[proofplan]
The proof constructs a value-preserving correspondence between admissible Bolza pairs and admissible augmented Mayer triples. Starting from a Bolza pair, the additional state $z$ is defined as the accumulated running cost, so the fundamental theorem for absolutely continuous functions gives the augmented differential equation. Conversely, any admissible augmented triple has $z$ equal to that same accumulated cost because $z(t_0)=0$ and $\dot z=L$ almost everywhere. Thus the two feasible problems have exactly the same objective values, and minimizers correspond.
[/proofplan]
[step:Send each Bolza admissible pair to an augmented Mayer admissible triple]
Let $(x,u) \in \mathcal A$ be arbitrary. Define
\begin{align*}
\ell_{x,u}: [t_0,t_1] \to \mathbb{R}
\end{align*}
by
\begin{align*}
\ell_{x,u}(t) = L(t,x(t),u(t)).
\end{align*}
By the definition of $\mathcal A$, $\ell_{x,u} \in L^1([t_0,t_1],\mathcal{L}^1)$. Define
\begin{align*}
z_{x,u}: [t_0,t_1] \to \mathbb{R}
\end{align*}
by
\begin{align*}
z_{x,u}(t) = \int_{[t_0,t]} \ell_{x,u}(s)\, d\mathcal{L}^1(s).
\end{align*}
By the [Fundamental Theorem of Calculus](/theorems/632) for absolutely continuous functions applied to the $L^1$ function $\ell_{x,u}$, the function $z_{x,u}$ is absolutely continuous, satisfies $z_{x,u}(t_0)=0$, and satisfies
\begin{align*}
\dot{z}_{x,u}(t) = \ell_{x,u}(t) = L(t,x(t),u(t))
\end{align*}
for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$. The pair $(x,u)$ already satisfies the original dynamics and all original admissibility, endpoint, path, and control constraints. Hence $(x,z_{x,u},u)$ is admissible for the augmented Mayer problem.
[guided]
Start with an admissible Bolza pair $(x,u) \in \mathcal A$. The only missing variable in the Mayer formulation is the scalar bookkeeping state $z$, which should record the running cost accumulated up to time $t$. To make this precise, define
\begin{align*}
\ell_{x,u}: [t_0,t_1] \to \mathbb{R}
\end{align*}
by
\begin{align*}
\ell_{x,u}(t) = L(t,x(t),u(t)).
\end{align*}
The definition of $\mathcal A$ includes exactly the integrability hypothesis needed here: $\ell_{x,u} \in L^1([t_0,t_1],\mathcal{L}^1)$. Therefore its integral primitive is well-defined. Define
\begin{align*}
z_{x,u}: [t_0,t_1] \to \mathbb{R}
\end{align*}
by
\begin{align*}
z_{x,u}(t) = \int_{[t_0,t]} \ell_{x,u}(s)\, d\mathcal{L}^1(s).
\end{align*}
We now verify the augmented admissibility conditions one by one. First, the [Fundamental Theorem of Calculus](/theorems/632) for absolutely continuous functions says that the indefinite integral of an $L^1$ function is absolutely continuous and has derivative equal to the integrand almost everywhere. Applying this to $\ell_{x,u}$ gives that $z_{x,u}$ is absolutely continuous and
\begin{align*}
\dot{z}_{x,u}(t) = \ell_{x,u}(t)
\end{align*}
for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$. Substituting the definition of $\ell_{x,u}$ gives
\begin{align*}
\dot{z}_{x,u}(t) = L(t,x(t),u(t))
\end{align*}
for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$.
Second, evaluating the defining integral at the initial time gives
\begin{align*}
z_{x,u}(t_0) = \int_{[t_0,t_0]} \ell_{x,u}(s)\, d\mathcal{L}^1(s) = 0,
\end{align*}
because the interval $[t_0,t_0]$ has $\mathcal{L}^1$-measure zero. Finally, the augmented problem imposes on $x$ and $u$ exactly the same original admissibility, endpoint, path, and control constraints as $\mathcal A$, and these already hold because $(x,u) \in \mathcal A$. Thus $(x,z_{x,u},u)$ is admissible for the augmented Mayer problem.
[/guided]
[/step]
[step:Compute that the Mayer terminal cost equals the Bolza cost]
For the augmented triple constructed above, the terminal value of the new state is
\begin{align*}
z_{x,u}(t_1) = \int_{[t_0,t_1]} L(t,x(t),u(t))\, d\mathcal{L}^1(t).
\end{align*}
Therefore
\begin{align*}
\Psi(x(t_1),z_{x,u}(t_1)) = \Phi(x(t_1)) + z_{x,u}(t_1).
\end{align*}
Substituting the expression for $z_{x,u}(t_1)$ gives
\begin{align*}
\Psi(x(t_1),z_{x,u}(t_1)) = \Phi(x(t_1)) + \int_{[t_0,t_1]} L(t,x(t),u(t))\, d\mathcal{L}^1(t) = J[x,u].
\end{align*}
Thus every admissible Bolza pair produces an admissible augmented Mayer triple with the same objective value.
[/step]
[step:Recover the Bolza cost from any augmented Mayer admissible triple]
Let $(x,z,u)$ be admissible for the augmented Mayer problem. By its definition, $(x,u) \in \mathcal A$, the function $z: [t_0,t_1] \to \mathbb{R}$ is absolutely continuous, $z(t_0)=0$, and
\begin{align*}
\dot{z}(t) = L(t,x(t),u(t))
\end{align*}
for $\mathcal{L}^1$-a.e. $t \in [t_0,t_1]$. Since $(x,u) \in \mathcal A$, the map
\begin{align*}
\ell_{x,u}: [t_0,t_1] \to \mathbb{R}
\end{align*}
defined by
\begin{align*}
\ell_{x,u}(t) = L(t,x(t),u(t))
\end{align*}
belongs to $L^1([t_0,t_1],\mathcal{L}^1)$.
Applying the integral representation in the [Fundamental Theorem of Calculus](/theorems/632) for absolutely continuous functions to $z$, for every $\tau \in [t_0,t_1]$ we have
\begin{align*}
z(\tau) - z(t_0) = \int_{[t_0,\tau]} \dot{z}(s)\, d\mathcal{L}^1(s).
\end{align*}
Using $z(t_0)=0$ and $\dot z(s)=L(s,x(s),u(s))$ for $\mathcal{L}^1$-a.e. $s$ yields
\begin{align*}
z(\tau) = \int_{[t_0,\tau]} L(s,x(s),u(s))\, d\mathcal{L}^1(s)
\end{align*}
for every $\tau \in [t_0,t_1]$. In particular, taking $\tau=t_1$ gives
\begin{align*}
z(t_1) = \int_{[t_0,t_1]} L(s,x(s),u(s))\, d\mathcal{L}^1(s).
\end{align*}
Consequently
\begin{align*}
\Psi(x(t_1),z(t_1)) = \Phi(x(t_1)) + \int_{[t_0,t_1]} L(t,x(t),u(t))\, d\mathcal{L}^1(t) = J[x,u].
\end{align*}
Thus every admissible augmented Mayer triple projects to an admissible Bolza pair with the same objective value.
[/step]
[step:Conclude equality of optimal values and correspondence of minimizers]
Let $\mathcal V_B$ denote the set of objective values attained by admissible Bolza pairs:
\begin{align*}
\mathcal V_B = \{J[x,u] : (x,u) \in \mathcal A\}.
\end{align*}
Let $\mathcal V_M$ denote the set of objective values attained by admissible augmented Mayer triples:
\begin{align*}
\mathcal V_M = \{\Psi(x(t_1),z(t_1)) : (x,z,u) \text{ is augmented Mayer admissible}\}.
\end{align*}
The first construction shows $\mathcal V_B \subset \mathcal V_M$, and the reverse recovery shows $\mathcal V_M \subset \mathcal V_B$. Hence
\begin{align*}
\mathcal V_B = \mathcal V_M.
\end{align*}
Therefore the two problems have the same optimal value, namely the common infimum of this set of feasible objective values, with the convention that the infimum of the empty set is $+\infty$.
It remains to prove the minimizer statement. Suppose $(x,u) \in \mathcal A$ minimizes $J$, and define
\begin{align*}
z(t) = \int_{[t_0,t]} L(s,x(s),u(s))\, d\mathcal{L}^1(s).
\end{align*}
The first step shows that $(x,z,u)$ is augmented Mayer admissible, and the value identity gives
\begin{align*}
\Psi(x(t_1),z(t_1)) = J[x,u].
\end{align*}
For any augmented Mayer admissible triple $(\tilde{x},\tilde{z},\tilde{u})$, the third step gives
\begin{align*}
\Psi(\tilde{x}(t_1),\tilde{z}(t_1)) = J[\tilde{x},\tilde{u}].
\end{align*}
Since $(x,u)$ minimizes $J$ over $\mathcal A$,
\begin{align*}
J[x,u] \leq J[\tilde{x},\tilde{u}].
\end{align*}
Thus
\begin{align*}
\Psi(x(t_1),z(t_1)) \leq \Psi(\tilde{x}(t_1),\tilde{z}(t_1)),
\end{align*}
so $(x,z,u)$ minimizes the augmented Mayer problem.
Conversely, suppose $(x,z,u)$ minimizes the augmented Mayer problem. The third step shows $(x,u) \in \mathcal A$ and
\begin{align*}
\Psi(x(t_1),z(t_1)) = J[x,u].
\end{align*}
For any $(\tilde{x},\tilde{u}) \in \mathcal A$, define the map
\begin{align*}
\tilde{z}: [t_0,t_1] \to \mathbb{R}
\end{align*}
by
\begin{align*}
\tilde{z}(t) = \int_{[t_0,t]} L(s,\tilde{x}(s),\tilde{u}(s))\, d\mathcal{L}^1(s).
\end{align*}
The first step shows that $(\tilde{x},\tilde{z},\tilde{u})$ is augmented Mayer admissible, and the second step gives
\begin{align*}
\Psi(\tilde{x}(t_1),\tilde{z}(t_1)) = J[\tilde{x},\tilde{u}].
\end{align*}
Minimality of $(x,z,u)$ for the augmented Mayer problem implies
\begin{align*}
J[x,u] = \Psi(x(t_1),z(t_1)) \leq \Psi(\tilde{x}(t_1),\tilde{z}(t_1)) = J[\tilde{x},\tilde{u}].
\end{align*}
Since $(\tilde{x},\tilde{u}) \in \mathcal A$ was arbitrary, $(x,u)$ minimizes $J$ over $\mathcal A$. This proves the stated equivalence of minimizers.
[/step]
