[proofplan] We first pass the pointwise robust Lyapunov inequality to every admissible disturbed closed-loop trajectory, obtaining an almost-everywhere scalar differential inequality for $V(x(t))$. Properness and positive definiteness of $V$ provide compact sublevel sets and uniform bounds over initial balls. For practical stability, the Lyapunov value decreases at rate at least $\varepsilon$ whenever it lies above the $V$-height of the residual region $\{\alpha(|x|)\leq c_{\bar d}+\varepsilon\}$, which gives finite entrance time and forward completeness. When $c_{\bar d}=0$, the same inequality gives a strict global Lyapunov decrease away from the origin, and the robust equilibrium condition ensures that the origin is an equilibrium for every admissible disturbance. [/proofplan] [step:Pass the robust Lyapunov inequality to admissible trajectories] Define the closed-loop drift with frozen disturbance value as the map \begin{align*} F: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n,\qquad F(x,\delta)=f(x)+g(x)k(x)+p(x)\delta. \end{align*} Fix a Lebesgue-measurable disturbance $d: [0,\infty)\to\mathbb{R}^m$ with $|d(t)|\leq \bar d$ for $\mathcal{L}^1$-a.e. $t\geq 0$, and let $x:[0,T_{\max})\to\mathbb{R}^n$ be a maximal Caratheodory solution of $\dot{x}(t)=F(x(t),d(t))$. By the definition of Caratheodory solution in the statement, $x$ is absolutely continuous on compact subintervals of $[0,T_{\max})$ and satisfies $\dot{x}(t)=F(x(t),d(t))$ for $\mathcal{L}^1$-a.e. $t\in[0,T_{\max})$. Since $V$ is $C^1$, the chain rule for absolutely continuous curves gives, for $\mathcal{L}^1$-a.e. such $t$, \begin{align*} \frac{d}{dt}V(x(t))=\nabla V(x(t))\cdot F(x(t),d(t)). \end{align*} At every such $t$ for which $|d(t)|\leq \bar d$, the supremum hypothesis applied at $x(t)$ gives \begin{align*} \frac{d}{dt}V(x(t))\leq -\alpha(|x(t)|)+c_{\bar d}. \end{align*} Thus this scalar differential inequality holds for $\mathcal{L}^1$-a.e. $t\in[0,T_{\max})$. [guided] The disturbance $d$ is only assumed measurable, so the statement explicitly uses Caratheodory solutions: $x$ is absolutely continuous on every compact time interval and satisfies the differential equation for $\mathcal{L}^1$-a.e. time. This is exactly the regularity needed to differentiate $V(x(t))$ almost everywhere. We define the closed-loop vector field with a frozen disturbance value by \begin{align*} F: \mathbb{R}^n \times \mathbb{R}^m \to \mathbb{R}^n,\qquad F(x,\delta)=f(x)+g(x)k(x)+p(x)\delta. \end{align*} For a fixed admissible disturbance $d: [0,\infty)\to\mathbb{R}^m$, the trajectory satisfies $\dot{x}(t)=F(x(t),d(t))$ for $\mathcal{L}^1$-a.e. $t$. Since $V:\mathbb{R}^n\to[0,\infty)$ is continuously differentiable and $x$ is absolutely continuous on compact intervals, the chain rule for absolutely continuous curves gives \begin{align*} \frac{d}{dt}V(x(t))=\nabla V(x(t))\cdot F(x(t),d(t)) \end{align*} for $\mathcal{L}^1$-a.e. $t$. Now we use the robustness built into the hypothesis. For almost every $t$, the disturbance value satisfies $|d(t)|\leq \bar d$, so $d(t)$ is one of the $\delta$ values over which the supremum is taken. Therefore \begin{align*} \nabla V(x(t))\cdot F(x(t),d(t))\leq \sup_{\delta\in\mathbb{R}^m,\ |\delta|\leq \bar d}\nabla V(x(t))\cdot F(x(t),\delta). \end{align*} The assumed Lyapunov inequality at the point $x(t)$ then yields \begin{align*} \frac{d}{dt}V(x(t))\leq -\alpha(|x(t)|)+c_{\bar d} \end{align*} for $\mathcal{L}^1$-a.e. $t\in[0,T_{\max})$. This is the key reduction: all admissible disturbed trajectories satisfy the same scalar Lyapunov inequality, independent of the particular measurable disturbance. [/guided] [/step] [step:Use properness to make the residual Lyapunov height finite] Fix $\varepsilon>0$ and define the residual set \begin{align*} A_{\varepsilon}:=\{y\in\mathbb{R}^n:\alpha(|y|)\leq c_{\bar d}+\varepsilon\}. \end{align*} Because $\alpha\in\mathcal{K}_{\infty}$ is continuous, strictly increasing, and unbounded, $A_{\varepsilon}$ is a closed bounded subset of $\mathbb{R}^n$, hence compact. Define the finite constant \begin{align*} M_{\varepsilon}:=\max_{y\in A_{\varepsilon}}V(y). \end{align*} If $V(z)>M_{\varepsilon}$ for some $z\in\mathbb{R}^n$, then $z\notin A_{\varepsilon}$, and hence \begin{align*} \alpha(|z|)>c_{\bar d}+\varepsilon. \end{align*} Consequently, whenever $V(x(t))>M_{\varepsilon}$, the almost-everywhere differential inequality gives \begin{align*} \frac{d}{dt}V(x(t))\leq -\varepsilon \end{align*} for $\mathcal{L}^1$-a.e. such $t$. [/step] [step:Obtain the uniform entrance time into the practical sublevel set] Let $R>0$ and define \begin{align*} M_R:=\max_{|y|\leq R}V(y). \end{align*} The maximum is finite because $V$ is continuous and $\{y\in\mathbb{R}^n:|y|\leq R\}$ is compact. Set \begin{align*} T(R,\varepsilon):=\frac{\max\{M_R-M_{\varepsilon}-\varepsilon,0\}}{\varepsilon}. \end{align*} For a fixed solution, define the absolutely continuous scalar map \begin{align*} w:[0,T_{\max})\to[0,\infty),\qquad w(t)=V(x(t)). \end{align*} If $|x(0)|\leq R$, then $w(0)\leq M_R$. On every interval on which $w(t)>M_{\varepsilon}+\varepsilon$, one also has $w(t)>M_{\varepsilon}$, and the preceding step gives $w'(t)\leq -\varepsilon$ for $\mathcal{L}^1$-a.e. such $t$. Integrating this almost-everywhere inequality over that interval gives \begin{align*} w(t)\leq w(s)-\varepsilon(t-s) \end{align*} whenever $s<t$ and $w(q)>M_{\varepsilon}+\varepsilon$ for all $q\in(s,t)$. Hence the solution enters the sublevel set \begin{align*} \{z\in\mathbb{R}^n:V(z)\leq M_{\varepsilon}+\varepsilon\} \end{align*} no later than time $T(R,\varepsilon)$, provided the solution exists up to that time. The same integral argument proves positive invariance of this sublevel set. Suppose, to the contrary, that after an entrance time there are $a<b$ with $w(a)\leq M_{\varepsilon}+\varepsilon$ and $w(b)>M_{\varepsilon}+\varepsilon$. By continuity of $w$, choose $s\in[a,b)$ such that $w(s)=M_{\varepsilon}+\varepsilon$ and $w(q)>M_{\varepsilon}+\varepsilon$ for every $q\in(s,b]$. Integrating $w'(q)\leq -\varepsilon$ for $\mathcal{L}^1$-a.e. $q\in(s,b)$ gives \begin{align*} w(b)\leq w(s)-\varepsilon(b-s)<M_{\varepsilon}+\varepsilon, \end{align*} which contradicts the choice of $b$. Therefore, for all $t\geq T(R,\varepsilon)$ in the interval of existence, \begin{align*} V(x(t))\leq M_{\varepsilon}+\varepsilon=\max_{\alpha(|y|)\leq c_{\bar d}+\varepsilon}V(y)+\varepsilon. \end{align*} This estimate is uniform over all admissible disturbances and all initial conditions with $|x(0)|\leq R$. [/step] [step:Exclude finite escape by compact sublevel bounds] It remains to show that $T_{\max}=\infty$. Let $\tau\in(0,T_{\max})$. Define $w:[0,T_{\max})\to[0,\infty)$ by $w(t)=V(x(t))$. The integral argument from the previous step gives the following bound on the whole interval $[0,\tau]$: \begin{align*} w(t)\leq \max\{w(0),M_{\varepsilon}+\varepsilon\} \end{align*} for every $t\in[0,\tau]$. Indeed, while $w(t)>M_{\varepsilon}+\varepsilon$, the inequality $w'(t)\leq-\varepsilon$ holds for $\mathcal{L}^1$-a.e. such $t$, so $w$ cannot increase above its initial value on any interval contained in that region; and the no-exit argument already proved that once $w$ reaches $M_{\varepsilon}+\varepsilon$, it cannot cross upward through that level. Thus $x([0,\tau])$ is contained in the sublevel set \begin{align*} K:=\{z\in\mathbb{R}^n:V(z)\leq \max\{V(x(0)),M_{\varepsilon}+\varepsilon\}\}. \end{align*} This set is compact because $V$ is proper and continuous. We now verify the hypotheses of the Caratheodory continuation criterion on this compact state set. The map $(t,z)\mapsto F(z,d(t))$ is measurable in $t$ for each fixed $z\in K$, because $d$ is Lebesgue-measurable and $F$ is continuous. For each compact neighbourhood $K'\subset\mathbb{R}^n$ of $K$, the locally Lipschitz maps $f$, $g$, $p$, and $k$ are bounded on $K'$, and the maps $z\mapsto f(z)+g(z)k(z)$ and $z\mapsto p(z)$ have finite Lipschitz constants on $K'$. Since $|d(t)|\leq\bar d$ for $\mathcal{L}^1$-a.e. $t$, there are constants $C_{K'}>0$ and $L_{K'}>0$, depending only on $K'$ and $\bar d$, such that for $\mathcal{L}^1$-a.e. $t$ and all $z_1,z_2\in K'$, \begin{align*} |F(z_1,d(t))-F(z_2,d(t))|\leq L_{K'}|z_1-z_2| \end{align*} and \begin{align*} |F(z_1,d(t))|\leq C_{K'}. \end{align*} The constant functions $t\mapsto L_{K'}$ and $t\mapsto C_{K'}$ belong to $L^1([0,S],\mathcal{L}^1)$ for every finite $S>0$. Hence the Caratheodory vector field has measurable time dependence, locally Lipschitz state dependence, and local $L^1$ growth bounds on finite intervals. The standard continuation criterion for Caratheodory ordinary differential equations therefore rules out finite maximal time while the trajectory remains in a compact subset of the state space. Since the preceding compact bound holds on every $[0,\tau]$ with $\tau<T_{\max}$, finite escape is impossible, and $T_{\max}=\infty$. Combining this with the previous step proves forward completeness and the asserted uniform ultimate bound. [/step] [step:Derive uniform global asymptotic stability in the zero residual case] Assume now that $c_{\bar d}=0$ and that \begin{align*} F(0,\delta)=0 \end{align*} for every $\delta\in\mathbb{R}^m$ with $|\delta|\leq \bar d$. Then the origin is an equilibrium for every admissible disturbance, because $F(0,d(t))=0$ for $\mathcal{L}^1$-a.e. $t$. Since $V$ is continuous, positive definite, and proper, the standard comparison-functions lemma for proper positive definite functions gives functions $\underline{\alpha},\overline{\alpha}\in\mathcal{K}_{\infty}$ such that \begin{align*} \underline{\alpha}(|z|)\leq V(z)\leq \overline{\alpha}(|z|) \end{align*} for every $z\in\mathbb{R}^n$. Concretely, these functions may be obtained by regularizing the radial lower and upper envelopes of $V$ on spheres and closed balls; positive definiteness gives a positive lower envelope away from $0$, and properness makes the upper envelope finite and unbounded. Along every admissible trajectory, \begin{align*} \frac{d}{dt}V(x(t))\leq -\alpha(|x(t)|) \end{align*} for $\mathcal{L}^1$-a.e. $t$. In particular, $V(x(t))$ is nonincreasing. For uniform stability, let $\rho>0$ be given. Since $\underline{\alpha}(\rho)>0$ and $\overline{\alpha}(s)\to0$ as $s\downarrow0$, choose $\delta_{\rho}>0$ such that $\overline{\alpha}(\delta_{\rho})<\underline{\alpha}(\rho)$. If $|x(0)|\leq\delta_{\rho}$, then for all $t\geq0$, \begin{align*} \underline{\alpha}(|x(t)|)\leq V(x(t))\leq V(x(0))\leq \overline{\alpha}(|x(0)|)<\underline{\alpha}(\rho), \end{align*} so $|x(t)|<\rho$. For uniform attractivity on an initial ball $\{z\in\mathbb{R}^n:|z|\leq R\}$, fix $\rho>0$. Define \begin{align*} \eta_\rho:=\frac{1}{2}\underline{\alpha}(\rho)>0. \end{align*} Also define the annular sublevel set \begin{align*} E_{R,\rho}:=\{z\in\mathbb{R}^n:\eta_\rho\leq V(z)\leq\overline{\alpha}(R)\}. \end{align*} If $E_{R,\rho}=\varnothing$, then every solution with $|x(0)|\leq R$ satisfies $V(x(0))\leq\overline{\alpha}(R)<\eta_\rho$. Since $V(x(t))$ is nonincreasing, $V(x(t))<\eta_\rho$ for every $t\geq0$, so the attractivity conclusion holds with $T_R(\rho):=0$. It remains to consider the case $E_{R,\rho}\neq\varnothing$. The set $E_{R,\rho}$ is compact because it is a closed subset of the proper sublevel set $\{z\in\mathbb{R}^n:V(z)\leq\overline{\alpha}(R)\}$. It does not contain the origin, since $V(0)=0<\eta_\rho$. Therefore the continuous positive definite function $z\mapsto\alpha(|z|)$ has a positive minimum on $E_{R,\rho}$. Define \begin{align*} \mu_R(\rho):=\min\{\alpha(|z|):z\in E_{R,\rho}\}>0. \end{align*} If $|x(0)|\leq R$, then $V(x(0))\leq\overline{\alpha}(R)$, and monotonicity gives $V(x(t))\leq\overline{\alpha}(R)$ for all $t\geq0$. Hence whenever $V(x(t))\geq\eta_\rho$, one has $x(t)\in E_{R,\rho}$ and therefore $\alpha(|x(t)|)\geq\mu_R(\rho)$. The differential inequality yields \begin{align*} \frac{d}{dt}V(x(t))\leq -\mu_R(\rho) \end{align*} for $\mathcal{L}^1$-a.e. $t$ such that $V(x(t))\geq\eta_\rho$. Set \begin{align*} T_R(\rho):=\frac{\max\{\overline{\alpha}(R)-\eta_\rho,0\}}{\mu_R(\rho)}. \end{align*} The same absolutely continuous comparison and no-exit argument used for the practical sublevel set shows that $V(x(t))<\eta_\rho$ for all $t\geq T_R(\rho)$. Indeed, before entering $\{z\in\mathbb{R}^n:V(z)<\eta_\rho\}$ the scalar function $t\mapsto V(x(t))$ decreases at rate at least $\mu_R(\rho)$, and once it reaches the boundary value $\eta_\rho$, the same inequality prevents an upward crossing through that boundary. Finally, in either case, $V(x(t))<\eta_\rho=\frac{1}{2}\underline{\alpha}(\rho)$ implies \begin{align*} \underline{\alpha}(|x(t)|)\leq V(x(t))<\underline{\alpha}(\rho), \end{align*} so $|x(t)|<\rho$. Hence for every $R>0$ and every $\rho>0$ there exists $T_R(\rho)\geq0$ such that $|x(0)|\leq R$ implies $|x(t)|<\rho$ for all $t\geq T_R(\rho)$, uniformly over all admissible disturbances. Thus the origin is uniformly globally attractive. Uniform stability and uniform global attractivity together give global asymptotic stability of the origin, uniformly over all Lebesgue-measurable disturbances satisfying $|d(t)|\leq\bar d$ for $\mathcal{L}^1$-a.e. $t\geq0$. [/step]