A sufficient condition for the singularity of two measures I meet this problem in the proof of the Feldman-Hajek theorem from Stochastic Equations in infinite dimensions written by Da Prato.
Two probability measures $\gamma$ and $\mu$ are singular  iff there exists a Borel set $A$ such that $\gamma(A)=1,\mu(A^c)=0$.
Given two probability measures $\gamma$ and $\mu$ on a topological measurable space $(H,\mathcal{B}(H))$ where $\mathcal{B}(H)$  stands for the Borel $\sigma$-field of $H$.
Assume there exists a sequence of random variables $\{\xi_n\}$ on $(H,\mathcal{B}(H))$ such that
$\int_H \xi_n^2\mu(dx)=1$ for every $n\in N$ and $$\lim_{n\rightarrow \infty}\int_H \xi_n^2\gamma(dx)=0.$$
Now it comes the confusing part. The author says that there exists a sequence $\gamma_k \uparrow \infty$ and $n_k\uparrow \infty$ such that
$$\mu(\{x\in H:\lim_{n\rightarrow\infty}|\gamma_k\xi_{n_k}(x)|=+\infty\})=1$$
$$\gamma(\{x\in H:\lim_{n\rightarrow\infty}|\gamma_k\xi_{n_k}(x)|=0\})=1$$
And thus $\gamma$ and $\mu$ are singular.
I wonder how the series $\gamma_k $ are constructed. Clearly they have to be randomly chosen and measurable!
———————————————————————————————————————————————————
Thanks @LostStatistician18 for reminding me of some conditions I may have missed.
I find $\xi_{n}$ should be symmetrically normally distributed both under $\mu$ and $\gamma$.
 A: I feel like a condition might be missing here. It seems that what needs to happen here is that with $I_n = \int \xi_n^2 d\gamma$, we can choose a subsequence such that for some $0<\delta< 1/2$, $\sum_{k=1}^\infty I_{n_k}^{1-2\delta} < \infty$. It follows by Chebyshev's inequality that with $\gamma_k = I_{n_k}^\delta$, $\gamma( |\xi_{n_k}|> \epsilon/\gamma_k) \le I_{n_k}/ (\gamma_k^2\epsilon^2)$, which is summable, so the second equality $\gamma( \lim_{k\to \infty}|\gamma_k \xi_{n_k}| =0)=1$ follows from the Borel-Cantelli theorem.
The problem to me now seems to be that you need some sort of condition about uniform positivity (measured by $\mu$) of the random variables $\xi_n$. For example, a counter example to this statement as is comes from taking $(H,\mathcal{B})$ to be the unit interval with the standard Borel sets, $\mu$ to be point mass of size 1/2 at say 1/4, combined with Lebesgue measure on all subsets of $[1/2,1]$, and $\gamma$ standard Lebesgue measure. Clearly $\mu$ and $\gamma$ are not singular. Then define the functions $\xi_n$ to be a "tent" (polygonal) function with height $\sqrt{2}$ at $1/4$, and with base of size $2/n$. Then $\xi_n$ satisfies the conditions of the result, but for any sequence $\gamma_k$ tending to infinity and any subsequence $n_k$, $\mu ( \lim_{k \to \infty} |\gamma_k \xi_{n_k}| = \infty ) = 1/2$, since this latter probability just comes from the point mass at 1/4.
Sorry this is not an answer, but this was too long to write as a comment.
A: Let $ \gamma_k=k^2 $ and
\begin{equation*}
 n_k=\begin{cases}1,\quad&k=1,\\
    \min\{n>n_{k-1}:\int_H\xi^2_n(x) \gamma(dx)\le\frac1{k^6}\},\quad&k\ge 2,
\end{cases}
\end{equation*}
then
\begin{gather*}
\int_H\gamma_k^2\xi^2_{n_k}(x) \gamma(dx)\le\frac1{k^2},\\
\sum_{k\ge1}\int_H\gamma_k^2\xi^2_{n_k}(x) \gamma(dx)<\infty,\\
\gamma\Big(\Big\{x\in H: \sum_{k\ge1}\gamma_k^2\xi^2_{n_k}(x)<\infty \Big\} \Big)=1, 
\end{gather*}
and
\begin{equation*}
\gamma(\{x\in H: \lim_{k\to\infty}|\gamma_k\xi_{n_k}(x)|=0 \} )=1. \tag{1}
\end{equation*}
For $ \mu $, on the other hand, we have
\begin{gather*}
\mu(\{x\in H:|\gamma_k\xi_{n_k}(x)|<a\} )
=\sqrt{\frac2{\pi}}\int_0^{a/{\gamma_k}}e^{-x^2/2}\,dx\le\sqrt{\frac2{\pi}}\frac{a}{\gamma_k},\\
\sum_{k\ge1}\mu(\{x\in H:|\gamma_k\xi_{n_k}(x)|<a\} )<\infty,\quad \forall a>0,\\
\mu (\{x\in H:|\gamma_k\xi_{n_k}(x)|<a \quad\text{i.o. in $ k $}\} )=0,\quad 
\text{by Borel-Cantelli Lemmma}\quad\forall a>0.
\end{gather*}
Hence,
\begin{equation*}
 \mu (\{x\in H: \lim_{k\to\infty}|\gamma_k\xi_{n_k}(x)|=+\infty \} )=1.
 \tag{2}
\end{equation*}
(1) and (2) are the desired.
A: If $Z$ is standard normal, then
$$
\Bbb E[\exp(-b|Z|)]\sim {C\over b},\qquad b\to+\infty,
$$
for some constant $C>0$.
Now select a subsequence $(n_k)$ such that such that $\Bbb E_\gamma[\xi_{n_k}^2]\le k^{-4}$ (where $\Bbb E_\gamma$ denotes integration with respect to $\gamma$) and take $\gamma_k=k^{1.1}$.
You have
$$
\sum_k \Bbb E_\mu[\exp(-\gamma_k|\xi_k|)]<\infty,
$$
because $\sum_k \gamma^{-1}_k<\infty$.
This implies that $\sum_ke^{-\gamma_k|\xi_k|}$ converges  $\mu$-a.s., and so $\lim_k\gamma_k|\xi_k|=+\infty$, $\mu$-a.s. On the other hand,
$$
\Bbb E_\gamma\left[\sum_k\gamma_k^2\xi_k^2\right]\le\sum_k k^{-1.8}<\infty,
$$
and so $\sum_k\gamma_k^2\xi_k^2<\infty$, $\gamma$-a.s., and therefore $\lim_n\gamma_k|\xi_k|=0$, $\gamma$-a.s.
This argument doesn't seem to need the $\xi_k$ to be Gaussian. It would be enough for all the $\xi_k$ to have the same distribution under $\mu$ with $\mu\{x:\xi_k(x)=0\}=0$, in addition to the condition on second moments under  $\gamma$.
