[proofplan]
The proof is an identification of the two matrix-valued integrands. We write the controllability Gramian of the dual pair using the standard controllability-Gramian formula, then simplify the double transposes $(C^\top)^\top$ and $(A^\top)^\top$. After this simplification, the integrand is exactly the defining integrand of the observability Gramian.
[/proofplan]
[step:Write the dual controllability Gramian with matching dimensions]
Since $A \in \mathbb{R}^{n \times n}$ and $C \in \mathbb{R}^{p \times n}$, the dual input matrix is $C^\top \in \mathbb{R}^{n \times p}$. The dual pair is therefore $(A^\top, C^\top)$, where $A^\top \in \mathbb{R}^{n \times n}$ is the state matrix and $C^\top$ is the input matrix. Its controllability Gramian over $[0,T]$ is
\begin{align*}
W_c^{\mathrm{dual}}(T) := \int_0^{\!T}e^{tA^\top} C^\top (C^\top)^\top e^{t(A^\top)^\top} \, d\mathcal{L}^1(t).
\end{align*}
The integrand is an $n \times n$ matrix for each $t \in [0,T]$, because
\begin{align*}
e^{tA^\top} \in \mathbb{R}^{n \times n}, \quad C^\top \in \mathbb{R}^{n \times p}, \quad (C^\top)^\top \in \mathbb{R}^{p \times n}, \quad e^{t(A^\top)^\top} \in \mathbb{R}^{n \times n}.
\end{align*}
[/step]
[step:Simplify the transpose identities inside the integrand]
For every real matrix, taking the transpose twice returns the original matrix. Applying this to $C$ and $A$ gives
\begin{align*}
(C^\top)^\top = C
\end{align*}
and
\begin{align*}
(A^\top)^\top = A.
\end{align*}
Substituting these identities into the dual controllability Gramian gives
\begin{align*}
W_c^{\mathrm{dual}}(T) = \int_0^{\!T}e^{tA^\top} C^\top C e^{tA} \, d\mathcal{L}^1(t).
\end{align*}
[guided]
The only algebraic issue is to make sure that the dual controllability Gramian has the same integrand as the observability Gramian. The transpose operation satisfies the involution identity: for every real matrix $M$, one has $(M^\top)^\top = M$. We apply this identity first to the matrix $C \in \mathbb{R}^{p \times n}$, obtaining
\begin{align*}
(C^\top)^\top = C.
\end{align*}
We also apply the same identity to the state matrix $A \in \mathbb{R}^{n \times n}$, obtaining
\begin{align*}
(A^\top)^\top = A.
\end{align*}
Now substitute these two equalities into the defining formula for the dual controllability Gramian:
\begin{align*}
W_c^{\mathrm{dual}}(T) = \int_0^{\!T}e^{tA^\top} C^\top (C^\top)^\top e^{t(A^\top)^\top} \, d\mathcal{L}^1(t).
\end{align*}
Replacing $(C^\top)^\top$ by $C$ and replacing $(A^\top)^\top$ by $A$ gives
\begin{align*}
W_c^{\mathrm{dual}}(T) = \int_0^{\!T}e^{tA^\top} C^\top C e^{tA} \, d\mathcal{L}^1(t).
\end{align*}
This is the point of passing to the dual pair: the input factor $C^\top (C^\top)^\top$ becomes the observation factor $C^\top C$, while the terminal exponential $e^{t(A^\top)^\top}$ becomes $e^{tA}$.
[/guided]
[/step]
[step:Identify the resulting integral with the observability Gramian]
By the definition of the observability Gramian of $(C,A)$ over $[0,T]$,
\begin{align*}
W_o(T) = \int_0^{\!T}e^{tA^\top} C^\top C e^{tA} \, d\mathcal{L}^1(t).
\end{align*}
The expression obtained for $W_c^{\mathrm{dual}}(T)$ is the same matrix-valued integral. Therefore
\begin{align*}
W_c^{\mathrm{dual}}(T) = W_o(T).
\end{align*}
This proves that the observability Gramian of $(C,A)$ over $[0,T]$ is exactly the controllability Gramian of the dual pair $(A^\top,C^\top)$ over the same time interval.
[/step]
