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Let $\gamma_\infty$ denote the product Gaussian measure on $\mathbb{R}^\mathbb{N}$. Which $a,b \geq 0$ satisfy that for every Borel set $K\subseteq \mathbb{R}^\mathbb{N}$ of positive measure, $a K + b y$ has positive measure for $\gamma_\infty$-a.e. $y$?

The knee-jerk answer is ``only $a = 1$, $b=0$.'' The Cameron-Martin Theorem tells us that $x \mapsto x + y$ is absolutely continuous exactly when $y$ is in $\ell^2$ (a set of $\gamma_\infty$-measure $0$). Similar arguments apply to linear transformations like those above.

This does not answer the question, however. In fact Solecki and I have recently proved that if $a \in \mathbb{R}$ and $K \subseteq \mathbb{R}^\mathbb{N}$ is Borel and of positive measure, then $\gamma_\infty(\sqrt{1+b^2} K + b K) > 0$ for $\gamma_\infty$-a.e. $y$ (i.e. a sufficient conditions is that $a^2 = b^2 +1$). See http://arxiv.org/abs/1201.3947 for the preprint. It is not difficult to prove that a necessary requirement for a positive answer is that $a^2 + b^2 -2ab \leq 1 \leq a^2 + b^2 + 2ab$. This leaves some discrepancy, however, and that is the question at hand.

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Would you be so kind to add few more comments, for educating us. When you define measure - measureable sets are those which are coincide with R except finite number of places ? Correct ? How this measure is related to Wiener measure ? What means "a.e." ? –  Alexander Chervov Feb 1 '12 at 20:25
    
@Alexander: The space can be defined equivalently by saying that the coordine functions on R^N are identically and independently distributed (i.i.d.) and N(0,1) (gaussian on R with mean 0, variance 1). a.e. means "almost everywhere" which means except for a set of measure 0. I didn't use "measurable" anywhere; the Borel sets are the sigma-algebra generated by the open sets; these are all measurable. (The measurable sets are those which are a union of a Borel set and a measure 0 set.) But in the above question, you could replace "Borel" with "compact" (in the product topology). –  Justin Moore Feb 2 '12 at 13:23

1 Answer 1

The answer is that $(a,b)$ must satisfy $a^2 = b^2 + 1$. It is possible to verify (see the preprint above) that for any $b$ and Borel $K \subseteq \mathbb{R}^{\mathbb{N}}$, that $$\gamma_\infty (K) = \int \gamma_\infty (\sqrt{1+b^2} K + b y) d \gamma_\infty (y).$$ By replacing $K$ by $(a/\sqrt{1+b^2})K$ in this equation, we obtain $$\gamma_\infty (\frac{a}{\sqrt{1+ b^2}}K) = \int \gamma_\infty (a K + b y) d \gamma_\infty (y).$$ Let $K$ be all $x$ in $\mathbb{R}^{\mathbb{N}}$ such that $$ \lim_{n \to \infty} \left(\frac{1}{n} \sum_{i< n} x_n^2 \right)^{\frac{1}{2}} = 1. $$ Then $\gamma_\infty(K) = 1$ and therefore $(a/\sqrt{1+b^2}) K$ has measure $0$ unless $a^2 = b^2 + 1$. Thus if $a^2 \ne b^2 + 1$, then $\int \gamma_\infty (a K + b y) d \gamma_\infty (y) = 0$ and hence the integrand vanishes for almost every $y$.

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