Proof of The Extended Good-Set Stopping Time of the Realized Control: Clipped-Out Time and Energy Hitting Bounds

Throughout, "the good-set lemma" is the 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, "the flow stability lemma" is the flow stability lemma, and "the toolkit" is the integral toolkit on a compact interval. All notation is that of the statement.

Claim 1. Fix $t\in[0,T]$. By claim 3 of the progressive lemma, the restriction of each component $\hat{\alpha}^\kappa$ to $[0,t]\times\Omega$ is measurable with respect to $\mathcal{B}_{[0,t]}\otimes\mathcal{G}_t$ and the Borel $\sigma$-algebra of the real line. Each map $(s,\omega)\mapsto A^\kappa_s$ on $[0,t]\times\Omega$ is measurable with respect to the same $\sigma$-algebras: for a Borel set $E$ of the real line, the preimage is $\bigl((A^\kappa)^{-1}(E)\cap[0,t]\bigr)\times\Omega$, a member of $\mathcal{B}_{[0,t]}\otimes\mathcal{G}_t$ by the definition of the product $\sigma$-algebra, since $A^\kappa$ is measurable and $[0,t]$ is a Borel set. The map $(s,\omega)\mapsto|\hat{\alpha}(s,\omega)-A_s|$ on $[0,t]\times\Omega$ is then measurable by measurability of continuous functions of measurable maps, being the composition of the $2m$ measurable real maps above with the map $(x,y)\mapsto|x-y|$ on $\mathbb{R}^m\times\mathbb{R}^m$, which is sequentially continuous: if $x_n\to x$ and $y_n\to y$ componentwise then $\bigl||x_n-y_n|-|x-y|\bigr|\le|(x_n-y_n)-(x-y)|\to0$ by the triangle inequality for the Euclidean norm. Consequently

$$E^{(t)}=\{(s,\omega)\in[0,t]\times\Omega:\ |\hat{\alpha}(s,\omega)-A_s|>\delta\}\in\mathcal{B}_{[0,t]}\otimes\mathcal{G}_t ,$$

and the restriction of $I^{\mathrm{out}}$ to $[0,t]\times\Omega$, the indicator of $E^{(t)}$, is measurable. As $t\in[0,T]$ was arbitrary, the family $(I^{\mathrm{out}}_s)_{s\in[0,T]}$ is progressively measurable with respect to $(\mathcal{G}_t)_{t\in[0,T]}$; its values are $0$ and $1$ by definition.

By claim 4 of the progressive measurability toolkit, applied to this progressively measurable family, which is bounded by $1$, the family $\mathcal{O}=(\mathcal{O}_t)_{t\in[0,T]}$ of its indefinite integrals is progressively measurable with respect to $(\mathcal{G}_t)_{t\in[0,T]}$, each $\mathcal{O}_t$ is $\mathcal{G}_t$-measurable, and $|\mathcal{O}_t-\mathcal{O}_{t_0}|\le t-t_0$ for $0\le t_0\le t\le T$. Since $\mathcal{G}_r\subseteq\mathcal{F}^{\mathrm{sys}}_r$ for every $r$ (claim 1 of the progressive lemma), every set of $\mathcal{B}_{[0,r]}\otimes\mathcal{G}_r$ lies in $\mathcal{B}_{[0,r]}\otimes\mathcal{F}^{\mathrm{sys}}_r$, so $\mathcal{O}$ is progressively measurable with respect to $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$ as well. Monotonicity: by claim 2 of the toolkit (zero extension), $\mathcal{O}_{t_0}(\omega)=\int_{[0,T]}\mathbf{1}_{[0,t_0]}(s)I^{\mathrm{out}}_s(\omega)\,ds$ and $\mathcal{O}_t(\omega)=\int_{[0,T]}\mathbf{1}_{[0,t]}(s)I^{\mathrm{out}}_s(\omega)\,ds$; the integrands are ordered pointwise ($\mathbf{1}_{[0,t_0]}\le\mathbf{1}_{[0,t]}$ and $I^{\mathrm{out}}\ge0$), so $\mathcal{O}_{t_0}(\omega)\le\mathcal{O}_t(\omega)$ by the monotonicity clause of linearity and monotonicity of the integral. Thus $0\le\mathcal{O}_t-\mathcal{O}_{t_0}\le t-t_0$; every path is nondecreasing, and continuous on $[0,T]$ (given $\eta>0$, $|s-r|\le\eta$ forces $|\mathcal{O}_s-\mathcal{O}_r|\le\eta$), with $\mathcal{O}_0=0$ (the convention for $t=0$) and $\mathcal{O}_T\le T-0=T$. Finally, at every $(s,\omega)$ one has $\delta^2I^{\mathrm{out}}_s(\omega)\le|\hat{\alpha}(s,\omega)-A_s|^2$ (if $I^{\mathrm{out}}_s(\omega)=1$ then $|\hat{\alpha}(s,\omega)-A_s|>\delta$; otherwise the left side is $0$), so by linearity and monotonicity of the integral, $\delta^2\mathcal{O}_t(\omega)\le\mathcal{E}_t(\omega)$ for every $t$ and $\omega$.

Claim 2. First we record the hitting value: if $H_{\mathcal{O}}(\omega)\neq\emptyset$ then $\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)\ge\theta_{\mathrm{out}}$. Indeed, every $t\in H_{\mathcal{O}}(\omega)$ satisfies $t\ge\tau_{\mathrm{out}}(\omega)$ (a greatest lower bound is a lower bound) and $\theta_{\mathrm{out}}\le\mathcal{O}_t(\omega)\le\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)+(t-\tau_{\mathrm{out}}(\omega))$ by claim 1, so $t\ge\tau_{\mathrm{out}}(\omega)+\theta_{\mathrm{out}}-\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)$; the number $\tau_{\mathrm{out}}(\omega)+\theta_{\mathrm{out}}-\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)$ is therefore a lower bound of $H_{\mathcal{O}}(\omega)$, hence at most the greatest lower bound $\tau_{\mathrm{out}}(\omega)$, which gives $\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)\ge\theta_{\mathrm{out}}$. In particular, since $\tau_{\mathrm{out}}(\omega)<T$ forces $H_{\mathcal{O}}(\omega)\neq\emptyset$, one gets $\mathcal{O}_{\tau_{\mathrm{out}}}\ge\theta_{\mathrm{out}}$ on $\{\tau_{\mathrm{out}}<T\}$ and, by monotonicity of paths, $\{\tau_{\mathrm{out}}<T\}\subseteq\{\mathcal{O}_T\ge\theta_{\mathrm{out}}\}$.

Fix $q\in[0,T)$. If $\mathcal{O}_q(\omega)\ge\theta_{\mathrm{out}}$ then $q\in H_{\mathcal{O}}(\omega)$, so $\tau_{\mathrm{out}}(\omega)\le q$. Conversely, if $\tau_{\mathrm{out}}(\omega)\le q<T$ then $H_{\mathcal{O}}(\omega)\neq\emptyset$, so by the hitting value and path monotonicity $\mathcal{O}_q(\omega)\ge\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)\ge\theta_{\mathrm{out}}$. Hence $\{\tau_{\mathrm{out}}\le q\}=\{\mathcal{O}_q\ge\theta_{\mathrm{out}}\}\in\mathcal{G}_q$, the function $\mathcal{O}_q$ being $\mathcal{G}_q$-measurable by claim 1. For $q=T$, $\{\tau_{\mathrm{out}}\le T\}=\Omega\in\mathcal{G}_T$, since $\tau_{\mathrm{out}}$ takes values in $[0,T]$. So $\tau_{\mathrm{out}}$ is a stopping time of $(\mathcal{G}_t)_{t\in[0,T]}$, and of $(\mathcal{F}^{\mathrm{sys}}_t)_{t\in[0,T]}$ because $\mathcal{G}_q\subseteq\mathcal{F}^{\mathrm{sys}}_q$ for every $q$.

If $t<\tau_{\mathrm{out}}(\omega)$ and one had $\mathcal{O}_t(\omega)\ge\theta_{\mathrm{out}}$, then $t\in H_{\mathcal{O}}(\omega)$ and $\tau_{\mathrm{out}}(\omega)\le t$, a contradiction; so $\mathcal{O}_t(\omega)<\theta_{\mathrm{out}}$. For the stopped bound put $u=\min(t,\tau_{\mathrm{out}}(\omega))$. If $u<\tau_{\mathrm{out}}(\omega)$ then $\mathcal{O}_u(\omega)<\theta_{\mathrm{out}}$. Otherwise $u=\tau_{\mathrm{out}}(\omega)$; if $u=0$ then $\mathcal{O}_u(\omega)=0<\theta_{\mathrm{out}}$; if $u>0$, suppose $\mathcal{O}_u(\omega)>\theta_{\mathrm{out}}$ and put $\eta=\bigl(\mathcal{O}_u(\omega)-\theta_{\mathrm{out}}\bigr)/2>0$. Since $\mathcal{O}_u(\omega)\le u$ by claim 1, $\eta<u$, so $s=u-\eta\in(0,u)$; then $s<\tau_{\mathrm{out}}(\omega)$ gives $\mathcal{O}_s(\omega)<\theta_{\mathrm{out}}$, while claim 1 gives $\mathcal{O}_u(\omega)\le\mathcal{O}_s(\omega)+\eta<\theta_{\mathrm{out}}+\eta=\mathcal{O}_u(\omega)-\eta$, a contradiction. Hence $\mathcal{O}_{\min(t,\tau_{\mathrm{out}}(\omega))}(\omega)\le\theta_{\mathrm{out}}$ in every case.

Claim 3. By claim 1 of the good-set lemma, $\tau_Y$ and $\tau_{\mathcal{E}}$ are stopping times of both filtrations, and by claim 2 so is $\tau_{\mathrm{out}}$; by claim 1 of the stopping-time toolkit, applied twice for each filtration, $\tau^*=\min(\tau,\tau_{\mathrm{out}})$ is a stopping time of both. For every $t$, $\{t<\tau^*\}=\Omega\setminus\{\tau^*\le t\}\in\mathcal{G}_t$. If $t<\tau^*(\omega)$ then $t<\tau(\omega)$, so $Y_t(\omega)<\varepsilon_Y$ and $\mathcal{E}_t(\omega)<c_{\mathcal{E}}$ by claim 2 of the good-set lemma, and $t<\tau_{\mathrm{out}}(\omega)$, so $\mathcal{O}_t(\omega)<\theta_{\mathrm{out}}$ by claim 2 above.

Stopped bounds: $\min(t,\tau^*(\omega))\le\min(t,\tau(\omega))$ and every path of $\mathcal{E}$ is nondecreasing (claim 5 of the progressive lemma), so $\mathcal{E}_{\min(t,\tau^*(\omega))}(\omega)\le\mathcal{E}_{\min(t,\tau(\omega))}(\omega)\le c_{\mathcal{E}}$ by claim 2 of the good-set lemma; likewise $\min(t,\tau^*(\omega))\le\min(t,\tau_{\mathrm{out}}(\omega))$ and paths of $\mathcal{O}$ are nondecreasing (claim 1), so $\mathcal{O}_{\min(t,\tau^*(\omega))}(\omega)\le\theta_{\mathrm{out}}$ by claim 2. For $Y$: assume $|x_0-S^*_0|<\varepsilon_Y$ and put $u=\min(t,\tau^*(\omega))\le\tau_Y(\omega)$. If $u<\tau_Y(\omega)$: were $Y_u(\omega)\ge\varepsilon_Y$, then $u\in H_Y(\omega)$ and $\tau_Y(\omega)\le u$, a contradiction; so $Y_u(\omega)<\varepsilon_Y$. If $u=\tau_Y(\omega)$ and $u=0$: $Y_0(\omega)=|\Phi_0(\omega)-S^*_0|=|x_0-S^*_0|<\varepsilon_Y$, since $\Phi_0(\omega)=x_0$ by claim 3 of the causality lemma. If $u=\tau_Y(\omega)>0$: every $s\in[0,u)$ satisfies $s<\tau_Y(\omega)$, hence $Y_s(\omega)<\varepsilon_Y$ as just argued; the path $s\mapsto Y_s(\omega)$ is continuous on $[0,T]$ (claim 4 of the causality lemma); were $Y_u(\omega)>\varepsilon_Y$, continuity at $u$ would furnish $\rho>0$ such that every $s\in[0,T]$ with $|s-u|\le\rho$ has $|Y_s(\omega)-Y_u(\omega)|<Y_u(\omega)-\varepsilon_Y$, and the point $s=\max(0,u-\rho)<u$ would satisfy $Y_s(\omega)>\varepsilon_Y$, a contradiction. So $Y_{\min(t,\tau^*(\omega))}(\omega)\le\varepsilon_Y$.

Finally, since $\tau,\tau_{\mathrm{out}}$ take values in $[0,T]$, $\{\tau^*<T\}=\{\tau<T\}\cup\{\tau_{\mathrm{out}}<T\}$. By finite subadditivity of $P$ (for events $A,B$: $P(A\cup B)=P(A)+P(B\setminus A)\le P(A)+P(B)$, by additivity and monotonicity of the measure $P$), claim 3 of the good-set lemma, the inclusion $\{\tau_{\mathrm{out}}<T\}\subseteq\{\mathcal{O}_T\ge\theta_{\mathrm{out}}\}$ of claim 2, and monotonicity of $P$:

$$P(\tau^*<T)\le P(\tau<T)+P(\tau_{\mathrm{out}}<T)\le P\bigl(\overline{Y}\ge\varepsilon_Y\bigr)+P\bigl(\mathcal{E}_T\ge c_{\mathcal{E}}\bigr)+P\bigl(\mathcal{O}_T\ge\theta_{\mathrm{out}}\bigr).$$

The first two sets on the right are in $\mathcal{G}_T$ by claim 3 of the good-set lemma, and $\{\mathcal{O}_T\ge\theta_{\mathrm{out}}\}\in\mathcal{G}_T$ since $\mathcal{O}_T$ is $\mathcal{G}_T$-measurable.

Claim 4. The components of $A$ are measurable by hypothesis, every value of $A$ lies in $\mathcal{A}$, and $|A_s|\le R$ for all $s$, so $s\mapsto|A_s|^2$ is measurable and bounded by $R^2$, hence integrable over $[0,T]$ (monotonicity against the constant $R^2$, whose integral is $R^2T$ by linearity and monotonicity of the integral); thus $A$ is square-integrable in the sense of claim 1 of the $L^2$ definition and determines an element $\zeta_A$ of $\mathcal{U}_{\mathcal{A}}$ of which it is an admissible representative in the sense of claim 2 of the flow stability lemma.

By claim 2 of the flow stability lemma, $S(x_0,\zeta_A)$ is the map furnished by claim 1 of the existence and uniqueness theorem for the generalized mean-field trajectory applied to the initial value $x_0$ and the admissible representative $A$; by that claim there is exactly one continuous map $x:[0,T]\to\mathbb{R}^l$ with $x^\gamma_t=x^\gamma_0+\int_{[0,t]}\hat{b}^\gamma(x_s,A_s)\,ds$ for all $t$ and $\gamma$, $\hat{b}$ being the projected drift. The map $S^*$ is continuous (componentwise, by the setting), takes values in $\Delta^l$, and satisfies the displayed equation with $b$ in place of $\hat{b}$ by hypothesis; since $\hat{b}(x,a)=b(x,a)$ for $x\in\Delta^l$ (claim 6 of the affine-rate lemma), the integrands coincide, so $S^*$ satisfies the $\hat{b}$-equation and, by uniqueness, $S^*=S(x_0,\zeta_A)$.

Now fix $t\in[0,T]$ and $\omega\in\Omega$, and define $\xi^{(t,\omega)}:[0,T]\to\mathcal{A}$ by $\xi^{(t,\omega)}(s)=\hat{\alpha}(s,\omega)$ for $s\le t$ and $\xi^{(t,\omega)}(s)=A_s$ for $s>t$. Each component equals $\mathbf{1}_{[0,t]}\hat{\alpha}^\kappa(\cdot,\omega)+\mathbf{1}_{(t,T]}A^\kappa$, a sum of products of measurable functions (the path components of $\hat{\alpha}(\cdot,\omega)$ being measurable by claim 3 of the realized-control lemma, indicators of Borel subsets of $[0,T]$ being measurable, and sums and products of measurable real functions being measurable by measurability of continuous functions of measurable maps); every value lies in $\mathcal{A}$ and is bounded by $R$ in norm, so as for $A$ above the path is an admissible representative of an element $\xi^{(t,\omega)}$ of $\mathcal{U}_{\mathcal{A}}$ (same-symbol convention).

The paths $\xi^{(t,\omega)}$ and $\hat{\alpha}(\cdot,\omega)$ agree at every point of $[0,t]$ (claim 3 of the realized-control lemma provides that the latter is an admissible representative of $\hat{\alpha}(\omega)$), so the set where they differ within $[0,t]$ is empty, hence $\lambda_{[0,T]}$-null, and claim 1 of the causality lemma gives

$$\Phi_t(\omega)=S_t\bigl(x_0,\hat{\alpha}(\omega)\bigr)=S_t\bigl(x_0,\xi^{(t,\omega)}\bigr).$$

Apply claims 3 and 4 of the flow stability lemma with base point $x_0$ and base control $\zeta_A$, whose flow is $S(x_0,\zeta_A)=S^*$, and with perturbed data $x_0'=x_0$, $\xi'=\xi^{(t,\omega)}$. For $\gamma\in\{1,\dots,l\}$ and $r\in[0,T]$,

$$g^\gamma_r\bigl(\xi^{(t,\omega)}\bigr)=\int_{[0,T]}\mathbf{1}_{[0,r]}(s)\,\bigl(\xi^{(t,\omega)}(s)-A_s\bigr)\cdot b^\gamma_1(S^*_s)\,ds .$$

The integrand vanishes for $s>t$, and for $s\le t$ its absolute value is at most $K_2\,|\hat{\alpha}(s,\omega)-A_s|$ by claim 1 of the flow stability lemma, both $\hat{\alpha}(s,\omega)$ and $A_s$ lying in $\mathcal{A}$. Hence, by linearity and monotonicity of the integral (from $\pm h\le|h|$ pointwise one gets $|\int h\,ds|\le\int|h|\,ds$ for an integrable real integrand $h$),

$$\bigl|g^\gamma_r\bigl(\xi^{(t,\omega)}\bigr)\bigr|\le K_2\,J_t(\omega),\qquad J_t(\omega)=\int_{[0,T]}\mathbf{1}_{[0,t]}(s)\,\bigl|\hat{\alpha}(s,\omega)-A_s\bigr|\,ds ,$$

the integrand of $J_t(\omega)$ being measurable (as in claim 1, for the fixed $\omega$) and bounded by $2R$, hence integrable. Taking $G=K_2J_t(\omega)$ in claim 4 of the flow stability lemma yields, for every $r\in[0,T]$, $\bigl|S_r(x_0,\xi^{(t,\omega)})-S^*_r\bigr|\le e^{\Lambda_bT}\bigl(0+\sqrt{l}\,K_2J_t(\omega)\bigr)$; at $r=t$, with the causality identity above, this is the first asserted inequality

$$Y_t(\omega)=\bigl|\Phi_t(\omega)-S^*_t\bigr|\le e^{\Lambda_bT}\sqrt{l}\,K_2\,J_t(\omega).$$

It remains to bound $J_t(\omega)$ by $\sqrt{T}\,\mathcal{E}_t(\omega)^{1/2}$. Write $f(s)=\mathbf{1}_{[0,t]}(s)|\hat{\alpha}(s,\omega)-A_s|\ge0$, so that $J_t(\omega)=\int_{[0,T]}f\,ds$ and, by claim 2 of the toolkit (zero extension), $\int_{[0,T]}f^2\,ds=\mathcal{E}_t(\omega)$. By claim 4 of the toolkit, applied to $f$ with $g=1$ on the interval $[0,T]$, $J_t(\omega)^2=\bigl(\int_{[0,T]}f\,ds\bigr)^2\le T\int_{[0,T]}f^2\,ds=T\,\mathcal{E}_t(\omega)$, so $J_t(\omega)\le\sqrt{T\,\mathcal{E}_t(\omega)}$, the nonnegative square root being nondecreasing and multiplicative. Hence $Y_t(\omega)\le e^{\Lambda_bT}\sqrt{l}\,K_2\sqrt{T}\,\mathcal{E}_t(\omega)^{1/2}=C_S\,\mathcal{E}_t(\omega)^{1/2}$, using that the nonnegative square root is multiplicative and nondecreasing (uniqueness of nonnegative square roots).

Claim 5. Each of the numbers $\varepsilon_Y^2/C_S^2$ (when $C_S>0$), $c_{\mathcal{E}}$ and $\delta^2\theta_{\mathrm{out}}$ is positive, so $m^*>0$. The family $\mathcal{E}$ is progressively measurable with respect to $(\mathcal{G}_t)_{t\in[0,T]}$ (claim 5 of the progressive lemma) and $\tau^*$ is a stopping time of that filtration (claim 3), so by claim 4 of the stopping-time toolkit the sampled function $\omega\mapsto\mathcal{E}_{\tau^*(\omega)}(\omega)$ is measurable, hence a random variable; its values lie in $[0,4R^2T]$ by claim 5 of the progressive lemma.

Let $\omega$ satisfy $\tau^*(\omega)<T$. The minimum $\tau^*(\omega)$ equals at least one of $\tau_Y(\omega)$, $\tau_{\mathcal{E}}(\omega)$, $\tau_{\mathrm{out}}(\omega)$.

Case $\tau^*(\omega)=\tau_Y(\omega)<T$. Then $H_Y(\omega)\neq\emptyset$ and $\tau_Y(\omega)$ is its greatest lower bound. We show $Y_{\tau_Y(\omega)}(\omega)\ge\varepsilon_Y$. Every $s<\tau_Y(\omega)$ has $Y_s(\omega)<\varepsilon_Y$ (as in claim 3). Suppose $Y_{\tau_Y(\omega)}(\omega)<\varepsilon_Y$. By continuity of the path of $Y$ (claim 4 of the causality lemma) at $\tau_Y(\omega)$ there is $\rho>0$ such that every $s\in[0,T]$ with $|s-\tau_Y(\omega)|\le\rho$ has $|Y_s(\omega)-Y_{\tau_Y(\omega)}(\omega)|<\varepsilon_Y-Y_{\tau_Y(\omega)}(\omega)$, hence $Y_s(\omega)<\varepsilon_Y$; combined with the pre-$\tau_Y$ bound, $Y_s(\omega)<\varepsilon_Y$ for every $s\in[0,\min(T,\tau_Y(\omega)+\rho)]$. If $\tau_Y(\omega)+\rho\ge T$ this gives $H_Y(\omega)=\emptyset$, contradicting $H_Y(\omega)\neq\emptyset$; otherwise every element of $H_Y(\omega)$ exceeds $\tau_Y(\omega)+\rho$, so $\tau_Y(\omega)+\rho$ is a lower bound of $H_Y(\omega)$ exceeding its greatest lower bound $\tau_Y(\omega)$, again a contradiction. Hence $Y_{\tau_Y(\omega)}(\omega)\ge\varepsilon_Y$. By claim 4, $\varepsilon_Y\le Y_{\tau_Y(\omega)}(\omega)\le C_S\,\mathcal{E}_{\tau_Y(\omega)}(\omega)^{1/2}$. If $C_S=0$ this is impossible (as $\varepsilon_Y>0$), so this case does not occur; if $C_S>0$, squaring (the square root being nondecreasing) gives $\mathcal{E}_{\tau^*(\omega)}(\omega)=\mathcal{E}_{\tau_Y(\omega)}(\omega)\ge\varepsilon_Y^2/C_S^2\ge m^*$.

Case $\tau^*(\omega)=\tau_{\mathcal{E}}(\omega)<T$. Then $H_{\mathcal{E}}(\omega)\neq\emptyset$. Every path of $\mathcal{E}$ satisfies $0\le\mathcal{E}_t-\mathcal{E}_{t_0}\le4R^2(t-t_0)$ (claim 5 of the progressive lemma), so the argument used for the hitting value in claim 2, with $\mathcal{O}$, $\theta_{\mathrm{out}}$, the Lipschitz constant $1$ replaced by $\mathcal{E}$, $c_{\mathcal{E}}$, $4R^2$, applies verbatim when $R>0$ and yields $\mathcal{E}_{\tau_{\mathcal{E}}(\omega)}(\omega)\ge c_{\mathcal{E}}$; when $R=0$ the process $\mathcal{E}$ is identically $0<c_{\mathcal{E}}$, $H_{\mathcal{E}}(\omega)=\emptyset$, and the case does not occur. Hence $\mathcal{E}_{\tau^*(\omega)}(\omega)\ge c_{\mathcal{E}}\ge m^*$.

Case $\tau^*(\omega)=\tau_{\mathrm{out}}(\omega)<T$. By the hitting value of claim 2, $\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)\ge\theta_{\mathrm{out}}$, so by claim 1, $\mathcal{E}_{\tau^*(\omega)}(\omega)\ge\delta^2\mathcal{O}_{\tau_{\mathrm{out}}(\omega)}(\omega)\ge\delta^2\theta_{\mathrm{out}}\ge m^*$.

In every case $\mathcal{E}_{\tau^*(\omega)}(\omega)\ge m^*$, proving $\{\tau^*<T\}\subseteq\{\mathcal{E}_{\tau^*}\ge m^*\}$. By monotonicity of $P$ and Markov's inequality applied to the nonnegative random variable $\mathcal{E}_{\tau^*}$ at level $m^*>0$,

$$P(\tau^*<T)\le P\bigl(\mathcal{E}_{\tau^*}\ge m^*\bigr)\le\frac{\mathbb{E}\bigl[\mathcal{E}_{\tau^*}\bigr]}{m^*}. \qquad\blacksquare$$