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I want to construct a Borel map from the unit interval to the Hilbert cube $f: [0,1] \to [0, 1]^\mathbb N$ so that \begin{equation} \lambda \left(f^{-1}\left( \prod_{i \in \mathbb N} E_i\right)\right)= \prod_{i \in \mathbb N} \lambda(E_i) \end{equation} for $\lambda$ the Lebesgue measure on the interval, $E_i \subseteq [0, 1]$ Borel, and $E_i = [0, 1]$ for all but finitely many indices. This gives the construction of the product measure without appealing to the Kolmogorov extension theorem (c.f. Tao's An Introduction to Measure Theory for the Kolmogorov approach to infinite product spaces). In the general case, I want to find a Borel map $f: [0, 1] \to \mathbb R^{\mathbb N}$ so that \begin{equation} \lambda \left(f^{-1}\left( \prod_{i \in \mathbb N} E_i\right)\right)= \prod_{i \in \mathbb N} \mu_i(E_i) \end{equation} for Radon probability measures $\mu_i$ on $\mathbb R$.

My initial thought was to try to encode the Hilbert cube into the dyadic intervals $[1/2^{n + 1}, 1/2^n]$, e.g. map these into the edges of the Hilbert cube, and try to construct measure preserving maps \begin{equation*} [0, 1] \to \bigsqcup_{n \in \mathbb N} [0, 1] \to [0, 1]^{\mathbb N}. \end{equation*} The first map isn't too bad, however the second is more nebulous. The thought was that this has something to do with independent events in $[0, 1]$ representing a rectangle in the Hilbert cube, e.g. $A \times B \times [0, 1] \times \cdots$ gets pulled back to $A \cap B$.

My second thought was to construct a space-filling curve in spirit of showing the $d$-dimensional Lebesgue measure $\lambda_d$ can be realised as the pushforward of $\lambda$ (c.f. the discussion here for existence of a space filling curve and here which states the Hilbert and Peano curves are measure preserving space filling curves).

The second seems a bit unwieldy but an approach that can work. The first seems more succinct but I can't get the details right. Moreover, the first seems more easy to generalize, i.e. if we replace $\lambda$ on the right hand side of our initial equation with Radon probability measures $\mu_i$ on $\mathbb R$ and the map into the Hilbert cube with a map $f: [0, 1] \to \mathbb R^{\mathbb N}$ by considering the cumulative distribution functions of $\mu_i$.

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    If you let $X_n \in {0,1}$ denote the $n$-th digit in the binary expansion of $x \sim U([0,1))$, then the $X_n$ are iid uniformly distributed. This means that if $\Bbb{N} =\biguplus I_n$ is a disjoint partition with countably infinite sets $I_n$, and if you pick bijections $\iota_n : I_n \to \Bbb{N}$ and set $Y_n := \sum_{\ell =1}^\infty 2^{-\ell} X_{\iota_n^{-1}(\ell)}$, then the map $x \mapsto (Y_n(x))_n$ does what you want. – PhoemueX Apr 14 '20 at 05:43
  • By the way, my greetings with the anniversary. – Alex Ravsky Apr 22 '20 at 12:05

1 Answers1

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After much thought and cobbling everything I could find online and in the literature, I have finally arrived at a solution. We first show that every Radon probability measure $\mu$ on the real line is the pushforward of the Lebesgue measure on the unit interval. Set $g: [0, 1] \to \overline{\mathbb R}$ to be the quantile function \begin{equation} g(x) = \inf \big\{ p \in \mathbb R : \mu((-\infty, p]) \geq x) \big\}. \end{equation} Note that $g$ is non-decreasing and, since the cumulative distribution function of $\mu$ is non-decreasing right continuous, the infimum is achieved. Thus, it satisfies $a < g(x)$ if and only if $\mu((-\infty, a]) < x$, and $g(x) \leq b$ if and only if $x \leq \mu((-\infty, b])$. In particular, \begin{equation} g^{-1} ((a, b]) = (\mu((-\infty, a]), \mu((-\infty, b])], \end{equation} so $g$ is Borel and satisfies $\mu = \lambda \circ g^{-1}$ on the half-open intervals. Arguing by $\pi$-$\lambda$, we conclude $g$ is indeed the desired pushforward map. Let $g_n : [0, 1] \to \mathbb R$ be the pushforward map corresponding to $\mu_n$, modified at the endpoints so that $g_n (0) = g_n (1) = 0$ in case they are infinite. This only adds finitely many points so the previous result still holds. Define the product map $\Psi: [0, 1]^{\mathbb N} \to {\mathbb R}^{\mathbb N}$ by $\{x_n\}_{n \in \mathbb N} \mapsto \{ g_n (x_n) \}_{n \in \mathbb N}$; notice that it pre-images rectangles to rectangles, \begin{equation} \Psi^{-1} \left( \prod_{n \in \mathbb N} E_n \right) = \prod_{n \in \mathbb N} g_n^{-1} (E_n) \end{equation} for any $E_n \subseteq \mathbb R$. This shows that $\Psi$ is Borel measurable and pushes the product Lebesgue measure onto the product of the Radon measures $\bigotimes_n \mu_n$. It remains then to find a Borel map $\Phi: [0, 1] \to [0, 1]^{\mathbb N}$ which pushes the Lebesgue measure forward to the product Lebesgue measure. Setting $f = \Psi \circ \Phi$ gives the result; \begin{equation} \lambda \left( f^{-1} \left( \prod_{n \in \mathbb N} E_n\right) \right) = \lambda \left( \Phi^{-1} \left(\prod_{n \in \mathbb N} g_n^{-1} (E_n) \right) \right) = \prod_{n \in \mathbb N} \lambda(g_n^{-1} (E_n)) = \prod_{n \in \mathbb N} \mu_n (E_n) \end{equation} whenever $\prod_n E_n \in \prod_n \mathcal B(\mathbb R)$. Using a uniformly distributed space filling curve or the binary expansion argument as indicated by PhoemueX in the comments furnishes the desired $\Phi$.