I'm trying to get a good handle on the spectral theory of random signals, but there seem to be two different definitions of the power spectrum whose relation is not clear to me. On the one hand, we know that the autocovariance function of a stationary random signal has a spectral decomposition
$$ \begin{align*} C_{XX}(\tau) &= \int_{-\infty}^{+\infty} e^{i2\pi \xi \tau} \, F(d\xi) \\ &= \int_{-\infty}^{+\infty} e^{i2\pi \xi \tau} f(\xi) \, d\xi + \sum_{j=-\infty}^{+\infty} e^{i2\pi \xi_j \tau} \, p(\xi_j) \,, \end{align*} $$
where $F$ is the spectral distribution which is then separated into continuous and discrete components using Lebesgue decomposition, yielding the power spectral density $f(\xi)$. This is the definition found in e.g. Koopmans, Brockwell & Davis, etc. On the other hand, Wikipedia and many textbooks give the definition
$$ f(\xi) = \lim_{T \to \infty} \vert \hat{X}_T(\xi) \vert^2 \,, $$
where $X_T$ denotes the truncation of $X(t)$ to the interval $[-T/2, T/2]$. I'm having real trouble seeing how these two formulations coincide. Complicating matters even further, I've also encounter the following variational on the latter definition in the Wikipedia article on Brownian noise, where the PSD is taken to be given by the square of the Fourier transform (I use their notation for consistency):
$$ S(\omega) = \vert \mathcal{F}\left\lbrace W(t)) \right\rbrace(\omega) \vert^2 \,. $$
I don't know what the Fourier transform of Brownian motion would even mean. How to tie this all together and iron out the kinks?
