Let \(\nu, \mu\) be signed measures, and show that \[ \nu \perp \mu \text{ and } \nu \ll \abs{ \mu} \implies \nu = 0 .\]
Solution
1. There is a measurable \(A\) with \(|\mu|(A) = 0\) and \(|\nu|(A^c) = 0\).
This is the definition of \(\nu \perp \mu\): there is a partition \(X = A \sqcup A^c\) with \(A\) null for \(\mu\) and \(A^c\) null for \(\nu\), and a set is null for a signed measure exactly when it has total variation measure \(0\).
2. \(|\nu|(A) = 0\).
\(\nu \ll |\mu|\) means \(|\nu|(E) = 0\) whenever \(|\mu|(E) = 0\), equivalently \(\nu(E) = 0\) for every measurable \(E\) with \(|\mu|(E) = 0\). Apply it to \(E = A\) from step 1.
3. Q.E.D.
By steps 1 and 2, \(|\nu|(X) = |\nu|(A) + |\nu|(A^c) = 0\). Since \(|\nu(E)| \le |\nu|(E)\) for every \(E\), \(\nu = 0\).