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A question regarding notation of the radon nikodym derivative: If the derivative $\frac{d\nu}{d\mu}:=f$ exists, some authors write something like $d\nu = f d\mu$, i.e., they treat the derivative as if it was an actual fraction and plug in $f d\mu$ for $d\nu$ etc.

My problem, however: I dont know what this is expression is supposed to mean since $\frac{d\nu}{d\mu}$ is not an actual fraction of course but only the notation for a specific function. So what does the expression $d\nu = f d\mu$ mean?

joriki
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guest1
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    I would say it's a shorthand for $\int d\nu = \int fd\mu$, just omitted the integral. – Kaira Feb 21 '23 at 15:14
  • This is a great question, but also one which has been asked many times before on this site. See discussion in answers here, or here for instance. – postylem Feb 21 '23 at 15:41
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    Does this answer your question? Question on integral, notation and Nikodym derivative. Also see here –  Feb 21 '23 at 16:14
  • thanks for the suggestions. However, I dont find the answers in these threads necessarily satisfying. In particular, I still dont see how it is justified to say $\int g d\nu=\int g f d\mu$. I.e., in that example ppl who use that notation suggest that they would simply substitute the infinitesimals via $d\nu = f d\mu$ but again, this shouldnt be allowed because there is no reason we should be allowed to treat the radon nikodym derivative $f=\frac{d\nu}{d\mu}$ like an actual fraction. – guest1 Feb 24 '23 at 14:27
  • So looking at the wikipedia of radon nikodym, if g is integrable and $\nu$ is absolutely continuous w.r.t. to $\mu$, then $\int g d\nu = \int g \frac{d\nu}{d\mu}d\mu$. This is actually the theorem that ppl use implicitely. But it requires $g$ to be integrable which ppl seem not to care about when using the abusive notation $d\nu = f d\mu$ and substitute it as they wish. – guest1 Feb 24 '23 at 14:40

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