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Show that $\pi(E)=sup\lbrace \mu(A): A\subseteq E, A\in\mathbf{X} \rbrace$ is a measure on $\mathbf{X}$.

$\mu$ is a charge on $\mathbf{X}$ ($\sigma$-algebra), let $\pi$ be defined for $E\in\mathbf{X}$.

I have difficulty showing the $\sigma$-algebra countable additive property.

Could you please help.

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Found on wikipedia: A charge is a finitely additive, signed measure. What is X, an algebra? a sigma-algebra? It seems to me, if $\mu$ is a positive charge, then $\pi = \mu$ so this will not gain countable additivity for us. –  GEdgar Mar 2 '14 at 13:43
@GEdgar Bartle uses the term for general countably additive finite signed measures in his "The Elements of Integration and Lebesgue Measure". So it should be enough to use the Hahn decomposition theorem. –  Michael Greinecker Mar 3 '14 at 0:11
So, user should specify. I would vote to put it "on hold" until he does, but that is not permitted when there is a bounty. –  GEdgar Mar 3 '14 at 14:41

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