Question about Radon Nikodym

Question

If $\mu$ is absolutely continuous with respect to $\upsilon$, then a theorem is $\frac{d|\mu|}{d\upsilon}=\left|\frac{d\mu}{d\upsilon}\right|$. Let's say that $\lambda$ is the Lebesgue measure and $\upsilon(A)=\int_Ah \, d\lambda$, where $\frac{d\upsilon}{d\lambda}=h$. If we define $\mu=|\upsilon|$, then $|\int_Ah \, d\lambda|=\mu(A)=\int_A|h| \, d\lambda$ by the theorem. This doesn't seem correct though. Could someone explain what is wrong.

Answer 1

The mistake is $\mu(A)=|\int_Ah\,d\lambda |$. You need to go back to the definition of $|\nu |$. It is not defined by $|\nu|(A)=|\nu(A)|$. Rather, $|\nu |$ is defined as $\nu^++\nu^-$, where $\nu^+(A)=\nu(A\cap P)$ and $\nu^-(A)=-\nu(A\cap N)$, where $(P,N)$ is the Hahn decomposition of $\nu$.