The Natural Probability on M(Ω)

Two weeks ago, Dr. Winston Ewert announced at Uncommon Descent a kind of open mike. He put up a page at Google Moderator and asked for questions. Unfortunately, not many took advantage of this offer, but I added three questions from the top of my head. The experience made me revisit the paper A General Theory of Information Cost Incurred by Successful Search again, and when I tried - as usual - to construct simple examples, I run into further questions - so, here is another one:

In their paper, the authors W. Dembski, W. Ewert, and R. Marks (DEM) talk about something they call the natural probability:

Processes that exhibit stochastic behavior arise from what may be called a natural probability. The natural probability characterizes the ordinary stochastic behavior of the process in question. Often the natural probability is the uniform probability. Thus, for a perfect cube with distinguishable sides composed of a rigid homogenous material (i.e., an ordinary die), the probability of any one of its six sides landing on a given toss is 1/6. Yet, for a loaded die, those probabilities will be skewed, with one side consuming the lion’s share of probability. For the loaded die, the natural probability is not uniform.

This natural probability on the search space translates through their idea of lifting to the space of measures $\mathbf{M}(\Omega)$:

As the natural probability on $\Omega$, $\mu$ is not confined simply to $\Omega$ lifts to $\mathbf{M}(\Omega)$, so that its lifting, namely $\overline{\mu}$, becomes the natural probability on $\mathbf{M}(\Omega)$ (this parallels how the uniform probability $\mathbf{U}$, when it is the natural probability on $\Omega$, lifts to the uniform probability $\overline{\mathbf{U}}$ on $\mathbf{M}(\Omega)$, which then becomes the natural probability for this higher-order search space).

As usual, I look at an easy example: a loaded coin which always shows head. So $\Omega=\{H,T\}$ and $\mu=\delta_H$ is the natural measure on $\Omega$. What happens on $\mathbf{M}(\Omega)= \{h\cdot\delta_H + t\cdot\delta_T|0 \le h,t \le 1; h+t=1 \}$? Luckily, $$(\mathbf{M}(\{H,T\}),\mathbf{U}) \cong ([0,1],\lambda).$$ Let's jump the hoops:

1. The Radon-Nikodym derivative of $\delta_H$ with respect to $\mathbf{U}$ is $f(H) = \frac{d\delta_H}{d\mathbf{U}}(H) = 2$, $f(T) = \frac{d\delta_H}{d\mathbf{U}}(T) = 0$
2. Let $\theta \in \mathbf{M}(\{H,T\})$, i.e., $\theta= h\delta_H + t\delta_T$. Then$$\overline{f}{(\theta)} = \int_{\Omega} f(x)d\theta(x)$$ $$=f(H)\cdot\theta(\{H\}) + f(T) \cdot\theta(\{T\})$$ $$=2 \cdot h$$

Here, I have the density of my natural measure on $\mathbf{M}(\Omega)$ with regard to $\overline{\mathbf{U}}$, $$d\overline{\delta_H}(h\cdot\delta_H + t\cdot\delta_T) = 2 \cdot h \cdot d\overline{\mathbf{U}}(h\cdot\delta_H + t\cdot\delta_T).$$ But what is it good for? For the uniform probability, DEM showed the identity $$\mathbf{U}=\int_{\mathbf{M}(\Omega)}\theta d\overline{\mathbf{U}} .$$ Unfortunately, for $\int_{\mathbf{M}(\Omega)}\theta d\overline{\delta_H}$, I get nothing similar: $$\int_{\mathbf{M}(\Omega)}\theta d\overline{\delta_H} = \frac{2}{3}\delta_H + \frac{1}{3}\delta_T$$

So, again, what does this mean? Wouldn't the Dirac delta function be a more natural measure on $\mathbf{M}(\Omega)$?

I hope that Dr. Winston Ewert reacts to all of the questions before Google Moderator shuts down for good on June 30, 2015...