The greatest lower bound - Signed Measures Let $\mu,\nu$ be two finite signed measures on a measurable space $(X,\mathcal A )$. The function $\mu\wedge\nu$ is given by$$(\mu\wedge\nu)(E)=\inf\{\mu(A)+\nu(E\setminus A):~A\in\mathcal A, A\subset E\}, ~\forall E\in\mathcal A.$$
Show that:$$\mu\wedge\nu=\frac12(\mu+\nu-|\mu-\nu|),$$
where $|\mu-\nu|$ is the total variation of $\mu-\nu$.
I already show that $\mu\wedge\nu\ge\frac12(\mu+\nu-|\mu-\nu|)$: Indeed, Denotes $\lambda=\frac12(\mu+\nu-|\mu-\nu|)$. Then we have
$$\lambda=\frac12(\mu+\nu-|\mu-\nu|)\le\frac12(\mu+\nu-\mu+\nu|)=\nu.$$
Similarly, $\lambda\le\mu.$
For any $E\in \mathcal A$, fixed $\varepsilon>0$, there exists $A\in\mathcal A, A\subset E$ such that
$$(\mu\wedge\nu)(E)+\varepsilon\ge\mu(A)+\nu(E\setminus A)\ge\lambda(A)+\lambda(E\setminus A)=\lambda(E).$$
This inequality holds for $\varepsilon>0$ arbitrary, thus we get 
$$\mu\wedge\nu\ge\lambda= \frac12(\mu+\nu-|\mu-\nu|).$$
For the conversely inequality, I guess that if $\gamma$ is a signed measure on $(X,\mathcal A)$ such that $\gamma\le\mu$ and $\gamma\le\nu$ then
$$\gamma\le\frac12(\mu+\nu-|\mu-\nu|).$$
But I have no idea to do this. Can someone please help me. thanks
 A: It seems to me that if you've already got the Jordan decomposition, then you should be able to show this directly.
\begin{align}
& \frac{1}{2}\big(\mu(E) + \nu(E) - |\mu - \nu|(E)\big) = \frac{1}{2}\big(\mu(E) + \nu(E) - [(\mu - \nu)^+(E) + (\mu - \nu)^-(E)]\big) \\
&= \frac{1}{2}\big(\mu(E) + \nu(E) - [\sup\{\mu(A) - \nu(A) : \mathcal{A} \ni A \subset E \} - \inf\{\mu(A) - \nu(A) : \mathcal{A} \ni A \subset E \}]\big) \\
&= \frac{1}{2}\big(\mu(E) + \inf\{\nu(A) - \mu(A) : \mathcal{A} \ni A \subset E \} + \nu(E) + \inf\{\mu(A) - \nu(A) : \mathcal{A} \ni A \subset E \}\big) \\
&= \frac{1}{2}\big(\inf\{\mu(E) + \nu(A) - \mu(A) : \mathcal{A} \ni A \subset E \} + \inf\{\nu(E) + \mu(A) - \nu(A) : \mathcal{A} \ni A \subset E \}\big) \\
&= \frac{1}{2}\big(\inf\{\mu(E \setminus A) + \nu(A) : \mathcal{A} \ni A \subset E \} + \inf\{\nu(E \setminus A) + \mu(A) : \mathcal{A} \ni A \subset E \}\big)\\
&=  \frac{1}{2}(\nu \wedge \mu)(E) + \frac{1}{2}(\mu \wedge \nu)(E) = (\mu \wedge \nu)(E).
\end{align}
