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It is easy to prove that such a measure cannot exist by showing a contradiction in the properties it must have (e.g. space must have probability 1 while being countably additive and shift-invariant). However, someone has shown me a construction, and I cannot find a satisfying direct answer for why it won't work:

Identify each natural by its dyadic expansion: \begin{equation} x \simeq \dots0001b_{N-1}\dots b_2b_1. \end{equation}

Now take $\Omega=\{\text{binary sequences}\}$ and consider the product measure where each letter is iid with $\mu(b_n=1)=0.5$, forming some probability space $(\Omega,\mathcal{A},\mu)$. Now condition on: \begin{align} S:=\{(b_{n})_n\text{ nonzero finitely often}\} \end{align} (in limit, approximating $S$ with positive-probability sets) and use the limiting measure.

Why, directly, is it impossible to come up with a reasonable way of conditioning on $S$?

I expect the Borel-Kolmogorov paradox is unresolvable here: there is a net of nontrivial[1] sub-$\sigma$-algebras of $\mathcal{A}$ which approximate $\mathcal{A}\cap S$[2], but whose conditional measures do not converge.

This depends on the precise meaning of a net of sigma-algebras "approximating" another. Can anyone demonstrate that for any reasonable topology on sigma-algebras, any net that approximates $\mathcal{A}\cap S$ cannot form a limiting measure?


[1]: nontrivial here means the sigma-algebra has sets with probability in $]0,1[$. If a sigma-algebra is nontrivial then conditioning on it makes sense.

[2]: One way some particular sigma-algebras can converge to another is to take $(A_k)$ a net of measurable, positive-probability sets convergent to $S$ (that is, $\lim\sup A_k=\lim\inf A_k=S$). Then it is reasonable to say $\mathcal{A}\cap A_k:=\{A\cap A_k |A\in \mathcal{A}\}$ approaches $\mathcal{A}\cap S$, since all their sets converge. Maybe there is a more general idea of sigma-algebra convergence, or better, a topology on sigma-algebras.

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    $\begingroup$ The measure $S$ has probability zero; conditioning on an event of measure zero will involve dividing by zero. $\endgroup$ – Lord Shark the Unknown Aug 27 '17 at 10:51
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    $\begingroup$ @LordSharktheUnknown: Not true. Naively applying the simplest definition of conditional probability would result in division by zero, but more sophisticated constructions exist. The problem here is that it's not evident that one can apply any of the more sophisticated methods, and indeed the impossibility proof implies that one cannot. $\endgroup$ – Hurkyl Aug 27 '17 at 10:52
  • $\begingroup$ Even if you could produce and check a countably infinite number of equally probable infinite binary sequences in a finite time, finding one with a finite number of $1$s would have probability $0$ in finite time. Any short-cut would remove the uniformity of the distribution $\endgroup$ – Henry Aug 27 '17 at 11:35
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    $\begingroup$ Abstractly, conditioning on an event of probability zero is nonsense. Occasionally one can make sense of it. But you certainly have not done the work for that here. And (as you say at the beginning) it cannot be done in this case. $\endgroup$ – GEdgar Aug 27 '17 at 11:37
  • $\begingroup$ @GEdgar, yes but my question is to demonstrate that no such work can be done. $\endgroup$ – enthdegree Aug 27 '17 at 21:59

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