[proofplan]
The argument is the linearity of both the derivative and the right-hand side of the state equation. First we check that the linear combination of absolutely continuous trajectories is again absolutely continuous, so it is an admissible candidate trajectory. Then we restrict to the full-measure set where both original differential equations hold and both derivatives exist, compute the derivative of the linear combination there, and use the linearity of the matrices $A$ and $B$. Finally, evaluating at $t=0$ gives the asserted initial condition.
[/proofplan]
[step:Show the linear combination is an admissible absolutely continuous trajectory]
Define the function $x: [0,T] \to \mathbb{R}^n$ by
\begin{align*}
x(t)=\alpha x_1(t)+\beta x_2(t).
\end{align*}
Since $x_1$ and $x_2$ are absolutely continuous and finite linear combinations of absolutely continuous $\mathbb{R}^n$-valued functions are absolutely continuous, $x$ is absolutely continuous on $[0,T]$.
Also define the control $u: [0,T] \to \mathbb{R}^m$ by
\begin{align*}
u(t)=\alpha u_1(t)+\beta u_2(t).
\end{align*}
Because $L^2([0,T];\mathbb{R}^m)$ is a real [vector space](/page/Vector%20Space), $u \in L^2([0,T];\mathbb{R}^m)$.
[guided]
We need first to check that the proposed trajectory is a legitimate object of the same kind as the original trajectories. Define $x: [0,T] \to \mathbb{R}^n$ by
\begin{align*}
x(t)=\alpha x_1(t)+\beta x_2(t).
\end{align*}
The hypotheses say that $x_1$ and $x_2$ are absolutely continuous functions from $[0,T]$ into $\mathbb{R}^n$. Absolute continuity is stable under multiplication by real scalars and addition, so the finite linear combination $\alpha x_1+\beta x_2$ is absolutely continuous. Hence $x$ is absolutely continuous on $[0,T]$.
The input appearing in the new state equation is the corresponding linear combination of the two inputs. Define $u: [0,T] \to \mathbb{R}^m$ by
\begin{align*}
u(t)=\alpha u_1(t)+\beta u_2(t).
\end{align*}
Since $u_1,u_2 \in L^2([0,T];\mathbb{R}^m)$ and $L^2([0,T];\mathbb{R}^m)$ is closed under finite real linear combinations, we have $u \in L^2([0,T];\mathbb{R}^m)$. Thus both the proposed trajectory and the proposed control have the regularity required by the statement.
[/guided]
[/step]
[step:Differentiate on the common full-measure set where both state equations hold]
For each $i \in \{1,2\}$, let $E_i \subset [0,T]$ be a set of full $\mathcal{L}^1$-measure on which $x_i$ is differentiable and
\begin{align*}
\dot{x}_i(t)=Ax_i(t)+Bu_i(t).
\end{align*}
Set $E=E_1 \cap E_2$. Since the intersection of two full-measure subsets of $[0,T]$ has full $\mathcal{L}^1$-measure, $E$ has full $\mathcal{L}^1$-measure in $[0,T]$.
For every $t \in E$, both derivatives $\dot{x}_1(t)$ and $\dot{x}_2(t)$ exist. By linearity of the derivative at such a point,
\begin{align*}
\dot{x}(t)=\alpha \dot{x}_1(t)+\beta \dot{x}_2(t).
\end{align*}
Using the two state equations at $t \in E$, this becomes
\begin{align*}
\dot{x}(t)=\alpha(Ax_1(t)+Bu_1(t))+\beta(Ax_2(t)+Bu_2(t)).
\end{align*}
[/step]
[step:Use linearity of the matrices to identify the new state equation]
Fix $t \in E$. Since $A:\mathbb{R}^n \to \mathbb{R}^n$ and $B:\mathbb{R}^m \to \mathbb{R}^n$ are linear maps, we have
\begin{align*}
\alpha(Ax_1(t)+Bu_1(t))+\beta(Ax_2(t)+Bu_2(t))=A(\alpha x_1(t)+\beta x_2(t))+B(\alpha u_1(t)+\beta u_2(t)).
\end{align*}
By the definitions of $x$ and $u$, this is
\begin{align*}
\dot{x}(t)=Ax(t)+Bu(t)=Ax(t)+B(\alpha u_1(t)+\beta u_2(t)).
\end{align*}
Since $E$ has full $\mathcal{L}^1$-measure, the differential equation holds for $\mathcal{L}^1$-a.e. $t \in [0,T]$.
[/step]
[step:Evaluate the linear combination at the initial time]
Using the definition of $x$ at $t=0$ and the initial conditions for $x_1$ and $x_2$, we obtain
\begin{align*}
x(0)=\alpha x_1(0)+\beta x_2(0)=\alpha x_{1,0}+\beta x_{2,0}.
\end{align*}
Thus $x$ is an absolutely continuous solution of the stated linear state equation with the stated initial condition.
[/step]
