So, this answer is going to need installments. There needs to be a few chapters to lead up to it. Before getting to the definition of white noise, we gotta begin with stochastic processes (a fancy name for "random processes"), specifically processes that are strictly stationary, more specifically Markov processes.
I am going to make lotsa assumptions at the outset to make my life easier:
- This is about finite power signals and not about finite energy signals: $$ 0 \ < \ \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} \big|x(t)\big|^2 \ \mathrm{d}t \ < \ \infty $$
- So then the appropriate definition of cross-correlation is: $$ R_{xy}(\tau) \triangleq \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} x(t+\tau)y^*(t) \ \mathrm{d}t $$
- Autocorrelation of $x(t)$ is simply $R_{xx}(\tau)$ and $R_{xx}(0)$ is the mean power of $x(t)$.
- $x(t)$ is ergodic in every sense which means that any expressible average or mean w.r.t. time (like those expressed above) can be expressed as a probabilistic average or mean. This is the Expected value (sometimes called "Expectation operator"): $$ \operatorname{E}\Big\{ g(x) \Big\} \triangleq \int\limits_{-\infty}^{\infty} g(\alpha) p_x(\alpha) \ \mathrm{d}\alpha $$ where $p_x(\alpha)$ is the probability density function (p.d.f.) of the random variable $x$ and $g(\cdot)$ is some well-defined function. A random process $x(t)$ sampled at any given time $t$ is a random variable.
So (with another assumption that $x(t)$ and $t$ are both real) the implication that ergodicity has on the autocorrelation is:
$$\begin{align}
R_{xx}(\tau) &= \lim_{T \to \infty} \frac{1}{T}\int\limits_{-\frac{T}2}^{\frac{T}2} x(t+\tau)x(t) \ \mathrm{d}t \qquad \qquad & \text{(average w.r.t. time)}\\
\\
&= \operatorname{E}\Big\{ x(t+\tau)x(t) \Big\} \qquad \qquad & \text{(probabilistic average)} \\
\\
&= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)x(t)}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta\\
\end{align}$$
where $p_{x(t+\tau)x(t)}(\alpha,\beta)$ is the joint p.d.f. for the two random variables $x(t+\tau)$ and $x(t)$.
$ R_{xx}(0) = \operatorname{E}\Big\{ \big|x(t)\big|^2 \Big\}$ is the mean-square of $x(t)$ and, if the mean $\operatorname{E}\Big\{ x(t) \Big\}=0$, the mean-square and variance are equal.
Now let's assume, additionally, that this random process is Gaussian and zero mean:
$$ p_{x(t)}(\alpha) = \frac{1}{\sqrt{2 \pi} \sigma_x} e^{-\frac12 \left(\frac{\alpha}{\sigma_x}\right)^2} $$
Being zero-mean, then the mean-square, $\operatorname{E}\Big\{ \big|x(t)\big|^2 \Big\}$, and variance, $\sigma_x^2$, are the same and larger than zero.
Additionally, we'll assume a Markov process. So the value of the previous states may affect the current state. The value of the state at $x(t)$ may have an effect on the probability of the state at a later time, $x(t+\tau)$.
As a judiciously-arranged example, consider this class of Markov process. Given the known earlier value $x(t)$, the dependent p.d.f. for the later random value $x(t+\tau)$ is:
$$\begin{align}
p_{x(t+\tau)}\big(\alpha|x(t)\big) &= \frac{1}{\sqrt{2 \pi \left(\sigma_x^2 - R_{xx}(\tau)\right)}} e^{-\frac12 \frac{\left(\alpha-x(t) \sigma_x^{-2} R_{xx}(\tau)\right)^2}{\sigma_x^2 - R_{xx}(\tau)}} \\
\\
&= \frac{1}{\sqrt{2 \pi} \sigma(\tau)} e^{-\frac12 \left(\frac{\alpha-\mu(\tau)}{\sigma(\tau)}\right)^2} \\
\end{align}$$
where the dependent (on $x(t)$ and $\tau$) mean and variance are:
$$\begin{align}
\mu(\tau) &= x(t) \sigma_x^{-2} R_{xx}(\tau) \\
\\
\sigma(\tau) &= \sqrt{\sigma_x^2 - R_{xx}(\tau)} \\
\end{align}$$
I think that the joint p.d.f. is related to the conditional p.d.f. as:
$$\begin{align}
p_{x(t+\tau)x(t)}(\alpha,\beta) &= p_{x(t+\tau)}\big(\alpha|\beta \big) \cdot p_{x(t)}(\beta) \\
\\
&= \frac{1}{\sqrt{2 \pi \left(\sigma_x^2 - R_{xx}(\tau)\right)}} e^{-\frac12 \frac{\left(\alpha-\beta\sigma_x^{-2}R_{xx}(\tau)\right)^2}{\sigma_x^2 - R_{xx}(\tau)}} \ \cdot \ \frac{1}{\sqrt{2 \pi} \sigma_x} e^{-\frac12 \left(\frac{\beta}{\sigma_x}\right)^2} \\
\end{align}$$
Then the autocorrelation of this Markov random process is:
$$\begin{align}
R_{xx}(\tau) &= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)x(t)}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta \\
\\
&= \int\limits_{-\infty}^{\infty}\int\limits_{-\infty}^{\infty} \alpha\beta \ p_{x(t+\tau)}\big(\alpha|\beta \big) p_{x(t)}(\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta \\
\\
&= \int\limits_{-\infty}^{\infty} \beta \, p_{x(t)}(\beta) \ \left[ \int\limits_{-\infty}^{\infty} \alpha \ p_{x(t+\tau)}\big(\alpha|\beta \big) \ \mathrm{d}\alpha \right] \ \mathrm{d}\beta \\
\\
&= \int\limits_{-\infty}^{\infty} \beta \, p_{x(t)}(\beta) \ \Big[ \beta \sigma_x^{-2} R_{xx}(\tau) \Big] \ \mathrm{d}\beta \\
\\
&= \left[ \int\limits_{-\infty}^{\infty} \beta^2 \, p_{x(t)}(\beta) \ \mathrm{d}\beta \right] \ \sigma_x^{-2} R_{xx}(\tau) \\
\\
&= \Big[ \sigma_x^2 \Big] \, \sigma_x^{-2} R_{xx}(\tau) \\
\\
&= R_{xx}(\tau) \qquad \qquad \checkmark \\
\end{align}$$
This confirms the choice of dependent mean $\mu(\tau) = x(t) \sigma_x^{-2} R_{xx}(\tau)$.
The choice of dependent variance $\sigma^2(\tau)$ is, perhaps, less critical but we want $\sigma(0)=0$, and when $R_{xx}(\tau)=0$ we want $p_{x(t+\tau)}\big(\alpha|x(t)\big)$ to be independent of $x(t)$ which would mean that $\mu(\tau)=0$ and $\sigma^2(\tau)=\sigma_x^2$. Perhaps, the dependent variance should be $\sigma^2(\tau)=\sigma_x^2-|R_{xx}(\tau)|$. I dunno.
If $R_{xx}(\tau) = \sigma_x^2 \, e^{-|4 \nu \tau|}$ and $\nu > 0$, then this might be the Ornstein-Uhlenbeck process (essentially white noise filtered through a first-order RC low-pass filter) and it might approach Brownian motion (essentially white noise passed through an integrator) as $\nu \to 0$. I would say it approaches white noise as $\nu \to \infty$.
If, instead, $R_{xx}(\tau) = \sigma_x^2 \operatorname{sinc}(2 \nu \tau)$, it might be what I call "bandlimited white noise" (and the bandlimits are $\pm \nu$). Again, send $\nu \to \infty$ and you have a definition for white noise.
So the power spectrum is
$$ S_{xx}(f) \triangleq \mathscr{F}\Big\{ R_{xx}(\tau) \Big\} $$
For the Ornstein-Uhlenbeck process, then
$$R_{xx}(\tau) = \sigma_x^2 \, e^{-|4 \nu \tau|}$$
and
$$ S_{xx}(f) = \frac{\sigma_x^2}{2\nu} \ \frac{1}{1 + \left( \frac{\pi f}{2 \nu} \right)^2} $$
For the bandlimited white noise process, then
$$R_{xx}(\tau) = \sigma_x^2 \operatorname{sinc}(2 \nu \tau)$$
and
$$ S_{xx}(f) = \frac{\sigma_x^2}{2\nu} \ \Pi \left( \frac{f}{2\nu} \right) $$
Both of these have $S_{xx}(0) = \frac{\sigma_x^2}{2\nu}$ and an effective noise bandwidth (one-sided) of $\nu$ (two-sided bandwidth is $2\nu$). If the variance $\sigma_x^2$ is fixed to a non-zero value, the value of the power spectrum $S_{xx}(f)$ goes to zero (for all $f$) in the limit as $\nu \to \infty$.
If $S_{xx}(0)$ were to remain constant, say $S_{xx}(0) = \frac{\eta}{2}$, then the variance is $\sigma_x^2 = \eta \nu$ and would increase without bound as the effective noise bandwidth, $\nu$, goes to infinity. Note that, in both processes, as the noise bandwidth $\nu$ increases without bound, the autocorrelation becomes a dirac impulse in the limit.
Ornstein-Uhlenbeck to white noise:
$$\begin{align}
R_{xx}(\tau) &= \lim_{\nu \to \infty} \eta \nu \, e^{-|4 \nu \tau|} \\
\\
&= \frac{\eta}{2} \delta(\tau) \\
\\
\end{align}$$
bandlimited white noise to white noise:
$$\begin{align}
R_{xx}(\tau) &= \lim_{\nu \to \infty} \eta \nu \, \operatorname{sinc}(2 \nu \tau) \\
\\
&= \frac{\eta}{2} \delta(\tau) \\
\\
\end{align}$$
And the power spectra are both
$$ S_{xx}(f) = \frac{\eta}{2} \qquad \qquad \forall f \in \mathbb{R} $$
if $S_{xx}(0) = \frac{\eta}{2}$ is held to a constant and bandlimit $\nu$ goes to $\infty$. This is what decorrelates $x(t_1)$ from $x(t_2)$ if $t_1 \ne t_2$, but again, it requires infinite variance, $\sigma_x^2 \to \infty$.
Appendix:
Some definitions so that we can make sure we're all on the same page.
Continuous Fourier transform and inverse:
$$\begin{align}
\mathscr{F}\Big\{ x(t) \Big\} \triangleq X(f) &= \int\limits_{-\infty}^{\infty} x(t) \, e^{-j 2 \pi f t} \, \mathrm{d}t \\
\\
\mathscr{F}^{-1}\Big\{ X(f) \Big\} \triangleq x(t) &= \int\limits_{-\infty}^{\infty} X(f) \, e^{+j 2 \pi f t} \, \mathrm{d}f \\
\end{align}$$
Rectangular function (sometimes "$\operatorname{rect}(u)$"):
$$\Pi(u) \triangleq \begin{cases}
1 \qquad & \text{ if } |u| < \tfrac12 \\
\tfrac12 \qquad & \text{ if } |u| = \tfrac12 \\
0 \qquad & \text{ if } |u| > \tfrac12 \\
\end{cases}$$
The inverse Fourier transform of the rectangular function is:
$$ \mathscr{F}^{-1} \left\{ \Pi\left( \tfrac{f}{2\nu} \right) \right\} = 2\nu \, \operatorname{sinc}(2\nu t) $$
The sinc function:
$$\operatorname{sinc}(u) \triangleq \begin{cases}
\frac{\sin(\pi u)}{\pi u} \qquad & \text{ if } u \ne 0 \\
1 \qquad & \text{ if } u = 0 \\
\end{cases}$$
More definitions and notation:
The probability density function (p.d.f.) of a random variable $x$ is denoted as
$$ p_x(\alpha) \triangleq \lim_{\Delta x \to 0} \frac{1}{\Delta x} \operatorname{P} \Big\{ \alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \Big\} $$
where $\operatorname{P} \Big\{ \alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \Big\}$ is the probability that $x$ exists between $\alpha-\tfrac12 \Delta x$ and $\alpha+\tfrac12 \Delta x$.
The Expected value of some function of single random variable $x$ is
$$ \operatorname{E}\Big\{ g(x) \Big\} \triangleq \int\limits_{-\infty}^{\infty} g(\alpha) \, p_x(\alpha) \ \mathrm{d}\alpha $$
And for a function of two random variables, $x$ and $y$, the Expected value is
$$ \operatorname{E}\Big\{ g(x,y) \Big\} \triangleq \int\limits_{-\infty}^{\infty} \int\limits_{-\infty}^{\infty} g(\alpha,\beta) \, p_{xy}(\alpha,\beta) \ \mathrm{d}\alpha \, \mathrm{d}\beta $$
where $p_{xy}(\alpha,\beta)$ is the joint p.d.f. for random variables $x$ and $y$ and is defined as
$$ p_{xy}(\alpha, \beta) \triangleq \lim_{\Delta x \to 0} \lim_{\Delta y \to 0} \frac{1}{\Delta x} \frac{1}{\Delta y} \ \operatorname{P} \left\{ \begin{matrix}
\alpha-\tfrac12 \Delta x < x < \alpha+\tfrac12\Delta x \\
\text{and} \\
\beta-\tfrac12 \Delta y < y < \beta+\tfrac12\Delta y \\
\end{matrix} \right\} $$