How to formalize notion of "maximally dominated/dominating" probability? I want to choose a probability measure on the measurable space $(\Omega, \mathcal{F})$ that dominates, or is dominated by, "as many probability measures as possible." The question is how to formalize the phrase that I placed in scare quotes. 
I will give some simple examples to help show what I have in mind. ($P$ dominates $Q$ means $P(A)=0$ implies $Q(A)=0$, and we write $Q \ll P$.)
"Maximally dominating" probabilities. Suppose $(\Omega, \mathcal{F})$ is countable and assume $\mathcal{F}$ is the powerset of $\Omega$. If $P$ is a probability that assigns the singletons in $\mathcal{F}$ positive measure, then $P$ dominates every probability $Q$ on $(\Omega, \mathcal{F})$. So probabilities that assign positive measure to singletons are "maximally" dominating, they dominate "as many probabilities as possible."
Of course, for general $(\Omega, \mathcal{F})$ this won't work. Instead, I thought we could define a probability $P$ to be maximally dominating if there is no $Q$ such that 
$$\{\mu: \mu \ll Q \} \supset \{\mu: \mu \ll P \}.$$ 
This definition seems to pass some simple sanity checks. For example, point masses $\delta_\omega$ are not maximally dominating on this definition since $\{\mu: \mu \ll \delta_\omega \} = \{ \delta_\omega\}$, but any $Q$ with $Q(\{\omega \})>0$ dominates $\delta_\omega$ and many other probabilities as well. But I am unsure how to proceed from here; I don't have an example of a maximally dominating measure.
"Maximally dominated" probabilities. In this case, I don't really have any ideas. Even for countable $(\Omega, \mathcal{F})$, we cannot hope to do what we did above: there's no probability that's dominated by every other probability (if we didn't require our measures to be probabilities, though, then the $0$ measure would be maximally dominated.)
To sum up,

Are there any known formalisms for the notions I'm trying to capture?

 A: You can construct a poset of (equivalence classes of) probability measures.  For $(\Omega,\mathcal{F})$, let $P(\Omega,\mathcal{F})$ be the set of probability measures on $(\Omega, \mathcal{F})$.  For two measures $\mu, \nu \in (\Omega,\mathcal{F})$, say that $\mu \sim \nu$ if $\mu \ll \nu$ and $\nu \ll \mu$.  Then we can consider $P(\Omega,\mathcal{F})/\sim$ as a partially ordered set with partial ordering $\ll$.  Then a "maximally dominating" measure is simply one that is maximal in this poset, and a "maximally dominated" measure is one that is minimal in this poset.  For instance, if $\Omega = \mathbb{Z}$ and $\mathcal{F} = 2^\mathbb{Z}$, then any probability measure supported on the entire set is maximal.
A: This is an addendum to Marcus M's answer. I will give a characterization of maximal/minimal measures with respect to the partial order induced by $\ll$.

Let's say a measure is strictly positive if its only null set is the empty set.
Then the following is true:

Let $(\Omega,\mathcal{F})$ be a measurable space. Then a probability measure on $(\Omega,\mathcal{F})$ is maximal if and only if it is strictly positive.

Proof:


*

*Let $\mu$ be strictly positive. Let $\nu$ be a probability measure such that $\mu \ll \nu$. Then $\nu$ is also strictly positive, so $\nu \sim \mu$. Thus $\mu$ is maximal.

*Assume that $\mu$ is not strictly positive. Then there is some non-empty set $A \in \mathcal{F}$ such that $\mu(A) =0$. Take a point $x \in A$ and take $\nu$ to be the Dirac measure centered at $x$.
Now, consider the probability measure $\rho = \mu/2 + \nu/2$. Then $\mu \ll \rho$, but $\rho$ and $\mu$ are not equivalent since $\rho(A) = 1/2$, while $\mu(A) =0$. Therefore, we have shown that $\mu$ is not maximal.

Now let's characterize the minimal measures. As suggested by Aduh in the comments, those are the measures that take only the values $0$ and $1$:

Let $(\Omega, \mathcal{F})$ be a measurable space. Then
   a measure on $(\Omega, \mathcal{F})$ is minimal if and only if it is $0-1$ valued.

Proof:


*

*$0-1$ valued $\Rightarrow$ minimal.


Assume that $\mu$ is not minimal. This means there exists a probability measure $\nu$ with $\nu \ll \mu$ and $\nu \nsim \mu$. Thus there is a measurable set $F$ such that $\nu(F) =0$ and $\mu(F) > 0$. Then $\nu(\Omega \setminus F) =1$ since $\nu$ is a probability. This implies that $\mu(\Omega \setminus F) > 0$ since $\nu \ll \mu$. Thus we obtain that
$$
\mu(F) + \mu( \Omega \setminus F) = 1
$$
with both $\mu(F)$ and $\mu( \Omega \setminus F)$ strictly positive. Thus $\mu$ cannot be $0-1$ valued.


*Minimal $\Rightarrow$ $0-1$ valued.


Assume $\mu$ is not $0-1$ valued. There is a measurable set $A$ with $0< \mu(A) <1$. The complement $A^c$ also satisfies $0 < \mu(A ^c) <1$. Define the measure $\nu$ to be the conditional probability with respect to $A$:
$$
\nu(F) = \frac{ \mu( F \cap A)}{\mu(A)}
$$
Then $\nu \ll \mu$ since $\mu(F) =0$ implies $\mu(F\cap A) =0$. However, $\nu$ and $\mu$ are not equivalent since $\nu( A^c) =0$ while $\mu(A^c) >0$. Therefore, $\mu$ is not minimal.
