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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!

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    $\begingroup$ $\mu(\{0\})=0$ but $v(\{0\})=1$. $\endgroup$ Commented Jul 3, 2023 at 23:33
  • $\begingroup$ There is not a thing as the Dirac delta function in the sense of a measurable function $\delta:\mathbb{R}\rightarrow\mathbb{R}$. The Dirac $\delta$ is a measure on sets of $\mathbb{R}$, that asking $1$ to every set containing $0$, and zero otherwise (It can also be considered as a generalized function of distribution). $\endgroup$ Commented Jul 4, 2023 at 0:05

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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 $\mu$.

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  • $\begingroup$ I'm sorry for confusing the concepts, but may I ask what is "point mass at 0" and why $\upsilon \perp \mu$? $\endgroup$ Commented Jul 4, 2023 at 0:34
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    $\begingroup$ @sam2018 The point mass (n.b. "Mass" is the German word for "measure", it's intuitive to view measures as assigning mass, this fits with the 'density' analogies too) refers to the "Dirac measure" (also denoted $\delta$) which is defined through $\delta=\chi_{\{0\}}$; the indicator function of a point (assigning mass to that point and nowhere else) $\endgroup$ Commented Jul 4, 2023 at 0:37
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    $\begingroup$ We have $\nu\perp\mu$ because we can partition $\Bbb R=A\sqcup B$ where $\mu|_B\equiv0$, $\nu|_A\equiv0$; choose $A=\Bbb R\setminus\{0\}$ and $B=\{0\}$ $\endgroup$ Commented Jul 4, 2023 at 0:39
  • $\begingroup$ So $\int_S f(x)\delta (x-x_0)d\mu(x) := \int_S f(x)d v(x) = f(x_0)\delta_{x_0}(S) = f(x_0) 1_S(x_0)$, and the leftmost expression is just a confusing/abuse of notation purely symbolic expression to represent what is on the right? $\endgroup$ Commented Nov 21, 2023 at 0:28
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    $\begingroup$ @lightxbulb yeah I guess so? $\delta$ is meaningful in things like distribution theory (apparently) but I don't ever work with it. From my view point, it's weird notation for "evaluation at a point" but take that with a pinch of salt because I never actually need to use it myself. It's convenient notation for point-mass measures too $\endgroup$ Commented Nov 21, 2023 at 0:38

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