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I am trying to calculate the Radon-Nikodym derivative for $\mu = m + \delta_0$ where $m$ is Lebesgue measure over a compact subset of $\mathbb R$ and $\delta_0$ is Dirac measure at $0$.

Clearly, $m \ll \mu$ and $\mu \perp \delta_0$. Therefore, the Radon-Nikodym derivative exists, as both measures are $\sigma-$finite.

Let $f$ be a function such that $m(E) = \int_E fd\mu$. What is $f$ explicitly? Can I say $ 1_{X \setminus\{0\}} = f$ ??

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2 Answers 2

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Yes, the Radon-Nikodym derivative is $1_{X\setminus\{0\}}$. Note that $1_{X\setminus\{0\}}\ne1$ ($\mu$-a.e.) as $0$ is an atom for $\mu$.

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  • $\begingroup$ Does it make sense to find Radon Nikodym derivative at a point for measures? $\endgroup$ Commented Apr 23, 2020 at 14:21
  • $\begingroup$ At an atom? Of course. $\endgroup$ Commented Apr 23, 2020 at 14:27
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The Radon-Nikodym derivative is $1_{X\setminus\{0\}}$. Note that $1_{X\setminus\{0\}}\ne1$ ($\mu$-a.e.) as $0$ is an atom for $\mu$. Add: Notice that $1_{X}$ is wrong, because μ({0})==1!=0, the R-N derivative is unique in the sense of μ-a.e.

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