# Proving the max/min of two finite signed measures

Let $\mu_1$ and $\mu_2$ be finite signed measures on the measurable space $(X, \mathcal A)$. Define signed measures $\mu_1 \vee \mu_2$ and $\mu_1 \wedge \mu_2$ on $(X, \mathcal A)$ by $\mu_1 \vee \mu_2 = \mu_1 + (\mu_2 - \mu_1)^+$ and $\mu_1 \wedge \mu_2 = \mu_1 - (\mu_1 - \mu_2)^+$.

(a) Show that $\mu_1 \vee \mu_2$ is the smallest of those finite signed measures $\nu$ that satisfy $\nu(A) \geq \mu_1(A)$ and $\nu(A) \geq \mu_2(A)$ for all $A \in \mathcal A$.

(b) Find and prove an analogous characterization of $\mu_1 \wedge \mu_2$.

When it comes to doing (a), I let $(P,N)$ be the Hahn decomposition of $X$ with respect to $\mu_2 - \mu_1$. Then $(\mu_1 \vee \mu_2)(A) = \mu_1 (A \cap N) + \mu_2 (A \cap P)$. I have no idea where to go from this point.

We want to show that $\nu(A)\geq \mu_1(A)$ and $\nu(A)\geq \mu_2(A)$ implies $\mu_1 \wedge \mu_2 \leq \nu$. $$\mu_1\wedge \mu_2 = \mu_1+(\mu_2-\mu_1)^+ = \begin{cases} \mu_2 \quad if\quad \mu_2(A) \geq \mu_1(A)\\ \mu_1 \quad if\quad \mu_1(A) \geq \mu_2(A)\\ \end{cases}$$ In both cases, $\mu_1 \wedge \mu_2 \leq \nu$ by assumptions. Note that function is maximum of $\mu_1,\mu_2$.

Definition is $f^+=\max(f,0)$. Hence $(f-g)^+=\max(f-g,0)$.

If at $x$, $\ f(x) \geq g(x)$. Then $(f-g)^+=f-g$.

If at $x$, $\ f(x) \leq g(x)$. Then $(f-g)^+ = 0$.

Now lets turn back to question, and look at the case $\mu_2(A)\geq \mu_1(A)$. $$\mu_1 + (\mu_2 - \mu_1)^+ = \mu_1 + \mu_2 - \mu_1 = \mu_2$$ Other case is similar.

• I guess my concern is: how do I show the second equality?
– Bib
Dec 10 '17 at 9:00
• I edited my answer, can you please check it. @Bib Dec 10 '17 at 9:18
• How do you get that $\mu^+=\max(\mu,0)$? In case it wasn't clear, my question was written in terms of the Jordan Decomposition: $\mu = \mu^+ - \mu^-$.
– Bib
Dec 10 '17 at 10:25
• Ah my bad, did not read carefully. $(\mu_1 \wedge \mu_2)(A) = \mu_1 (A \cap N) + \mu_2 (A \cap P) \leq \nu(A\cap N) + \nu(A\cap P) = \nu(A)$. Is that what you are looking for? Dec 10 '17 at 11:15