Suppose that $\mu$ is a signed measure on a $\sigma$-algerba of the subsets of X. Show that:
$\frac{1}{2}\lvert{\mu}\rvert(X) \leq \sup[{\vert{\mu(E)}\rvert: E \in \sigma-algebra}]$
also 1/2 is the biggest constant that the ineqality holds.
I'd used the defenition of total varaition but I don't know where this constant comes from.The definition of total varaition is about the $\Sigma\lvert{\mu(E)}\rvert$,but how this can be related to $\sup[\vert{\mu(E)}\rvert: E \in \sigma-algebra]$.
any help would be greatly appreciated.
By Hahn Decomposition Theorem we can write $\mu (E) =\mu^{+}(E)-\mu^{-}(E)=\mu (E \cap A)+\mu (E\cap A^{c})$ and $|\mu| (E) =\mu^{+}(E)-\mu^{-}(E)=\mu (E \cap A)-\mu (E\cap A^{c})$ for some measurable set $A$. Hence, $|\mu|(X) \leq |\mu(X\cap A)|+|\mu(X\cap A^{c})|\leq 2 \sup\{|\mu (F)|: F \in \mathcal F\}$.
Take $\mu (E)=\lambda (E \cap (-1,0))-\lambda (E \cap (0,1))$ for subsets of $(-1,1)$ to get an example where equality holds. Here $\lambda$ is the Lebesgue measure.
See https://en.wikipedia.org/wiki/Hahn_decomposition_theorem for Hahn Decomposition Theorem.