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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$ ??

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$ ??

I am trying to calculate 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, $m << \mu$ and $\mu \perp \delta_0 $ So, Radon Nikodym derivative exist, as both measures are $\sigma-$finite, Say $m(E) = \int_E fd\mu$.

Now what is this function $f$ explicitly?

Considering the set $A$ of all functions, $g : X -\{0\} \rightarrow [0, \infty]$ such that $\int_E g d\mu\le m(E)$ Can I say $ 1_{X -\{0\}} = f$ ??

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$ ??