Thread: Sup and inf of bounded variations

Suppose that v is a signed measure on $\displaystyle (X, \mathbb {M} ) $ and $\displaystyle E \subset \mathbb {M} $
Prove that:
a) $\displaystyle v^+(E) = sup \{ v(F) : F \in \mathbb {M} , F \subset E \} $

b) $\displaystyle v^-(E) = -inf \{ v(F) : F \in \mathbb {M} , F \subset E \} $

c) $\displaystyle \mid v \mid (E) = sup \{ \sum _1 ^n \mid v(E_j) \mid : n \in \mathbb {N} , E_1,...,E_n \ disjoint, \ \bigcup _1 ^n E_j = E \} $

Proof so far.

a) Now $\displaystyle v(E) = v^+(E)-v^-(E) $, so we have $\displaystyle v(E) \leq v^+(E) \ \ \ \forall E $
Since $\displaystyle v^+ \bot v^- $, we have $\displaystyle X = P \cup N $, P and N are disjoint, and N is $\displaystyle v^+ $-null, P is $\displaystyle v^-$-null.

If I pick $\displaystyle F \subset P $, then I will have $\displaystyle v(F)=v^+(F)-v^-(F)=v^+(F)$.

And that proves (a). (b) should be similar I think.

But I'm a bit lost on (c), mainly because of the sums there.

If I rewrite $\displaystyle \mid v \mid (E) = v^+(E)+v^-(E)=$$\displaystyle sup \{ v(F) : F \in \mathbb {M} , F \subset E \} -inf \{ v(F) : F \in \mathbb {M} , F \subset E \} $, will it help?

Thank you.