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Let $ N = \{1,2,3, \ldots \}$. We define $\varphi : N^N \mapsto [0,1]$ as $$ \varphi \left( (a_n)_{n \in_N} \right) = [0;a_1, a_2, \ldots ]$$ Where the expression on the right is a infinite continued fraction: $$[0;a_1,a_2,a_3,\dotsc] = \cfrac{1}{a_1+\cfrac{1}{a_2+\cfrac{1}{a_3+\cdots}}}$$ It is standard exercise that this is borel mapping (continuous in fact, see here).

We define the Gauss measure on $[0,1]$ as: $$ d\nu = \frac{1}{\ln(2)}\,\frac{dx}{1+x} $$

Let $\mu$ be a probability measure on $N$. By a product measure on $N^N$ i mean probabilistic measure $\widetilde { \mu }$ such that for any set of the form $F = F_1 \times \cdots \times F_n \times N \times N \times \cdots$ holds equality $$ \widetilde {\mu } (F) = \prod_{n=1}^n \mu (F_k) $$ Such a measure exists and is unique due to teorem about extending funtion on ring to a measure.

I am to prove that the Gauss measure $\nu$ is not a pushforward of any such measure $\mu$, meaning that for any $\mu$ the equality $$ \nu( A ) = \mu ( \varphi^{-1} [A])$$ cannot hold for every borel $A \subset [0,1]$.

I am looking for any help. I have no experience with infinite measure products. This is an exercise for my ergodic theory course.

I had obtained a tip that i don't understand:

Product measure has the property that basis clopen sets with disjoint supports are stochastically independent. Gauss measure does not have this property.

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wroobell
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    Don't complicate this with that "tip". Simply compute the measure on the projection of a single branch (you have infinitely many) and multiply two such numbers (for two different branches of even for the same branch). – John B Jun 23 '16 at 16:48
  • Can you be more specific? What is projection and what is branch? – wroobell Jun 23 '16 at 16:58
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    If you are not familiar with the approach that I have mentioned, compute $c_i=\nu(F_i\times\mathbb N^\mathbb N)$. This will be "the measure on the projection of a single branch". What you have to decide is whether $$c_ic_j=\nu(F_i\times F_j\times\mathbb N^\mathbb N)$$ and this amounts to a somewhat simple computation. – John B Jun 23 '16 at 17:03
  • I think your approach helped. I am gonna wait for my supervisors approval an then maybe publish it here. :) – wroobell Jun 23 '16 at 20:13
  • Thank you for your help, it worked. I posted the answer below. – wroobell Jul 05 '16 at 09:05

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Thanks to @Jonas i came up with a solution and my supervisor accepted it. Hence, i am posting it for completeness. Don't judge me if it is poorly written, i am translating it from Polish. ;)

From the source given in question we know that for any irrational $x \in [0,1]$ it correspond to some continued fraction $[0; a_1, a_2, \ldots ]$. Another thing that we know is that it is between points $$ x_n = [ 0; a_1, a_2, \ldots , a_n] \qquad x_{n+1} = [0; a_1, a_2, \ldots, a_{n+1} ] $$

To ease the notation lets denote for any $ \sigma \in N^{ < N}$ the set $ [[ \sigma ]] = \{ f \in N^N : \sigma \prec f \}$ and by $| \sigma|$ the length of $\sigma$.

Assume that for some product measure $\widetilde{ \mu}$ we have for all borel sets $A$ $$ \nu (A) = \widetilde{ \mu} \left( \varphi^{-1} (A) \right) $$

From the fact that $\varphi$ is a bijection from $N^N$ to $[0,1] \setminus Q$ and easy-to-check equalities $$[0;1] = 1 \qquad [0;2] = 1/2 \qquad [0;3] = 1/3 \qquad [0;1,2] = 2/3 \qquad [0;1,3] = 3/4 $$ we get the following equalities: $$ \varphi^{-1} [1/2, 1] = [[(1) ]] \qquad \varphi^{-1} [1/3 ,1/2] = [[(2) ]] \qquad \varphi^{-1} [2/3, 3/4] = [[(1,2) ]] $$

Hence, from the definition of product measure we have that $$ \widetilde{\mu} \left( [[(1,2)]] \right)= \widetilde{\mu} \left( [[(1)]] \right) \cdot \widetilde{\mu} \left( [[(2)]] \right)$$ By assertion that $\nu$ is a pushforward measure we should have $ \nu (2/3, 3/4) = \nu (1/2,1) \nu (1/3, 1/2)$. However, with the help of almighty Wolfram we can tell that this is not the case because: $$\nu (2/3, 3/4) = (\log 2)^{-1} ( \log 21 -\log 20) \qquad \nu (1/2,1) = (\log 2)^{-1} ( \log 4 -\log 3) \qquad \nu (1/3, 1/2) = (\log 2)^{-1} ( \log 9 -\log 8)$$ and $$( \log 4- \log 3)(\log 9 - \log 8) \neq \log 2 ( \log 21 - \log 20) $$ So we have contradiction.

wroobell
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