Prove that a total measure |$\mu$| is the smallest measure for the following Assume that we have a signed measure $\mu$ on a measurable space $(X,\Sigma)$ and a Hahn decomposition $(P,N)$ for $\mu$, then we define $|\mu| = \mu^{+} + \mu^{-}$ where $\mu^{+}(B) = \mu(B \cap P)$ and $\mu^{-}(B) = \mu(B \cap N)$ for each $B \in \Sigma$.
I have managed to show that $|\mu(B)| \leq |\mu|(B)$ for each $B \in \Sigma$, but would also like to show that $|\mu|$ is the smallest unsigned measure on $(X,\Sigma)$ that makes this possible.
So if I pick an unsigned measure $m$ on $(X,\Sigma)$ such that $|\mu(B)| \leq m(B)$ and need to show that $|\mu|(B) \leq m(B)$.
Since $m(B \cup P) \subseteq m(B)$ and $|\mu|(B) = \mu(B \cap P) - \mu(B \cap N)$, I guess I have to make use of this but I am not sure how to go further.
Any help would be greatly appreciated.   

$\mu ^{+}(B)=\mu(B \cap P) \leq |\mu(B \cap P)| \leq m(B \cap P)$ and $\mu ^{-}(B)=\mu(B \cap N) \leq |\mu(B \cap N)| \leq m(B \cap N)$. Hence $|\mu|(B)=\mu ^{+}(B)+\mu ^{-}(B)\leq m(B \cap P)+m(B \cap N)=m(B)$.
