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Sorry, I just need someone to clear up a confusion I have.

Let $(X,\mathcal{F},\mu)$ be any measure space, and let $\#$ be the counting measure on $X$. We have that $$\#(A) = 0 \implies |A| = 0 \implies A = \emptyset \implies \mu(A) = 0.$$

Thus, we get $\mu\ll\#$ and the Radon-Nikodym theorem gives a function $f$ such that $$\mu(A) = \int_A \ \mathrm{d}\#.$$ However, if $\mu$ is the Lebesgue measure on $\mathbb{R}$, then $$0 = \mu(\mathbb{Q}) = \int_{\mathbb{Q}}\ \mathrm{d}\# = \#(\mathbb{Q}) = \infty.$$

What am I doing wrong?

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    $\begingroup$ For any interval $I \in \mathbb{R}$ (or even countably infinite set), the counting measure is not $\sigma$-finite and hence you cannot apply Radon Nikodym Theorem. For any finite set, the derivate is simply $0$. $\endgroup$
    – sudeep5221
    Commented May 22 at 22:36
  • $\begingroup$ @sudeep5221 of course! Feel free to add that as an answer $\endgroup$
    – Sam
    Commented May 22 at 22:43

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The Radon-Nikodym Theorem requires the measure to be $\sigma$-finite. What you found is an example that shows that the hypothesis is crucial, as the counting measure is not $\sigma$-finite on any uncountable set. On a countable set, the Lebesgue measure is zero and so that RN derivative is zero.

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