Prove that $\mu \cup \nu $ is the largest signed measure that is less than or equal to both $\mu,\nu$.

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I am learning Signed Measures as a Part of my PhD coursework and I have got this exercise after just learning 1 page.

I dont know how to proceed.

Let $\mu ,\nu $ ne two finite measures on the measurable space $(S,\sum)$.

Define $\mu \cup \nu =\mu+(\nu-\mu)^+$ and $\mu \cap \nu =\mu -(\mu-\nu)^+$

Prove that $\mu \cup \nu $ is the largest signed measure that is less than or equal to both $\mu,\nu$.

How to show that $\mu \cup \nu $ is the largest signed measure?What is meant by that?

Should I take a measurable set $M$ and show that $\mu \cup \nu(M)$ is greater than all other measures?

Any help?Any resources can help me.

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The signed measures take values in $[-\infty,+\infty]$. That set has a linear order $\le$.

There then is a standard partial order on the signed measures on $(S,\Sigma)$:

$\mu_1 \le \mu_2$ iff $\forall A \in \Sigma: \mu_1(A) \le mu_2(A)$.

The largest signed measure that is $\le \mu, \nu$ is the measure $\lambda$ that obeys 2 properties:

$\lambda \le \mu$ and $\lambda \le \nu$ and for all other signed measures $\lambda'$ with $\lambda' \le \mu$ and $\lambda' \le \nu$ we have $\lambda' \ge \lambda$.

This is the standard definition of the infimum of two elements in a partial ordered set, and so I think you meant it to be for $\mu \cap \nu$, or more standard $\mu \land \nu$.

$\mu \cup \nu$ (really $\mu \lor \nu$) is the smallest signed measure $\ge$ both $\mu$ and $\nu$, the supremum.