Let $a \in \mathbb{R}$ and Dirac measure $\delta_a (A) = 0$ if $a \notin A$ and $\delta_a(A) = 1$ if $a \in A$, and let $\mu_1$ be the one-dimensional gaussian measure.
Let $\mu$ and $\nu$ be two $\sigma$ - finite measures over a measurable space $(E, \mathcal{B})$. We say that $\mu$ is absolutely continuous with respect to $\nu$, not. $\mu << \nu$, if $\forall A \in \mathcal{B}$ with $\nu (A) = 0$, then $\mu (A) = 0$.
I don't know to show that $\delta_a$ is not absolutely continuous with respect to $\mu_1$.
Suppose that $\delta_a << \mu_1$. From Radon - Nikodym Theorem follows that $\exists f : \mathbb{R} \rightarrow \mathbb{R}_+$ measurable so that $\delta_a(A) = \underset{A}{\int} f d\mu_1, \forall A \in \mathcal{B}(\mathbb{R})$. If $\delta_a(A) = 1$, then $\underset{A}{\int} f d \mu_1 = 1$. And then?
Thank you!
You don't need the Radon-Nikodym theorem here at all. Simply note that $\mu_1(\{a\})=0$ while $\delta_a(\{a\})=1$.