[proofplan] We define $f_X$ by the [Fourier series](/page/Fourier%20Series) of the autocovariance sequence. Absolute summability of $\gamma$ gives [uniform convergence](/page/Uniform%20Convergence) by the Weierstrass $M$-test, so the resulting function is continuous and $2\pi$-periodic. Termwise integration against the exponential $e^{ih\nu}$ recovers the autocovariance coefficients by orthogonality. Finally, the Wiener-Khinchin theorem supplies the spectral measure of the stationary process, and uniqueness of Fourier coefficients for finite measures identifies that measure with $f_X(\nu)\,d\mathcal{L}^1(\nu)$; positivity of the spectral measure then forces $f_X \geq 0$. [/proofplan]

[step:Define the candidate density by the autocovariance Fourier series]Define the partial sums $S_N: [-\pi,\pi] \to \mathbb{C}$, for $N \in \mathbb{N}$, by \begin{align*} S_N(\nu) = \frac{1}{2\pi}\sum_{h=-N}^{N}\gamma(h)e^{-ih\nu}. \end{align*} For each $h \in \mathbb{Z}$, the function $\varphi_h: [-\pi,\pi] \to \mathbb{C}$ defined by \begin{align*} \varphi_h(\nu) = \frac{1}{2\pi}\gamma(h)e^{-ih\nu} \end{align*} is continuous and satisfies \begin{align*} |\varphi_h(\nu)| \leq \frac{1}{2\pi}|\gamma(h)| \end{align*} for every $\nu \in [-\pi,\pi]$. Since $\sum_{h \in \mathbb{Z}}|\gamma(h)| < \infty$, the Weierstrass $M$-test implies that $\sum_{h \in \mathbb{Z}}\varphi_h$ converges uniformly on $[-\pi,\pi]$. Define $f_X: [-\pi,\pi] \to \mathbb{C}$ by \begin{align*} f_X(\nu) = \sum_{h \in \mathbb{Z}}\varphi_h(\nu) = \frac{1}{2\pi}\sum_{h \in \mathbb{Z}}\gamma(h)e^{-ih\nu}. \end{align*} As the uniform limit of continuous functions, $f_X$ is continuous. For every $h \in \mathbb{Z}$ and every $\nu \in \mathbb{R}$, \begin{align*} e^{-ih(\nu+2\pi)} = e^{-ih\nu}e^{-i2\pi h} = e^{-ih\nu}. \end{align*} Thus the same formula defines a $2\pi$-periodic [continuous function](/page/Continuous%20Function) on the circle.[/step]

[guided]The intended density is the Fourier series whose coefficients are the autocovariances. To make this a legitimate function, we first verify uniform convergence. For $N \in \mathbb{N}$, define the finite Fourier polynomial $S_N: [-\pi,\pi] \to \mathbb{C}$ by \begin{align*} S_N(\nu) = \frac{1}{2\pi}\sum_{h=-N}^{N}\gamma(h)e^{-ih\nu}. \end{align*} The $h$th summand is the function $\varphi_h: [-\pi,\pi] \to \mathbb{C}$ given by \begin{align*} \varphi_h(\nu) = \frac{1}{2\pi}\gamma(h)e^{-ih\nu}. \end{align*} This function is continuous because $\nu \mapsto e^{-ih\nu}$ is continuous. Also, for every $\nu \in [-\pi,\pi]$, \begin{align*} |\varphi_h(\nu)| = \frac{1}{2\pi}|\gamma(h)|\,|e^{-ih\nu}| = \frac{1}{2\pi}|\gamma(h)|. \end{align*} The majorizing numerical series \begin{align*} \sum_{h \in \mathbb{Z}}\frac{1}{2\pi}|\gamma(h)| \end{align*} converges by the assumed absolute summability of $\gamma$. Therefore the Weierstrass $M$-test gives uniform convergence of $\sum_{h \in \mathbb{Z}}\varphi_h$ on $[-\pi,\pi]$. This justifies defining $f_X: [-\pi,\pi] \to \mathbb{C}$ by \begin{align*} f_X(\nu) = \sum_{h \in \mathbb{Z}}\varphi_h(\nu) = \frac{1}{2\pi}\sum_{h \in \mathbb{Z}}\gamma(h)e^{-ih\nu}. \end{align*} Uniform convergence is the key point: it lets us pass continuity from the finite partial sums to the limit. Since each partial sum is continuous and the convergence is uniform, $f_X$ is continuous. Finally, for every $h \in \mathbb{Z}$ and every $\nu \in \mathbb{R}$, \begin{align*} e^{-ih(\nu+2\pi)} = e^{-ih\nu}e^{-i2\pi h} = e^{-ih\nu}, \end{align*} because $e^{-i2\pi h}=1$ for integer $h$. Therefore the same series defines a $2\pi$-periodic function.[/guided]

[step:Recover the autocovariances by termwise integration]Fix $k \in \mathbb{Z}$. Since $S_N \to f_X$ uniformly on $[-\pi,\pi]$ and the interval has finite [Lebesgue measure](/page/Lebesgue%20Measure), we may pass the limit through the integral: \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) = \lim_{N \to \infty}\int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu). \end{align*} For $N \geq |k|$, finite linearity of the [Lebesgue integral](/page/Lebesgue%20Integral) gives \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu) = \frac{1}{2\pi}\sum_{h=-N}^{N}\gamma(h)\int_{-\pi}^{\pi}e^{i(k-h)\nu}\,d\mathcal{L}^1(\nu). \end{align*} For each integer $\ell \in \mathbb{Z}$, the orthogonality relation for complex exponentials on $[-\pi,\pi]$ is as follows. If $\ell = 0$, then \begin{align*} \int_{-\pi}^{\pi} e^{i\ell\nu}\,d\mathcal{L}^1(\nu) = 2\pi. \end{align*} If $\ell \in \mathbb{Z}\setminus\{0\}$, then \begin{align*} \int_{-\pi}^{\pi} e^{i\ell\nu}\,d\mathcal{L}^1(\nu) = 0. \end{align*} Hence all terms in the preceding finite sum vanish except the term $h=k$, and therefore \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu) = \gamma(k) \end{align*} for every $N \geq |k|$. Passing to the limit yields \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) = \gamma(k). \end{align*} Since $k \in \mathbb{Z}$ was arbitrary, the inversion formula holds for every integer lag.[/step]

[guided]We fix an integer lag $k \in \mathbb{Z}$ and compute the $k$th Fourier coefficient of the candidate density. Since $S_N \to f_X$ uniformly on $[-\pi,\pi]$, the functions $e^{ik\nu}S_N(\nu)$ converge uniformly to $e^{ik\nu}f_X(\nu)$. The interval $[-\pi,\pi]$ has finite one-dimensional Lebesgue measure, so uniform convergence permits passage of the limit through the Lebesgue integral: \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) = \lim_{N \to \infty}\int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu). \end{align*} For $N \geq |k|$, the index $h=k$ appears in the finite sum defining $S_N$. By finite linearity of the Lebesgue integral, \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu) = \frac{1}{2\pi}\sum_{h=-N}^{N}\gamma(h)\int_{-\pi}^{\pi}e^{i(k-h)\nu}\,d\mathcal{L}^1(\nu). \end{align*} The role of orthogonality is to isolate the single coefficient whose index matches $k$. For each integer $\ell \in \mathbb{Z}$, if $\ell=0$, then \begin{align*} \int_{-\pi}^{\pi} e^{i\ell\nu}\,d\mathcal{L}^1(\nu) = 2\pi, \end{align*} and if $\ell \in \mathbb{Z}\setminus\{0\}$, then \begin{align*} \int_{-\pi}^{\pi} e^{i\ell\nu}\,d\mathcal{L}^1(\nu) = 0. \end{align*} Here we apply this with $\ell=k-h$. Therefore every term in the finite sum vanishes except the term $h=k$, and for every $N \geq |k|$ we obtain \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}S_N(\nu)\,d\mathcal{L}^1(\nu) = \frac{1}{2\pi}\gamma(k)2\pi = \gamma(k). \end{align*} Passing to the limit gives \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) = \gamma(k). \end{align*} Because $k \in \mathbb{Z}$ was arbitrary, the inversion formula holds for every integer lag.[/guided]

[step:Identify the spectral measure with the measure induced by the candidate density]Regard $[-\pi,\pi]$ as the standard representative interval for the torus $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$, with the endpoints identified. By the Wiener-Khinchin theorem for weakly stationary processes, there exists a finite positive Borel measure $\mu_X$ on $\mathbb{T}$ such that, for every $k \in \mathbb{Z}$, \begin{align*} \gamma(k) = \int_{-\pi}^{\pi} e^{ik\nu}\,d\mu_X(\nu). \end{align*} Define the finite complex Borel measure $\nu_X$ on $\mathbb{T}$ by \begin{align*} \nu_X(A) = \int_A f_X(\nu)\,d\mathcal{L}^1(\nu) \end{align*} for every Borel set $A \subset [-\pi,\pi]$, interpreted through this representative interval with endpoints identified. This measure is finite because $f_X$ is continuous on the compact interval $[-\pi,\pi]$, hence bounded. The previous step proves that the Fourier coefficients of $\nu_X$ agree with those of $\mu_X$: \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}\,d\nu_X(\nu) = \int_{-\pi}^{\pi} e^{ik\nu}\,d\mu_X(\nu) \end{align*} for every $k \in \mathbb{Z}$. By uniqueness of Fourier coefficients for finite Borel measures on the circle, we have $\nu_X = \mu_X$. Therefore $\mu_X$ is absolutely continuous with respect to $\mathcal{L}^1$ and has density $f_X$.[/step]

[guided]The Fourier coefficient computation shows that $f_X$ has the right coefficients, but the phrase “spectral density” refers to the spectral measure of the process. We now connect the constructed function to that measure. Regard $[-\pi,\pi]$ as the standard representative interval for the torus $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$, with the endpoints identified. The Wiener-Khinchin theorem for weakly stationary processes gives a finite positive Borel measure $\mu_X$ on $\mathbb{T}$ such that \begin{align*} \gamma(k) = \int_{-\pi}^{\pi} e^{ik\nu}\,d\mu_X(\nu) \end{align*} for every $k \in \mathbb{Z}$. The hypotheses needed here are precisely weak stationarity, which ensures that the autocovariance depends only on the lag $k$, and positive definiteness of the autocovariance sequence, which is part of the standard conclusion behind Wiener-Khinchin. Next define the measure produced by our candidate density. Let $\nu_X$ be the finite complex Borel measure on $\mathbb{T}$ given by \begin{align*} \nu_X(A) = \int_A f_X(\nu)\,d\mathcal{L}^1(\nu) \end{align*} for every Borel set $A \subset [-\pi,\pi]$, interpreted through the representative interval with endpoints identified. This is finite because $f_X$ is continuous on the compact interval $[-\pi,\pi]$, so there exists $M \geq 0$ such that $|f_X(\nu)| \leq M$ for every $\nu \in [-\pi,\pi]$, and hence \begin{align*} |\nu_X|([-\pi,\pi]) \leq \int_{-\pi}^{\pi}|f_X(\nu)|\,d\mathcal{L}^1(\nu) \leq 2\pi M < \infty. \end{align*} From the preceding step, the Fourier coefficients of $\nu_X$ are exactly the autocovariances: \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}\,d\nu_X(\nu) = \int_{-\pi}^{\pi}e^{ik\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) = \gamma(k). \end{align*} The Wiener-Khinchin measure has the same coefficients: \begin{align*} \int_{-\pi}^{\pi} e^{ik\nu}\,d\mu_X(\nu) = \gamma(k). \end{align*} Thus $\mu_X$ and $\nu_X$ have identical Fourier coefficients for every $k \in \mathbb{Z}$. By uniqueness of Fourier coefficients for finite Borel measures on the circle, the two measures are equal. Therefore the spectral measure is \begin{align*} d\mu_X(\nu) = f_X(\nu)\,d\mathcal{L}^1(\nu), \end{align*} which is exactly the assertion that $f_X$ is a spectral density.[/guided]

[step:Deduce real-valuedness and non-negativity from positivity of the spectral measure]It remains to verify that the continuous density is real-valued and non-negative. Define the continuous functions $u: [-\pi,\pi] \to \mathbb{R}$ and $v: [-\pi,\pi] \to \mathbb{R}$ by \begin{align*} u(\nu) = \operatorname{Re} f_X(\nu) \end{align*} and \begin{align*} v(\nu) = \operatorname{Im} f_X(\nu). \end{align*} Since $\mu_X$ is a positive measure and $\mu_X = \nu_X$, every Borel set $A \subset [-\pi,\pi]$ satisfies \begin{align*} \int_A f_X(\nu)\,d\mathcal{L}^1(\nu) = \mu_X(A) \in [0,\infty). \end{align*} Taking imaginary parts gives \begin{align*} \int_A v(\nu)\,d\mathcal{L}^1(\nu) = 0 \end{align*} for every Borel set $A \subset [-\pi,\pi]$. If $v(\nu_0) \neq 0$ for some $\nu_0 \in [-\pi,\pi]$, continuity of $v$ gives a relatively open interval $I \subset [-\pi,\pi]$ containing $\nu_0$ and a number $\varepsilon > 0$ such that either $v(\nu) \geq \varepsilon$ for every $\nu \in I$ or $v(\nu) \leq -\varepsilon$ for every $\nu \in I$. This would force \begin{align*} \left|\int_I v(\nu)\,d\mathcal{L}^1(\nu)\right| \geq \varepsilon\,\mathcal{L}^1(I) > 0, \end{align*} contradicting the preceding identity. Hence $v=0$ on $[-\pi,\pi]$, so $f_X$ is real-valued. Taking real parts gives \begin{align*} \int_A u(\nu)\,d\mathcal{L}^1(\nu) = \mu_X(A) \geq 0 \end{align*} for every Borel set $A \subset [-\pi,\pi]$. If $u(\nu_0)<0$ for some $\nu_0 \in [-\pi,\pi]$, continuity of $u$ gives a relatively open interval $I \subset [-\pi,\pi]$ containing $\nu_0$ and a number $\varepsilon > 0$ such that $u(\nu) \leq -\varepsilon$ for every $\nu \in I$. Then \begin{align*} \int_I u(\nu)\,d\mathcal{L}^1(\nu) \leq -\varepsilon\,\mathcal{L}^1(I) < 0, \end{align*} contradicting positivity of $\mu_X(I)$. Hence $f_X(\nu)=u(\nu) \geq 0$ for every $\nu \in [-\pi,\pi]$. Combining the construction of $f_X$, the coefficient recovery formula, and the identification of $\mu_X$ with $f_X(\nu)\,d\mathcal{L}^1(\nu)$ proves that $(X_t)_{t \in \mathbb{Z}}$ has the asserted continuous spectral density and that \begin{align*} \gamma(h) = \int_{-\pi}^{\pi} e^{ih\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) \end{align*} for every $h \in \mathbb{Z}$.[/step]

[guided]At this point we know that the complex measure induced by $f_X$ equals the spectral measure $\mu_X$. Since $\mu_X$ is positive, this equality must force the density to be real and non-negative. We prove both facts from the equality of measures. Define the real and imaginary parts of the density as continuous maps $u: [-\pi,\pi] \to \mathbb{R}$ and $v: [-\pi,\pi] \to \mathbb{R}$ by \begin{align*} u(\nu) = \operatorname{Re} f_X(\nu) \end{align*} and \begin{align*} v(\nu) = \operatorname{Im} f_X(\nu). \end{align*} For every Borel set $A \subset [-\pi,\pi]$, equality $\nu_X=\mu_X$ gives \begin{align*} \int_A f_X(\nu)\,d\mathcal{L}^1(\nu) = \mu_X(A). \end{align*} The right-hand side is a real non-negative number because $\mu_X$ is a positive measure. Taking imaginary parts therefore gives \begin{align*} \int_A v(\nu)\,d\mathcal{L}^1(\nu) = 0 \end{align*} for every Borel set $A \subset [-\pi,\pi]$. Why does this force $v$ to vanish pointwise, rather than only almost everywhere? The extra input is continuity. If $v(\nu_0) \neq 0$ at some point $\nu_0$, then by continuity there is a relatively open interval $I \subset [-\pi,\pi]$ containing $\nu_0$ and a number $\varepsilon>0$ such that either $v(\nu)\geq \varepsilon$ on $I$ or $v(\nu)\leq -\varepsilon$ on $I$. In either case, \begin{align*} \left|\int_I v(\nu)\,d\mathcal{L}^1(\nu)\right| \geq \varepsilon\,\mathcal{L}^1(I) > 0. \end{align*} This contradicts the identity $\int_I v\,d\mathcal{L}^1=0$. Thus $v=0$ everywhere, and $f_X$ is real-valued. Now take real parts in the same measure identity. For every Borel set $A \subset [-\pi,\pi]$, \begin{align*} \int_A u(\nu)\,d\mathcal{L}^1(\nu) = \mu_X(A) \geq 0. \end{align*} If $u(\nu_0)<0$ at some point, continuity gives a relatively open interval $I \subset [-\pi,\pi]$ containing $\nu_0$ and a number $\varepsilon>0$ such that $u(\nu)\leq -\varepsilon$ on $I$. Hence \begin{align*} \int_I u(\nu)\,d\mathcal{L}^1(\nu) \leq -\varepsilon\,\mathcal{L}^1(I)<0, \end{align*} which contradicts $\int_I u\,d\mathcal{L}^1=\mu_X(I)\geq 0$. Therefore $u\geq 0$ everywhere. Since $f_X=u+iv$ and $v=0$, we conclude that $f_X(\nu)\geq 0$ for every $\nu\in[-\pi,\pi]$. Together with the construction and coefficient recovery already proved, this shows that $(X_t)_{t \in \mathbb{Z}}$ has the asserted continuous spectral density and that \begin{align*} \gamma(h) = \int_{-\pi}^{\pi} e^{ih\nu}f_X(\nu)\,d\mathcal{L}^1(\nu) \end{align*} for every $h \in \mathbb{Z}$.[/guided]