Chapter 3: Problem 14
If \(\nu\) is an arbitrary signed measure and \(\mu\) is a \(\sigma\)-finite measure on \((X, \mathcal{M})\) such that \(\nu \ll \mu\), there exists an extended \(\mu\)-integrable function \(f: X \rightarrow[-\infty, \infty]\) such that \(d \nu=f d \mu\). Hints: a. It suffices to assume that \(\mu\) is finite and \(\nu\) is positive. b. With these assumptions, there exists \(E \in \mathcal{M}\) that is \(\sigma\)-finite for \(\nu\) such that \(\mu(E) \geq \mu(F)\) for all sets \(F\) that are \(\sigma\)-finite for \(\nu\). c. The Radon-Nikodym theorem applies on \(E\). If \(F \cap E=\mathscr{\text { , then either }}\) \(\nu(F)=\mu(F)=0\) or \(\mu(F)>0\) and \(|\nu(F)|=\infty\).
Short Answer
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Key Concepts
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