Calculating Radon Nikodym derivative

Question

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$ ??

score 7 · Accepted Answer · answered Dec 01 '13 at 19:06

7

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 $\mu$.

answered Dec 01 '13 at 19:06

Martin Argerami

217,281

Does it make sense to find Radon Nikodym derivative at a point for measures? – Manoj Kumar Apr 23 '20 at 14:21
At an atom? Of course. – Martin Argerami Apr 23 '20 at 14:27

score 2 · Answer 2 · answered Dec 24 '16 at 15:32

2

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.

answered Dec 24 '16 at 15:32

Weizheng Zhang

101

Calculating Radon Nikodym derivative

2 Answers2

Linked