[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 $t \in [t_0,t_1]$.
[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]
[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]
