I have some troubles with a problem about measures. This is the last missing subproblem of a bigger one, so I would be glad, if some of you can help me to finish this task.
Consider two measures $\mu, \nu$ and show, that there exists a greatest lower bound of these measures.
So according to the definition, a lower bound of two measures is another measure $\tau$ with $\tau \leq \mu$ and $\tau \leq \nu$, that holds $\forall \phi: \phi \leq \mu$ and $\phi \leq \nu \Rightarrow \phi \leq \tau$.
Due to our current topic is the Radon-Nikodym-Theorem my first idea was to find a proof via Lebesgue-decomposition, but I don't know anything about the absolut continuity of $\mu$ and $\nu$, so this doesn't seem to be the way to go.
Furthermore already tried to proof this via contraposition, but it also failed, so I'm slightly running out of ideas, what to try next.
Thank you very much!
EDIT: Following @Federico's suggestion:
Define $\lambda := \mu + \nu$. Then it holds $\mu, \nu \ll \lambda$. Using Radon-Nikodym-Theorem we get densities $m,n$ such that $\mu(A) = \int_A m d\lambda$ and $\nu(A) = \int_A n d\lambda$.
Now we define $\tau:=\inf(m,n) d\lambda$ and get $\tau \leq \mu$ because $\int_A \inf(m,n) d\lambda \leq \int_A m d\lambda$ and the same for $\nu$.
So we need to show, that $\forall \phi \leq \mu,~ \phi \leq \nu$ it holds $\phi \leq \tau$.
Due to $\phi$ is a measure and using the assumption, we get $0 \leq \phi \leq \mu, \nu$ which leads us to $\phi \ll \mu, \nu \ll \lambda$, so again using Radon-Nikodym, we get $\phi(A) = \int_A p d\lambda$.
So $\phi(A) = \int_A p d\lambda \leq \int_A m d\lambda$ and $\phi(A) = \int_A p d\lambda \leq \int_A m d\lambda$.
This holds for arbitrary $A$, so we conclude $\forall x \in \Omega: p \leq n$ and $p \leq m$.
So if $m$ or $n$ is the infimum, we got the statement. But what happens, if not? Can I conclude anything in this case?
Sketch (assuming $\mu$ and $\nu$ to be $\sigma$-finite.)
Define $\lambda = \mu+\nu$. Then $\mu \ll \lambda$ and $\nu \ll \lambda$. By Radon-Nikodym, there are measurable functions $m$ and $n$ such that $\mu=m\lambda$ and $\nu = n\lambda$.
Now define $\tau=(m\wedge n)\lambda$ and verify that indeed $\tau=\mu\wedge\nu$.