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?

• 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 $0$. Commented May 22 at 22:36
• @sudeep5221 of course! Feel free to add that as an answer
– Sam
Commented May 22 at 22:43

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.