You can't talk about "the" Gaussian measure on an infinite-dimensional Hilbert space, for the same reason that you can't talk about a uniform probability distribution over all integers. It doesn't exist; see Richard's answer. However, there are a lot of non-uniform Gaussian measures on infinite dimensional Hilbert spaces.

Consider the measure on $\mathbb{R}^\infty$ where the $j$th coordinate is a Gaussian with mean 0 and variance $\sigma_j^2$, where $\sum_{j=1}^{\infty} \sigma_j^2 < \infty$ (and different coordinates are independent). This is almost surely bounded in the $\ell_2$ metric, and any projection onto a finite-dimensional space has a Gaussian distribution. The squared length of a vector drawn from this measure is a sum of squares of Gaussians, and so follows some kind of generalized $\chi$-square distribution. If I knew more about generalized $\chi$-square distributions, I might be able to tell you what the measure of the unit ball was.

This kind of Gaussian distribution is very important in quantum optics. In fact, in quantum optics, a thermal state is Gaussian, so "the" Gaussian measure actually makes some sense.

aGaussian measure on $X$, which is a perfectly well-defined notion; but the answers so fare are written as if the OP asked abouttheGaussian measure, which is not. Of course, the answer to this will depend on the mean and covariance of $\mu$. – Mark Meckes Aug 22 '10 at 23:48