In Folland's real analysis, it writes
Nonexample: Let $\mu$ be Lebesgue measure and $\upsilon$ the point mass at 0 on $(\mathbb{R}, \mathcal{B}_{\mathbb{R}})$. Clearly $\upsilon \perp \mu$. The nonexistent Radon-Nikodym derivative $dv/d\mu$ is popularly known as the Dirac $\delta$-function.
I'm confused why is the Radon-Nikodym derivative $dv/d\mu$ said to be nonexistent? Isn't it just the Dirac $\delta$-function as the author later said?
Any help would be appreciated!
Suppose we consider $\delta$ as a measurable function $\Bbb R\to\overline{\Bbb R}$. How? The only reasonable way would be to say, $\delta:x\mapsto\begin{cases}+\infty&x=0\\0&x\neq0\end{cases}$. But then $\delta$ is a.e. zero and it's obvious that the induced measure $A\mapsto\int_A\delta\,\mathrm{d}\mu$ would just be the zero measure, which is not equal to $\nu$.
No Radon-Nikodym derivative exists because $\nu$ is not absolutely continuous w.r.t $\mu$.