MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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 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.

share|cite|improve this question
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

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 $y$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.