ only at the single point $t_0$: $\mathbf{1}_{(t_0,t^*]}h$ and $\mathbf{1}_{[t_0,t^*]}h$ agree off $\{t_0\}=[t_0,t_0]$, which is $\lambda$-null (the Lebesgue measure theorem), so their integrals over $[0,T]$ agree by claim 2 of the null-set integral lemma, and the latter equals the integral of the restriction over $[t_0,t^*]$ by claim 2 of the toolkit (zero extension). This gives the asserted bound, the square root being nondecreasing and multiplicative.

Claim 5. Each entry of $m^*_b$ is positive, so $m^*_b>0$. The family $\mathcal{E}$ is progressively measurable with respect to $(\mathcal{G}_t)_{t\in[0,T]}$ (claim 5 of the progressive lemma) and $\sigma^*$ is a $\mathcal{G}$-stopping time (claim 1), so $\omega\mapsto\mathcal{E}_{\sigma^*(\omega)}(\omega)$ is a random variable by claim 4 of the stopping-time toolkit, and so is $\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0}$; its values lie in $[0,4R^2T]$ by (F1).

Let $\omega$ satisfy $\sigma^*(\omega)<T$ and $Y_{t_0}(\omega)\le\varepsilon_1/(2C_a)$. The minimum $\sigma^*(\omega)$ equals one of the three clocks. If $\sigma^*(\omega)=\sigma_{\mathcal{E}}(\omega)<T$: by claim 3, $\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0}\ge c_{\mathcal{E}}\ge m^*_b$. If $\sigma^*(\omega)=\sigma_{\mathrm{out}}(\omega)<T$: by claim 3 and (F1), $\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0}\ge\delta^2(\mathcal{O}_{\sigma^*}-\mathcal{O}_{t_0})\ge\delta^2\theta_{\mathrm{out}}\ge m^*_b$. If $\sigma^*(\omega)=\sigma_Y(\omega)<T$: by claim 3 and claim 4 at $t=\sigma_Y(\omega)$,

so $C_aK_2\sqrt{T}(\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0})^{1/2}\ge\varepsilon_1/2$; if $K_2=0$ this is impossible ($\varepsilon_1>0$), so the case does not occur, and if $K_2>0$, squaring gives $\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0}\ge\varepsilon_1^2/(4C_a^2K_2^2T)\ge m^*_b$. This proves the inclusion.

Finally, let $D\in\mathcal{F}$ with $D\subseteq\{Y_{t_0}\le\varepsilon_1/(2C_a)\}$ and write $X=\mathcal{E}_{\sigma^*}-\mathcal{E}_{t_0}\ge0$. By the inclusion, $D\cap\{\sigma^*<T\}\subseteq D\cap\{X\ge m^*_b\}$, and pointwise $m^*_b\,\mathbf{1}_{D\cap\{X\ge m^*_b\}}\le\mathbf{1}_{D}X$, so by monotonicity and linearity of the expectation (linearity and monotonicity of the integral) and monotonicity of $P$,

Claim 6. Each entry of $m^{*,h}_b$ is positive ($h>0$), so $m^{*,h}_b>0$. The constant map $\omega\mapsto t_0+h$ is a stopping time of $(\mathcal{G}_t)_{t\in[0,T]}$, since $t_0+h\in[0,T]$ (from $0\le t_0$ and $h\le T-t_0$), and $\sigma^*$ is a stopping time of the same filtration (claim 1); hence $\sigma^*_h=\min(\sigma^*,t_0+h)$ is a stopping time of $(\mathcal{G}_t)_{t\in[0,T]}$ by claim 1 of the stopping-time toolkit. Exactly as in claim 5 — by progressive measurability of $\mathcal{E}$ (claim 5 of the progressive lemma) and claim 4 of the stopping-time toolkit — the function $X_h=\mathcal{E}_{\sigma^*_h}-\mathcal{E}_{t_0}$ is a random variable; and since $t_0\le\sigma^*_h\le t_0+h$ pointwise ($\sigma^*\ge t_0$ by claim 1), (F1) gives $0\le X_h\le4R^2\bigl(\sigma^*_h-t_0\bigr)\le4R^2h$.

Let $\omega$ satisfy $\sigma^*(\omega)<t_0+h$ and $Y_{t_0}(\omega)\le\varepsilon_1/(2C_a)$. Then $\sigma^*_h(\omega)=\sigma^*(\omega)$, so $X_h(\omega)=\mathcal{E}_{\sigma^*(\omega)}(\omega)-\mathcal{E}_{t_0}(\omega)$, and $\sigma^*(\omega)<t_0+h\le T$, so the hitting values of claim 3 are available. The minimum $\sigma^*(\omega)$ equals one of the three clocks. If $\sigma^*(\omega)=\sigma_{\mathcal{E}}(\omega)<T$: by claim 3, $X_h(\omega)\ge c_{\mathcal{E}}\ge m^{*,h}_b$. If $\sigma^*(\omega)=\sigma_{\mathrm{out}}(\omega)<T$: by claim 3 and (F1), $X_h(\omega)\ge\delta^2(\mathcal{O}_{\sigma^*}-\mathcal{O}_{t_0})\ge\delta^2\theta_{\mathrm{out}}\ge m^{*,h}_b$. If $\sigma^*(\omega)=\sigma_Y(\omega)<T$: by claim 3 and claim 4 at $t=\sigma_Y(\omega)$, together with $\sigma_Y(\omega)-t_0=\sigma^*(\omega)-t_0<h$ and the fact that the nonnegative square root is nondecreasing,

so $C_aK_2\sqrt{h}(X_h(\omega))^{1/2}\ge\varepsilon_1/2$; if $K_2=0$ this is impossible ($\varepsilon_1>0$), so the case does not occur, and if $K_2>0$, squaring gives $X_h(\omega)\ge\varepsilon_1^2/(4C_a^2K_2^2h)\ge m^{*,h}_b$. This proves the inclusion.

Finally, let $D\in\mathcal{F}$ with $D\subseteq\{Y_{t_0}\le\varepsilon_1/(2C_a)\}$. The set $\{\sigma^*<t_0+h\}$ is an event: for every positive integer $n$ with $1/n\le h$ one has $\{\sigma^*\le t_0+h-1/n\}\in\mathcal{G}_{t_0+h-1/n}\subseteq\mathcal{F}$, $\sigma^*$ being a stopping time (claim 1) and $t_0+h-1/n\in[0,T]$; and $\{\sigma^*<t_0+h\}$ is the union of these sets over such $n$: if $\sigma^*(\omega)<t_0+h$ then $t_0+h-\sigma^*(\omega)>0$, so by the Archimedean property there is a positive integer $n$ with $1/n\le t_0+h-\sigma^*(\omega)$, and this $n$ satisfies $1/n\le h$ because $\sigma^*(\omega)\ge t_0$ (claim 1); a countable union of events is an event, $\mathcal{F}$ being a $\sigma$-algebra of the underlying probability space. The same argument with $T$ in place of $t_0+h$ and $T-t_0$ in place of $h$ shows that the set $\{\sigma^*<T\}$ of claim 5 is an event (when $t_0=T$ it is empty, $\sigma^*$ being $[T,T]$-valued). By the inclusion just proved, $D\cap\{\sigma^*<t_0+h\}\subseteq D\cap\{X_h\ge m^{*,h}_b\}$, and pointwise $m^{*,h}_b\,\mathbf{1}_{D\cap\{X_h\ge m^{*,h}_b\}}\le\mathbf{1}_{D}X_h$, $X_h$ being nonnegative; so by monotonicity and linearity of the expectation (linearity and monotonicity of the integral) and monotonicity of $P$,