[proofplan] The proof first rewrites each absolutely continuous controlled trajectory in integral form and uses the linear growth hypothesis to obtain a scalar integral inequality for $|x_k(t)|$. An elementary Gronwall estimate, proved inside the argument to avoid any external dependence, gives a uniform bound independent of $k$ and $t$. The same linear growth estimate then bounds the derivatives $\dot{x}_k$ uniformly almost everywhere, and integrating this derivative bound over intervals gives a common Lipschitz modulus of continuity. [/proofplan]

[step:Convert the controlled ODE into a scalar integral inequality] Fix $k \in \mathbb{N}$. Since $x_k \in AC([t_0,t_1];\mathbb{R}^n)$ and $\dot{x}_k(t)=f(t,x_k(t),u_k(t))$ for $\mathcal{L}^1$-almost every $t \in [t_0,t_1]$, the integral characterization of absolute continuity gives, for every $t \in [t_0,t_1]$, \begin{align*} x_k(t)=x_0+\int_{t_0}^{t} f(s,x_k(s),u_k(s))\,d\mathcal{L}^1(s). \end{align*} Taking Euclidean norms and applying the triangle inequality for the Bochner integral in $\mathbb{R}^n$, we obtain \begin{align*} |x_k(t)| \leq |x_0|+\int_{t_0}^{t} |f(s,x_k(s),u_k(s))|\,d\mathcal{L}^1(s). \end{align*} The growth hypothesis applies because $s \in [t_0,t_1]$, $x_k(s) \in \mathbb{R}^n$, and $u_k(s) \in U_c$ for every $s$ at which $u_k$ is defined. Hence \begin{align*} |x_k(t)| \leq |x_0|+\int_{t_0}^{t} (a+b|x_k(s)|)\,d\mathcal{L}^1(s). \end{align*} Since $t-t_0 \leq T := t_1-t_0$, this implies \begin{align*} |x_k(t)| \leq |x_0|+aT+b\int_{t_0}^{t} |x_k(s)|\,d\mathcal{L}^1(s). \end{align*} [/step]

[step:Apply an elementary Gronwall estimate to bound every trajectory]Define $A := |x_0|+aT$. For the fixed index $k$, define $y_k: [t_0,t_1] \to [0,\infty)$ by $y_k(t) := |x_k(t)|$. The previous step gives \begin{align*} y_k(t) \leq A+b\int_{t_0}^{t} y_k(s)\,d\mathcal{L}^1(s) \end{align*} for every $t \in [t_0,t_1]$. We use the following elementary integral estimate. If $y: [t_0,t_1]\to [0,\infty)$ is integrable and satisfies \begin{align*} y(t) \leq A+b\int_{t_0}^{t} y(s)\,d\mathcal{L}^1(s) \end{align*} for every $t \in [t_0,t_1]$, then \begin{align*} y(t) \leq Ae^{b(t-t_0)} \end{align*} for every $t \in [t_0,t_1]$. Indeed, define $z: [t_0,t_1]\to [0,\infty)$ by \begin{align*} z(t):=A+b\int_{t_0}^{t} y(s)\,d\mathcal{L}^1(s). \end{align*} Then $z$ is absolutely continuous, $y(t)\leq z(t)$ for every $t$, and \begin{align*} z'(t)=by(t)\leq bz(t) \end{align*} for $\mathcal{L}^1$-almost every $t \in [t_0,t_1]$. Define $w: [t_0,t_1]\to \mathbb{R}$ by $w(t):=e^{-b(t-t_0)}z(t)$. Since $z$ is absolutely continuous and the exponential factor is smooth, $w$ is absolutely continuous, and for $\mathcal{L}^1$-almost every $t$, \begin{align*} w'(t)=e^{-b(t-t_0)}(z'(t)-bz(t))\leq 0. \end{align*} Therefore $w(t)\leq w(t_0)=A$ for every $t \in [t_0,t_1]$, so $z(t)\leq Ae^{b(t-t_0)}$. Since $y(t)\leq z(t)$, the estimate follows. Applying this estimate to $y_k$ gives \begin{align*} |x_k(t)|=y_k(t)\leq Ae^{b(t-t_0)}\leq Ae^{bT} \end{align*} for every $t \in [t_0,t_1]$. Thus, with \begin{align*} R:=(|x_0|+aT)e^{bT}, \end{align*} we have $|x_k(t)|\leq R$ for every $k \in \mathbb{N}$ and every $t \in [t_0,t_1]$.[/step]

[guided]The main point is to turn the vector-valued ODE into a one-dimensional inequality for the size of the trajectory. For a fixed $k \in \mathbb{N}$, set $y_k(t):=|x_k(t)|$. From the previous step, \begin{align*} y_k(t) \leq |x_0|+aT+b\int_{t_0}^{t} y_k(s)\,d\mathcal{L}^1(s). \end{align*} The constant term is independent of both $k$ and $t$, so define \begin{align*} A:=|x_0|+aT. \end{align*} Then \begin{align*} y_k(t) \leq A+b\int_{t_0}^{t} y_k(s)\,d\mathcal{L}^1(s). \end{align*} We now prove the needed Gronwall estimate directly. Suppose $y: [t_0,t_1]\to [0,\infty)$ is integrable and satisfies \begin{align*} y(t) \leq A+b\int_{t_0}^{t} y(s)\,d\mathcal{L}^1(s). \end{align*} The useful auxiliary function is the right-hand side: \begin{align*} z(t):=A+b\int_{t_0}^{t} y(s)\,d\mathcal{L}^1(s). \end{align*} This choice is natural because the inequality says $y(t)\leq z(t)$, while differentiating $z$ recovers $y$. Since $y$ is integrable, $z$ is absolutely continuous, and for $\mathcal{L}^1$-almost every $t \in [t_0,t_1]$, \begin{align*} z'(t)=by(t)\leq bz(t). \end{align*} To remove the factor $bz(t)$, multiply by the integrating factor $e^{-b(t-t_0)}$. Define $w: [t_0,t_1]\to \mathbb{R}$ by \begin{align*} w(t):=e^{-b(t-t_0)}z(t). \end{align*} The product of an absolutely [continuous function](/page/Continuous%20Function) and a smooth function is absolutely continuous, so $w$ is absolutely continuous. For $\mathcal{L}^1$-almost every $t$, \begin{align*} w'(t)=e^{-b(t-t_0)}(z'(t)-bz(t))\leq 0. \end{align*} Thus $w$ is nonincreasing on $[t_0,t_1]$, and hence \begin{align*} w(t)\leq w(t_0)=z(t_0)=A. \end{align*} Multiplying by $e^{b(t-t_0)}$ gives \begin{align*} z(t)\leq Ae^{b(t-t_0)}. \end{align*} Since $y(t)\leq z(t)$, we conclude \begin{align*} y(t)\leq Ae^{b(t-t_0)}. \end{align*} Applying this estimate to $y_k(t)=|x_k(t)|$ gives \begin{align*} |x_k(t)|\leq Ae^{b(t-t_0)}\leq Ae^{bT}. \end{align*} Therefore every trajectory remains inside the same closed Euclidean ball $\overline{B}(0,R)$, where \begin{align*} R:=(|x_0|+aT)e^{bT}. \end{align*} The important feature is that $R$ depends only on $a$, $b$, $t_0$, $t_1$, and $x_0$, not on the index $k$ or on the particular measurable control $u_k$.[/guided]

[step:Bound the derivatives uniformly almost everywhere] Let \begin{align*} L:=a+bR. \end{align*} For every $k \in \mathbb{N}$, the differential equation holds for $\mathcal{L}^1$-almost every $t \in [t_0,t_1]$. At such a time $t$, the growth estimate and the uniform bound from the previous step give \begin{align*} |\dot{x}_k(t)|=|f(t,x_k(t),u_k(t))|\leq a+b|x_k(t)|\leq a+bR=L. \end{align*} Thus $|\dot{x}_k(t)|\leq L$ for every $k \in \mathbb{N}$ and for $\mathcal{L}^1$-almost every $t \in [t_0,t_1]$. [/step]

[step:Integrate the derivative bound to obtain a common modulus of continuity] Fix $k \in \mathbb{N}$ and $s,t \in [t_0,t_1]$. If $s \leq t$, absolute continuity gives \begin{align*} x_k(t)-x_k(s)=\int_s^t \dot{x}_k(r)\,d\mathcal{L}^1(r). \end{align*} Taking norms, applying the triangle inequality for the Bochner integral, and using the almost everywhere derivative bound from the previous step, we get \begin{align*} |x_k(t)-x_k(s)|\leq \int_s^t |\dot{x}_k(r)|\,d\mathcal{L}^1(r)\leq \int_s^t L\,d\mathcal{L}^1(r)=L(t-s). \end{align*} If $t \leq s$, the same estimate with $s$ and $t$ interchanged gives \begin{align*} |x_k(t)-x_k(s)|\leq L(s-t). \end{align*} Therefore, for all $s,t \in [t_0,t_1]$, \begin{align*} |x_k(t)-x_k(s)|\leq L|t-s|. \end{align*} The constant $L=a+bR$ is independent of $k$, so the family $(x_k)_{k \in \mathbb{N}}$ has the common Lipschitz modulus $\omega(\delta):=L\delta$. Hence $(x_k)_{k \in \mathbb{N}}$ is equicontinuous on $[t_0,t_1]$. Together with the bound $|x_k(t)|\leq R$, this proves both asserted conclusions. [/step]