I know math people don't like the Dirac delta, so feel free to answer with your measure theory - I'll try my best to understand.
Suppose $x$ is a WSS stochastic process $\{x[n] : n \in \mathbb{Z}_{-\infty}^\infty \}$ with $x[n] \in \Re \; \forall \; n$.
It's a common result that the Fourier transform of the autocovariance function is a distribution of expected power over "frequency" and is called the power spectral density (PSD). But Papoulis also talks about the random variables that represent the Fourier transform of $x$ itself:
$$ X(f) = \sum_{n=-\infty}^{\infty}x[n]\exp(-2\pi fnj)$$
My first question: Is it true that $$\mathbb{E} \lim_{N\to\infty} \frac{1}{2N}|X(f)|^2 = S(f)$$ I suspect it is due to the fact that the expectation of the periodogram estimate approaches $S(f)$ as the number of samples it uses goes to infinity.
If that is true, it brings into play the Dirac delta squared and an odd relationship between limits and the Dirac delta... Papoulis says that if $x$ is a WSS process with PSD $S(f)$ then $$ \mathbb{E}X(u)X^*(v) = S(u)\delta(u-v)$$ So the expected energy at each frequency $u$ is a function $u \longmapsto S(u)\delta(0)$ so that it's basically the "power at $u$" times infinity, reflecting the nature that the energy is infinite (this makes sense) and that power was necessary idea to define a finite quantity.
Question 2: Does this mean that (if question 1 is true) $$ \lim_{N\to\infty}\frac{1}{N}\delta(0) = 1$$
Question 3: Suppose $x[n] = a$ for some nonrandom real $a$. Then the PSD is $\delta(f)a$ That implies the expected energy at $f$ is $a \delta(f)^2$. Is it possible to make sense of all of this?
I figured it all out. All the answers hinge on the definition $$\delta(f) := \sum_{n\in\mathbb{Z}}\exp(-2\pi fnj)$$
Then 1.) is true, because
$$\lim_{N\to\infty} \frac{1}{2N} \sum_{n=-N}^{N} \exp(-2 \pi fnj) = \begin{cases} 1 & f \in \mathbb{Z} \\ 0 & f \not\in \mathbb{Z} \end{cases} $$ Hence, when we use this result with our definition of $\mathbb{E}X(u)X(v)$, we find $$\mathbb{E} \lim_{N\to\infty} \frac{1}{2N}|X(f)|^2 = S(f)$$
Question 2 is also already answered (though as written, it equals $\frac{1}{2}$.) Question three is answered by interpreting $\delta$ as that diverging sum.