[proofplan]
We first prove that a representing measure produces a positive definite sequence by expanding the defining quadratic form under the integral. Conversely, a positive definite sequence defines a positive linear functional on trigonometric polynomials by prescribing its values on Fourier monomials. The Fejer-Riesz factorisation theorem gives positivity on non-negative trigonometric polynomials, hence a supremum-norm bound; after extending the functional to $C(\mathbb{T})$, the Riesz-Markov representation theorem produces the desired measure. Uniqueness follows because trigonometric polynomials are uniformly dense in $C(\mathbb{T})$, so a finite measure is determined by all of its Fourier coefficients.
[/proofplan]
[step:Show that Fourier coefficients of a positive measure are positive definite]
Let $\mathbb{T}:=\{z \in \mathbb{C}: |z|=1\}$ denote the unit circle, and let
\begin{align*}
\pi_{\mathbb{T}}: [-\pi,\pi] &\to \mathbb{T} \\
\lambda &\mapsto e^{i\lambda}
\end{align*}
be the quotient parametrisation identifying $-\pi$ and $\pi$. Suppose $\mu$ is a finite positive Borel measure on $\mathbb{T}$ and define $\gamma_\mu: \mathbb{Z}\to \mathbb{C}$ by
\begin{align*}
\gamma_\mu(h):=\int_{\mathbb{T}} z^h\,d\mu(z).
\end{align*}
For $h<0$, the expression $z^h$ means $\overline{z}^{\,|h|}$, since $|z|=1$.
Fix $m \in \mathbb{N}$, $h_1,\dots,h_m \in \mathbb{Z}$, and $c_1,\dots,c_m \in \mathbb{C}$. The function
\begin{align*}
P: \mathbb{T} &\to \mathbb{C} \\
z &\mapsto \sum_{j=1}^{m} c_j z^{h_j}
\end{align*}
is continuous and bounded. Since $\mu$ is finite, $|P|^2$ is $\mu$-integrable. Expanding the square and using linearity of the integral gives
\begin{align*}
\sum_{j=1}^{m}\sum_{k=1}^{m} c_j\overline{c_k}\,\gamma_\mu(h_j-h_k)
&=
\sum_{j=1}^{m}\sum_{k=1}^{m} c_j\overline{c_k}\int_{\mathbb{T}} z^{h_j-h_k}\,d\mu(z) \\
&=
\int_{\mathbb{T}} \sum_{j=1}^{m}\sum_{k=1}^{m} c_j\overline{c_k}z^{h_j}\overline{z^{h_k}}\,d\mu(z) \\
&=
\int_{\mathbb{T}} |P(z)|^2\,d\mu(z) \geq 0.
\end{align*}
Thus $\gamma_\mu$ is positive definite.
[/step]
[step:Define a positive functional on trigonometric polynomials]
Assume now that $\gamma: \mathbb{Z}\to\mathbb{C}$ is positive definite. Let $\mathcal{P}$ denote the complex vector space of trigonometric polynomials on $\mathbb{T}$, meaning functions
\begin{align*}
p: \mathbb{T} &\to \mathbb{C} \\
z &\mapsto \sum_{h\in F} a_h z^h,
\end{align*}
where $F\subset \mathbb{Z}$ is finite and $a_h\in\mathbb{C}$. The Laurent coefficients of such a polynomial are unique, so the following linear functional is well-defined:
\begin{align*}
L: \mathcal{P} &\to \mathbb{C} \\
\sum_{h\in F} a_h z^h &\mapsto \sum_{h\in F} a_h\gamma(h).
\end{align*}
For
\begin{align*}
p: \mathbb{T} &\to \mathbb{C} \\
z &\mapsto \sum_{j=1}^{m} a_j z^{h_j},
\end{align*}
we have
\begin{align*}
|p(z)|^2
=
\sum_{j=1}^{m}\sum_{k=1}^{m} a_j\overline{a_k}z^{h_j-h_k}.
\end{align*}
Therefore, by positive definiteness of $\gamma$,
\begin{align*}
L(|p|^2)
=
\sum_{j=1}^{m}\sum_{k=1}^{m} a_j\overline{a_k}\gamma(h_j-h_k)
\geq 0.
\end{align*}
By the Fejer-Riesz factorisation theorem, every non-negative trigonometric polynomial $r\in\mathcal{P}$ admits a factorisation $r=|q|^2$ for some $q\in\mathcal{P}$; this is a standard result not yet linked in the wiki. Hence
\begin{align*}
r(z)\geq 0 \ \text{for all } z\in\mathbb{T}
\quad\Longrightarrow\quad
L(r)\geq 0.
\end{align*}
[/step]
[step:Bound the functional by the supremum norm]
Define the supremum norm on $\mathcal{P}$ by
\begin{align*}
\|p\|_{C^0(\mathbb{T})}:=\sup_{z\in\mathbb{T}} |p(z)|.
\end{align*}
Since $\gamma$ is positive definite, applying the definition with $m=1$, $h_1=0$, and $c_1=1$ gives $\gamma(0)\geq 0$.
For $p\in\mathcal{P}$, the trigonometric polynomial
\begin{align*}
r_p: \mathbb{T} &\to \mathbb{R} \\
z &\mapsto \|p\|_{C^0(\mathbb{T})}^2-|p(z)|^2
\end{align*}
is non-negative on $\mathbb{T}$. By positivity of $L$ on non-negative trigonometric polynomials,
\begin{align*}
0\leq L(r_p)
=
\|p\|_{C^0(\mathbb{T})}^2L(1)-L(|p|^2)
=
\gamma(0)\|p\|_{C^0(\mathbb{T})}^2-L(|p|^2).
\end{align*}
Thus
\begin{align*}
L(|p|^2)\leq \gamma(0)\|p\|_{C^0(\mathbb{T})}^2.
\end{align*}
The map
\begin{align*}
B: \mathcal{P}\times\mathcal{P} &\to \mathbb{C} \\
(p,q) &\mapsto L(p\overline{q})
\end{align*}
is a positive semidefinite sesquilinear form, because $B(p,p)=L(|p|^2)\geq 0$. The Cauchy-Schwarz inequality for positive semidefinite sesquilinear forms gives
\begin{align*}
|L(p)|^2
=
|B(p,1)|^2
\leq B(p,p)B(1,1)
=
L(|p|^2)\gamma(0).
\end{align*}
Combining this with the previous estimate yields
\begin{align*}
|L(p)|^2\leq \gamma(0)^2\|p\|_{C^0(\mathbb{T})}^2,
\end{align*}
and therefore
\begin{align*}
|L(p)|\leq \gamma(0)\|p\|_{C^0(\mathbb{T})}.
\end{align*}
So $L$ is bounded with respect to the supremum norm.
[/step]
[step:Extend the functional to continuous functions and represent it by a measure]
By the complex Stone-Weierstrass theorem, the algebra $\mathcal{P}$ of trigonometric polynomials is uniformly dense in $C(\mathbb{T})$; this is a standard result not yet linked in the wiki. Since $L$ is bounded for the supremum norm, it extends uniquely to a bounded linear functional
\begin{align*}
\widetilde{L}: C(\mathbb{T}) &\to \mathbb{C}
\end{align*}
satisfying $\widetilde{L}(p)=L(p)$ for every $p\in\mathcal{P}$.
We verify that $\widetilde{L}$ is positive. Let $f\in C(\mathbb{T})$ satisfy $f(z)\geq 0$ for all $z\in\mathbb{T}$. Define
\begin{align*}
g: \mathbb{T} &\to [0,\infty) \\
z &\mapsto \sqrt{f(z)}.
\end{align*}
The function $g$ is continuous. By uniform density, choose a sequence $(p_n)_{n\in\mathbb{N}}$ in $\mathcal{P}$ such that
\begin{align*}
\|p_n-g\|_{C^0(\mathbb{T})}\to 0.
\end{align*}
Then $|p_n|^2\to g^2=f$ uniformly on $\mathbb{T}$. Since $\widetilde{L}$ is continuous and $L(|p_n|^2)\geq 0$ for every $n\in\mathbb{N}$,
\begin{align*}
\widetilde{L}(f)
=
\lim_{n\to\infty} \widetilde{L}(|p_n|^2)
=
\lim_{n\to\infty} L(|p_n|^2)
\geq 0.
\end{align*}
By the Riesz-Markov representation theorem for positive linear functionals on $C(\mathbb{T})$, a standard result not yet linked in the wiki, there exists a unique finite positive Borel measure $\nu$ on $\mathbb{T}$ such that
\begin{align*}
\widetilde{L}(f)=\int_{\mathbb{T}} f(z)\,d\nu(z)
\end{align*}
for every $f\in C(\mathbb{T})$. Taking $f(z)=z^h$ gives
\begin{align*}
\gamma(h)=L(z^h)=\widetilde{L}(z^h)=\int_{\mathbb{T}} z^h\,d\nu(z)
\end{align*}
for every $h\in\mathbb{Z}$.
[/step]
[step:Transfer the measure to $[-\pi,\pi]$ with identified endpoints]
Let
\begin{align*}
\pi_{\mathbb{T}}: [-\pi,\pi] &\to \mathbb{T} \\
\lambda &\mapsto e^{i\lambda}
\end{align*}
be the quotient map. Interpreting $[-\pi,\pi]/\{-\pi\sim\pi\}$ as $\mathbb{T}$ through $\pi_{\mathbb{T}}$, the measure $\nu$ is equivalently a finite positive Borel measure $\mu$ on the quotient circle. Since $z^h=e^{ih\lambda}$ when $z=e^{i\lambda}$, the representation becomes
\begin{align*}
\gamma(h)=\int_{[-\pi,\pi]/\{-\pi\sim\pi\}} e^{ih\lambda}\,d\mu(\lambda),
\end{align*}
for every $h\in\mathbb{Z}$.
[/step]
[step:Prove uniqueness from equality of Fourier coefficients]
Suppose $\mu_1$ and $\mu_2$ are finite positive Borel measures on $\mathbb{T}$ such that
\begin{align*}
\int_{\mathbb{T}} z^h\,d\mu_1(z)
=
\int_{\mathbb{T}} z^h\,d\mu_2(z)
\end{align*}
for every $h\in\mathbb{Z}$. By linearity, the two measures have the same integral against every trigonometric polynomial $p\in\mathcal{P}$:
\begin{align*}
\int_{\mathbb{T}} p(z)\,d\mu_1(z)
=
\int_{\mathbb{T}} p(z)\,d\mu_2(z).
\end{align*}
Since $\mathcal{P}$ is uniformly dense in $C(\mathbb{T})$ and both measures are finite, passing to uniform limits gives
\begin{align*}
\int_{\mathbb{T}} f(z)\,d\mu_1(z)
=
\int_{\mathbb{T}} f(z)\,d\mu_2(z)
\end{align*}
for every $f\in C(\mathbb{T})$. Finite Borel measures on a compact Hausdorff space are determined by their integrals against continuous functions, by the uniqueness part of the Riesz-Markov representation theorem. Therefore $\mu_1=\mu_2$ on $\mathbb{T}$, equivalently on $[-\pi,\pi]$ after identifying $-\pi$ and $\pi$.
This proves both existence and uniqueness of the representing measure, and completes the proof of the theorem.
[/step]
