Gaussian measure in Hilbert space It's part of proof of proposition 1.25 from An Introduction to Infinite-Dimensional Analysis, Giuseppe Da Prato, Springer, page 21.
Proposition:
Let $\mu = N_{a,Q}$ be a nondegenerate Gaussian measure in $H$, where $H$ is seperable Hilbert space. Then $\mu$ is full.
In proof we have 
$$ A_n = \Big\{ x\in H: \sum\limits_{k=1}^{n} x^2_k \leq \frac{r^2}{2} \Big\}$$
and that clearly $\mu(A_n) > 0$. Here is my question: 
Why is it clear that $\mu(A_n) > 0$?
We also have second set $B_n = \Big\{ x\in H: \sum\limits_{k=n+1}^{\infty} x^2_k < \frac{r^2}{2} \Big\}$, but in this case we are proving that $\mu(B_n)>0$ with help of Chebyshev inequality.
 A: Assume w.l.o.g. $a = 0$ and write $\mu = N_{Q} := N_{a, Q}$.
De Prato assumes that $N_{Q}$ is non-degenerate i.e. that $Q$ is injective. By definition of $N_{Q}$ as $N_{Q} = \times_{k = 1}^{\infty} N_{\lambda_k}$ where the $\lambda_k$ are the eigenvalues of $Q$ i.e. $Q = \sum_{k = 1}^{\infty} \lambda_k \langle \cdot, e_k \rangle e_k$ where the $e_k$ are the corresponding eigenbasis of $H$ w.r.t. $Q$. So in particular, the projection functionals $x \mapsto \langle x, e_k \rangle =: x_k$ are jointly Gaussian and uncorrelated, and thus independent.
Hence
\begin{align*}
\mu(A_n) &= \mu \left\{ x \in H : \sum_{k = 1}^n x_k^2 \leq \frac{r^2}{2} \right\} \\
&= \left( \mu \circ ( \langle \cdot, e_1 \rangle, \ldots, \langle \cdot, e_n \rangle )^{-1} \right) \left( \overline{B} \left(0,\frac{r}{\sqrt{2}} \right) \right) \\
&= \gamma_n \left( \overline{B} \left( 0,\frac{r}{\sqrt{2}} \right) \right) > 0
\end{align*}
where $\gamma_n$ is the standard Gaussian measure on $\mathbb{R}^n$ and $\overline{B}(0,r)$ denotes the closed unit ball around $0$ with radius $r$.
The fact that $Q$ is injective ensures that $\mu \circ ( \langle \cdot, e_1 \rangle, \ldots, \langle \cdot, e_n \rangle )^{-1}$ has density for any $n \in \mathbb{N}$ - if one of the $\langle \cdot, e_1 \rangle$ was degenerate, then the last equation would not be true. Note that in the proof it is crucial that $n \in \mathbb{N}$ is arbitrary so that one can show $\mu(B_n) > 0$. Think of how this would play out if $Q$ had finite-dimensional image.
