# Signed measure of subsets of zero-measured sets?

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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How about $K = \delta_{0} - \delta_1$, $B = \{0,1\}$ and $A_{0} = \{0\}$ and $A_{1} = \{1\}$, to give two silly examples? It would be helpful if you said what Rudin's prop. 6.8 says and what property (e) is since not everybody has Rudin at hand. – t.b. May 15 '11 at 21:16
Yikes, I meant (f). I'll correct this now! I've also added in the whole statement for those who don't own Rudin, apologies for the oversight. – paul May 15 '11 at 21:27
I still don't quite see the confusion. By hypothesis $\lambda_{2}$ is concentrated on some set $A$ on the complement which $\mu$ vanishes. As $\lambda_{1} \ll \mu$, we must have that $\lambda_{1}$ vanishes on the complement of $A$ as well. Maybe applying (e) that $|\lambda_{1}| \ll \mu$ helps understanding what's going on? – t.b. May 15 '11 at 21:39
Theo, I get everything that you're saying. It's just that in the proof of (f), we can infer from absolutely continuity that $\lambda_1(A) = 0$, and then Rudin says $\lambda_1(E) = 0$ for all $E \subset A$. But how can you conclude that last point if you assume that $\lambda_1$ is arbitrary? Everything else is crystal clear to me. – paul May 15 '11 at 21:42
Well, because $\mu(E) = 0$ for all $E \subset A$, as $\mu$ is positive. (Oops: I shouldn't have said on "the complement of which" - delete "the complement of" in my previous comment) – t.b. May 15 '11 at 21:44

## 1 Answer

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, $\lambda_{1} \perp \lambda_{2}$.

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