Show that $|\mu|$ is the smallest of those positive measures $\nu$ that satisfy $|\mu(A)| \le \nu(A)$.

Question

Let $\mu$ be a signed measure on a measurable space $(X, \mathcal{A})$, and let $|\mu|$ be the variation of $\mu$ (i.e., $|\mu| = \mu^+ + \mu^-$, where $\mu = \mu^+ - \mu^-$ is the Jordan decomposition of $\mu$).

Show that $|\mu|$ is the smallest of those positive measures $\nu$ that satisfy $$ |\mu(A)| \le \nu(A) $$ for each $A \in \mathscr{A}$.

Suppose that $\nu$ is a positive measure such that $|\mu(A)| \le \nu(A) \le |\mu|(A)$, for each $A \in \mathscr{A}$. Then $$ |\mu^+(A) - \mu^-(A)| \le \nu(A) \le \mu^+(A) + \mu^-(A). $$ I feel like this is the general idea, but it's not getting me anywhere.

Answer 1

The inequality $\vert \mu(A) \vert \leq \vert \mu \vert (A)$ is easy. Let us focus on the other part. In the Jordan decompositon we get that $\mu^+$ and $\mu^-$ are mutually singular. This means, that there exist disjoint measurable sets $A, B$ such that $X=A\cup B$ and $\mu^+(B)=0 =\mu^-(A)$. Then we compute $$ \vert \mu \vert (K) = \vert \mu \vert(K \cap A) + \vert \mu \vert (K \cap B) = \mu^+(K\cap A) + \mu^-(K \cap B) =\vert \mu (K \cap A ) \vert + \vert \mu (K \cap B) \vert \leq \nu(K\cap A) + \nu(K\cap B) = \nu(K). $$

Answer 2

We can write $\mu^{+}(A)=\mu (A\cap B)$ and $\mu^{-}(A)=-\mu (A\cap B^{c})$ for some $B$. Also $|\mu|=\mu^{+}+\mu^{-}$. Hence $|\mu|(A)=\mu^{+}(A)+\mu^{-}(A) =\mu(A\cap B)+\mu(A\cap B^{c})\leq \nu(A\cap B)+\nu(A\cap B^{c})=\nu (A)$.