Sup and inf of bounded variations

**tttcomrader** · Dec 12th 2009, 06:05 PM

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

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

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

Proof so far.

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

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

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

Thank you.

Thread: Sup and inf of bounded variations

LinkBack

Thread Tools

Display

Sup and inf of bounded variations

Similar Math Help Forum Discussions

Closure of a totaly bounded set is totally bounded

If f is a uniformly continuous function on a bounded set S, then f is a bounded on S

Any bounded seq. has a converging subseq. <=> every bounded monotone seq. converges

bounded variations

please help! variations

Search Tags