Discussion
Consider the following statements :
I. If \(v_{1}\), \(v_{2}\) and \(\mu\) are measures and \(v_{1} \perp \mu\), \(v_{2} \perp \mu\), then \(v_{1} + v_{2} \perp \mu\).
II. If \(v\) and \(\mu\) are measures such that \(v \ll \mu\) and \(v \perp \mu\), then \(v\) is identically zero.
Answer: D
Both I and II are true.
I: \(v_{1} \perp \mu\) gives a set \(A_{1}\) with \(\mu(A_{1}) = 0\) carrying all of \(v_{1}\), that is \(v_{1}(A_{1}^{c}) = 0\); similarly \(A_{2}\) for \(v_{2}\). Then \(A = A_{1} \cup A_{2}\) has \(\mu(A) = 0\) and \((v_{1} + v_{2})(A^{c}) = 0\), so \(v_{1} + v_{2} \perp \mu\).
II: \(v \perp \mu\) gives a set A with \(\mu(A) = 0\) and \(v(A^{c}) = 0\). Absolute continuity \(v \ll \mu\) then gives \(v(A) = 0\) too, so \(v(X) = v(A) + v(A^{c}) = 0\). A measure both absolutely continuous and singular with respect to \(\mu\) is zero; this is what makes the Lebesgue decomposition unique.
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