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Hey, I'm reading the proof of proposition 6.8 in Big Rudin, pg. 120. I'll just mention where my confusion lies: Suppose K is an arbitrary measure (complex, real, whatever). In the proof of property (f), Rudin says that if K(A) = 0, and F is a subset of A, then K(F) = 0. Rudin has forced that by construction in chapter 1 for positive measures, but why should it work for complex or signed measures? Does there exist two sets A,B such that A is a subset of B, measure of B is 0, but measure of A is non-zero (negative?), for some arbitrary measure?

EDIT: Rudin's prop 6.8, part (f) says: given two arbitrary measures $\lambda_1$, $\lambda_2$, a positive measure $\mu$. If $\lambda_1 \ll \mu$, and $\lambda_2 \perp \mu$, then $\lambda_1 \perp \lambda_2$.

Also, $\ll$ and $\perp$ denote absolute continuity and mutual singularity, respectively.

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    Okay, just a minute. Don't worry, this happens to all of us. :)2011-05-15

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Observe that $\mu$ is positive. Since $\mu$ is positive and $\mu \perp \lambda_{2}$, the latter is concentrated on a $\mu$-null set $A$. But as $\lambda_{1} \ll \mu$ and $\mu$ is positive, we conclude for all measurable $E \subset A$ that $\mu(E) = 0$, hence $\lambda_{1}(E) = 0$ by absolute continuity, and thus $\lambda_{1}$ is concentrated on the complement of $A$. In other words, $\lambda_{1} \perp \lambda_{2}$.