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Suppose $\mu$ and $\nu$ are two finite measures on $(S,\Sigma)$ and $\mu \wedge\nu =\mu - (\mu-\nu)^+$. I am trying to prove $\mu \wedge \nu$ is the largest signed measure that is less than or equal to both $\mu$ and $\nu$.

What I tried:

Let $W$ be a set of signed measures on $(S,\Sigma) $ such that for any $w \in W$, we have $w\leq \mu$ and $w \leq \nu$. I want to prove that $\sup W = \mu \wedge \nu$.

Is it good start? Is it a good idea to use Sup to solve this? And, I would really appreciate if you could give me some guide.

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  • $\begingroup$ Set-wise supremum of measures is not a measure except in trivial cases. So you cannot use sup. $\endgroup$ Jan 8, 2020 at 23:57

1 Answer 1

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Hint:

note that, from the definitions, we have

$\mu \wedge\nu(E) =\mu(E) - (\mu-\nu)^+(E)=\nu(E)$ if $\mu(E)\ge \nu(E)$

and

$\mu \wedge\nu(E) =\mu(E) - (\mu-\nu)^+(E)=\mu(E)$ if $\mu(E)\le \nu(E)$

Now let $\tau$ be any other signed measure such that $\tau\le \mu$ and $\tau\le \nu$.

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