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I have a question about this answer.

Is integration by substitution a special case of Radon–Nikodym theorem?

Author of the comment says: "The idea is that $f↦∫_{R^n}f∘Tdm$ corresponds to an absolutely continuous measure on $R^n$"

I don't understand it. Why the map that send function to the integral (number) is a measure? Isn't measure a function that takes sets and return numbers? What is the measure here?

Also where can I read proof of standard change of variable formula in $R^n$ using Radon-Nikodym theroem? I heard it is possible.

romperextremeabuser
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  • The measure of a set $A$ is the integral of its characteristic function $\chi_A$ (also known as the indicator function $I_A$). – Rob Arthan Oct 19 '23 at 21:15
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    Using Riesz’s theorem you can think of measures as being positive linear functionals on the space of continuous functions. See this answer of mine for a proof of the change of variables theorem in $R^n$ using the Radon-Nikodym theorem and the abstract change of variables theorem. – peek-a-boo Oct 19 '23 at 21:18
  • @peek-a-boo First of all thank you, I just finished reading about Radon-Nikodym theorem and Lebesgue differentiation theorem and I am now reading your answer and I got stuck.

    \begin{align} \int_{F[\Omega]}f,d\lambda&=\int_{\Omega}(f\circ F)\cdot d(F^\lambda)\tag{COV}\ &=\int_{\Omega}(f\circ F)\cdot \frac{d(F^\lambda)}{d\lambda},d\lambda\tag{Radon-Nikodym} \end{align}

    The Radon-Nikodym theorem says that there is a function which we name $\frac{dF^* \lambda}{d \lambda}$ such that $F^* \lambda (A)=\int_A \frac{dF^ *\lambda}{d \lambda} d \lambda$,

     – romperextremeabuser Oct 21 '23 at 20:01
  • Why is $d(F^* \lambda)=\frac{dF^* \lambda}{d \lambda} d \lambda$? Is it just a notation or there is deeper meaning in that? – romperextremeabuser Oct 21 '23 at 20:04
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    I was being a little fast there. Radon-Nikodym is only used to assert the existence of that function; and this is the most difficult part. Everything else is ‘elementary measure theory’. More explicitly, we have the following general and simple results: if $\mu$ is any measure on any set and $g\geq 0$ is a measurable function, then we can define a new measure $\nu(A):=\int_Ag,d\mu$ (of course we have absolute continuity here, and radon-Nikodym asserts the difficult converse). Now, with this measure $\nu$ at hand, we can ask how are $\nu$-integrals related to $\mu$-integrals. – peek-a-boo Oct 21 '23 at 20:12
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    See this answer of mine for the statement of the relationship between the two integrals (there I unfortunately used $g,f$ in the opposite roles). But anyway, this relationship is very easy to justify. So to summarize, the logic is as follows: first use Radon-Nikodym to assert existence of the function, then use this little measure-theory exercise to relate the integrals relative to the two different measures (and this relationship involves the Radon-Nikodym derivative). – peek-a-boo Oct 21 '23 at 20:19
  • After two whole days I understand everything, thank you – romperextremeabuser Oct 21 '23 at 22:57

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