The Radon-Nikodym Theorem says the following:
Theorem$\quad$Let $(X,\mathscr{A})$ be a measurable space, and let $\mu$ and $\nu$ be $\sigma$-finite positive measures on $(X,\mathscr{A})$. If $\nu$ is absolutely continuous with respect to $\mu$, then there is an $\mathscr{A}$-measurable function $g:X\to[0,+\infty)$ such that $\nu(A)=\int_Agd\mu$ holds for each $A$ in $\mathscr{A}$. The function is unique up to $\mu$-almost everywhere equality.
I want to show that the $\sigma$-finiteness assumption of $\mu$ cannot be omitted, but I got stuck:
If we let $X=[0,1]$, let $\mathscr{A}$ be the Borel $\sigma$-algebra of $[0,1]$, let $\mu$ be the counting measure on $(X,\mathscr{A})$, and let $\nu$ be the Lebesgue measure on $(X,\mathscr{A})$, then $\nu\ll\mu$. But I couldn't manage to show that there is no $\mathscr{A}$-measurable function $f$ such that $\nu(A)=\int_Afd\mu$ holds for all $A$ in $\mathscr{A}$. I wanted to prove it by way of contradiction, but I couldn't go anywhere.
Could someone please help me out? Thanks a lot in advance!
Reference$\quad$ Example 4.2.3 from Measure Theory by Donald Cohn.