[proofplan] We prove the criterion for an arbitrary fixed time horizon $T>0$. The key identity is that the quadratic form of the observability Gramian equals the time integral of the squared output generated by the initial state $x_0$. Since this output norm is continuous and nonnegative, the quadratic form vanishes exactly when the output vanishes on the whole interval $[0,T]$. Analyticity of the matrix exponential then promotes vanishing on $[0,T]$ to vanishing for all $t \geq 0$, which connects the Gramian condition to observability. [/proofplan]

[step:Express the Gramian quadratic form as the output energy] Fix $T>0$. For each vector $x_0 \in \mathbb{R}^n$, define the output trajectory \begin{align*} y_{x_0}: [0,T] \to \mathbb{R}^m,\quad y_{x_0}(t) = C e^{tA}x_0. \end{align*} Using the definition of $W_o(T)$ and linearity of the entrywise [Lebesgue integral](/page/Lebesgue%20Integral), we compute \begin{align*} x_0^\top W_o(T)x_0 = x_0^\top \left(\int_0^{\!T}e^{tA^\top}C^\top C e^{tA}\, d\mathcal{L}^1(t)\right)x_0. \end{align*} Because $x_0$ is independent of $t$, multiplication by $x_0^\top$ and $x_0$ passes through the finite-dimensional entrywise integral, giving \begin{align*} x_0^\top W_o(T)x_0 = \int_0^{\!T}x_0^\top e^{tA^\top}C^\top C e^{tA}x_0\, d\mathcal{L}^1(t). \end{align*} Since $x_0^\top e^{tA^\top} = (e^{tA}x_0)^\top$, the integrand is the Euclidean squared norm of $C e^{tA}x_0$: \begin{align*} x_0^\top e^{tA^\top}C^\top C e^{tA}x_0 = |C e^{tA}x_0|^2. \end{align*} Therefore \begin{align*} x_0^\top W_o(T)x_0 = \int_0^{\!T}|C e^{tA}x_0|^2\, d\mathcal{L}^1(t). \end{align*} [/step]

[step:Identify the kernel of the Gramian with zero output on $[0,T]$] For fixed $x_0 \in \mathbb{R}^n$, define \begin{align*} g_{x_0}: [0,T] \to [0,\infty),\quad t \mapsto |C e^{tA}x_0|^2. \end{align*} The map $g_{x_0}$ is continuous because $t \mapsto e^{tA}$ is continuous and matrix multiplication is continuous. By the previous step, \begin{align*} x_0^\top W_o(T)x_0 = \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t). \end{align*} If $C e^{tA}x_0 = 0$ for every $t \in [0,T]$, then $g_{x_0}=0$ on $[0,T]$, hence $x_0^\top W_o(T)x_0=0$. Conversely, suppose $x_0^\top W_o(T)x_0=0$. Then \begin{align*} \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t)=0. \end{align*} Since $g_{x_0}$ is continuous and nonnegative, it must vanish at every point of $[0,T]$. Indeed, if $g_{x_0}(t_0)>0$ for some $t_0 \in [0,T]$, continuity gives a subinterval $I \subset [0,T]$ with positive $\mathcal{L}^1$-measure and a constant $\varepsilon>0$ such that $g_{x_0}(t)\geq \varepsilon$ for every $t \in I$, forcing \begin{align*} \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t) \geq \int_I \varepsilon\, d\mathcal{L}^1(t) = \varepsilon\,\mathcal{L}^1(I)>0, \end{align*} a contradiction. Hence $C e^{tA}x_0=0$ for every $t \in [0,T]$.[/step]

[guided]Fix an initial state $x_0 \in \mathbb{R}^n$. The point of this step is to turn the algebraic condition $x_0^\top W_o(T)x_0=0$ into the dynamical condition that the measured output is zero throughout the observation interval. Define \begin{align*} g_{x_0}: [0,T] \to [0,\infty),\quad t \mapsto |C e^{tA}x_0|^2. \end{align*} This function is continuous because the matrix exponential depends continuously on $t$, and the maps given by multiplication by $C$, multiplication by $x_0$, and the Euclidean norm squared are continuous. From the Gramian identity already proved, \begin{align*} x_0^\top W_o(T)x_0 = \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t). \end{align*} If $C e^{tA}x_0=0$ for every $t \in [0,T]$, then $g_{x_0}(t)=0$ for every $t \in [0,T]$, so the integral is zero. The converse is the important direction. Suppose \begin{align*} x_0^\top W_o(T)x_0=0. \end{align*} Then \begin{align*} \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t)=0. \end{align*} A nonnegative [continuous function](/page/Continuous%20Function) on an interval can have integral zero only if it is identically zero. To verify this directly, assume instead that $g_{x_0}(t_0)>0$ for some $t_0 \in [0,T]$. By continuity, there are a number $\varepsilon>0$ and a subinterval $I \subset [0,T]$ with positive [Lebesgue measure](/page/Lebesgue%20Measure) such that $g_{x_0}(t)\geq \varepsilon$ for all $t \in I$. Therefore \begin{align*} \int_0^{\!T}g_{x_0}(t)\, d\mathcal{L}^1(t) \geq \int_I \varepsilon\, d\mathcal{L}^1(t) = \varepsilon\,\mathcal{L}^1(I)>0, \end{align*} contradicting the zero integral. Thus $g_{x_0}(t)=0$ for all $t \in [0,T]$, which is exactly \begin{align*} C e^{tA}x_0=0 \end{align*} for every $t \in [0,T]$.[/guided]

[step:Extend zero output from $[0,T]$ to all nonnegative times] Let $x_0 \in \mathbb{R}^n$. Define \begin{align*} h_{x_0}: [0,\infty) \to \mathbb{R}^m,\quad h_{x_0}(t) = C e^{tA}x_0. \end{align*} Each component of $h_{x_0}$ is a real analytic function of $t$, because \begin{align*} e^{tA} = \sum_{k=0}^{\infty} \frac{t^k A^k}{k!} \end{align*} with convergence in every finite-dimensional matrix norm for all $t \in \mathbb{R}$. If $h_{x_0}(t)=0$ for every $t \in [0,T]$, then each component of $h_{x_0}$ vanishes on the interval $[0,T]$, which contains points accumulating in $\mathbb{R}$. By the identity theorem for real analytic functions, each component vanishes on all of $[0,\infty)$. Hence \begin{align*} C e^{tA}x_0=0 \end{align*} for every $t \geq 0$. [/step]

[step:Derive positive definiteness from observability] Assume that $(C,A)$ is observable. Let $x_0 \in \mathbb{R}^n$ be nonzero. If $x_0^\top W_o(T)x_0=0$, then the kernel identification above gives \begin{align*} C e^{tA}x_0=0 \end{align*} for every $t \in [0,T]$. By the [analytic continuation](/page/Analytic%20Continuation) step, this implies \begin{align*} C e^{tA}x_0=0 \end{align*} for every $t \geq 0$. Observability then forces $x_0=0$, contradicting the choice of $x_0$. Moreover, for each $t \in [0,T]$, the matrix $e^{tA^\top}C^\top C e^{tA}$ is symmetric because its transpose equals itself. Linearity of the entrywise Lebesgue integral gives that $W_o(T)$ is symmetric. Therefore \begin{align*} x_0^\top W_o(T)x_0>0 \end{align*} for every nonzero $x_0 \in \mathbb{R}^n$, so $W_o(T)$ is positive definite. [/step]

[step:Derive observability from positive definiteness] Assume that $W_o(T)$ is positive definite. Let $x_0 \in \mathbb{R}^n$ satisfy \begin{align*} C e^{tA}x_0=0 \end{align*} for every $t \geq 0$. In particular, the same equality holds for every $t \in [0,T]$. By the Gramian identity, \begin{align*} x_0^\top W_o(T)x_0 = \int_0^{\!T}|C e^{tA}x_0|^2\, d\mathcal{L}^1(t)=0. \end{align*} Since $W_o(T)$ is positive definite, the equality $x_0^\top W_o(T)x_0=0$ implies $x_0=0$. Thus no nonzero initial state produces identically zero output on $[0,\infty)$, and therefore $(C,A)$ is observable. [/step]

[step:Conclude the criterion for every time horizon] The argument was carried out for an arbitrary fixed $T>0$. Hence, for each $T>0$, the pair $(C,A)$ is observable if and only if $W_o(T)$ is positive definite. Equivalently, $(C,A)$ is observable if and only if $W_o(T)$ is positive definite for every $T>0$. [/step]