How do I show that $\mu \wedge \nu \leq \mu, \nu$?

Question

I'm using Royden & Fitzpatrick for real analysis, and one of the problems defines (for signed measures $\mu$ and $\nu$) $\mu \wedge \nu = \frac{1}{2}(\mu + \nu - |\mu - \nu|)$ and $\mu \vee \nu = \mu + \nu - \mu \wedge \nu$, and requires us to show that $\mu \wedge \nu \leq \mu, \nu$ and that if $\eta$ is any other signed measure smaller than $\mu, \nu$, then $\eta \leq \mu \wedge \nu$. I've tried everything I can think of, but the best I've been able to come up with is $\mu \wedge \nu \leq |\nu|, |\mu|$ with the reverse triangle inequality. Can anyone point me in the right direction?

Answer 1

Let $A$ be an element of the corresponding $\sigma$-algebra $\Sigma$. Then for any $\varepsilon > 0$ we find disjoint $A_1,\ldots,A_n\subset A$ in $\Sigma$ such that $$ |\mu-\nu|(A) = \sum_{j=1}^n|\mu(A_j)-\nu(A_j)| + \varepsilon', $$ where $\varepsilon'\le\varepsilon$. Hence, \begin{align*} (\mu\wedge\nu)(A) &= \sum_{j=1}^n\big[\mu(A_j) + \nu(A_j) - |\mu(A_j)-\nu(A_j)|\big] - \varepsilon'\\ &= \sum_{j=1}^n\min\{\mu(A_j),\nu(A_j)\} - \varepsilon'. \end{align*} From this everything follows.