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I first note that a similar question was asked here: Calculating Radon Nikodym derivative, though the explicit steps used to calculate the derivative were not made clear.

Over measurable space $([0,1],\mathcal{B}_{[0,1]}), \mu:=\nu+\delta_{0}.$

$\forall A\in\mathcal{B}_{[0,1]}$ define $\nu(A):=|A|.$

Finally, $\delta_{0} $ is the dirac measure with criterion of measured set containing $0$ in order to evaluate to 1.

$\mu >>\nu$ is clear and therefore $d\nu/d\mu$ exists.

$d\nu/d\mu \overset{*}= \frac{\nu(dy)}{\mu(dy)}=\frac{|dy|}{|dy|+\delta_{0}(dy)} = 1/\frac{|dy|+\delta_{0}(dy)}{|dy|} =1/(1+\delta_{0}(dy)/|dy|)$. It is immediately apparent at this step, however, that this derivative will always be nonzero. Thus, I did something wrong in my calculation (and would like some help seeing where I went wrong). * is due to Notation when integrating with respect to a measure and is where I imagine I went wrong.

UPDATE:

It is clear that $\nu(A)=\int_{A}d\nu= \int_{A}\frac{d\nu}{d\mu}d\mu\Leftarrow \frac{d\nu}{d\mu}=\mathbf{1}_{\{X/0\}}$ to cancel out the dirac measure part in $\mu$ when it is not measure $0$, however, can the derivative be calculated from the differentials in some way or is it purely determined through the integral?

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  • $\begingroup$ The "differential" is just notation and is defined by the integral equation. $\endgroup$ Commented Mar 20, 2019 at 0:17
  • $\begingroup$ @user251257 Interesting... is this why SDEs are typically interpreted in the integral equation form en.wikipedia.org/wiki/…? $\endgroup$ Commented Mar 20, 2019 at 1:27
  • $\begingroup$ As stated in the wikipedia article, SDE is an integral equation. The "differential equation" is just informal notation. $\endgroup$ Commented Mar 20, 2019 at 1:32
  • $\begingroup$ @user251257 I see, thank you for pointing that out! I wasn't sure as to why SDEs were formally integral equations until now. $\endgroup$ Commented Mar 20, 2019 at 1:59

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Your analysis was basically correct, and will be made rigorous below. At $$ \frac1{1+\frac{\delta_0(dy)}{|dy|}} $$ look what happens when $dy$ is around $0$. Then $|dy|$ is infinitely small,while $\delta_0(dy)=1$. Therefore, the above is $\frac1{1+\frac10}=\frac1{\infty}=0$, so the derivative is equal to $0$ at $y=0$, as expected. You were think $1/\text{stuff}$ could not be zero, but you forgot about stuff $=\infty$.

To get a differential characterization of $\frac{d\nu}{d\mu}$, it can shown that for $\mu$ a.e. $x$, you have $$ \frac{d\nu}{d\mu}(x)=\lim_{\epsilon\to 0}\frac{\nu((x-\epsilon,x+\epsilon)\cap [0,1])}{\mu((x-\epsilon,x+\epsilon)\cap [0,1])}. $$ This is a generalization of the Lebesgue differentiation theorem. I do not know what the standard reference for this is, but chapter 6.3 of these notes has a proof.

Now, when $x\neq 0$, then both the numerator and denominator are equal for all small enough $\epsilon$, so the limit is $1$. However, when $x=0$, the numerator will tend to zero while the denominator will tend to $1$. This shows that $$ \frac{d\nu}{d\mu}=1_{(0,1]}\qquad \mu \;a.e. $$

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  • $\begingroup$ I see. This is surprising as I thought it would be the case that because $\mu$ is not absolutely continuous w.r.t. $\nu$ that perhaps $ (\frac{\nu(dy)}{\mu(dy)})^{-1}$ does not exist as in en.wikipedia.org/wiki/Radon%E2%80%93Nikodym_theorem#Properties. How is it then that my inversion of the original density was proper? $\endgroup$ Commented Mar 20, 2019 at 23:26
  • $\begingroup$ $(\frac{\nu(dy)}{\mu(dy)})^{-1}$ always exists when $\frac{\nu(dy)}{\mu(dy)}$ exists, though it might be infinite. It is only $\frac{\mu(dy)}{\nu(dy)}$ which may not exist. If $\frac{\mu(dy)}{\nu(dy)}$ exists, then it equals $(\frac{\nu(dy)}{\mu(dy)})^{-1}$. @BayesIsBae $\endgroup$ Commented Mar 21, 2019 at 0:08
  • $\begingroup$ That is very interesting! $\endgroup$ Commented Mar 21, 2019 at 2:30
  • $\begingroup$ Well, we are working over the measurable space $([0,1],\mathcal B_{[0,1]})$, so I was making sure all of the subsets actually lived in that space. $\endgroup$ Commented Apr 4, 2019 at 1:22
  • $\begingroup$ Sorry about that - I realized only after asking in a previous comment $\endgroup$ Commented Apr 4, 2019 at 1:24

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