Derive Outer Measure from $\sigma$-Algebra

Solution 1:

I'll show the answer is "yes" for finite, countably infinite, and some nice uncountable spaces $\Omega$.

Precisely, the pairs $(\Omega,\mathcal{F})$ that I'll show it for are the ones such that there exist some collection $\{E_\alpha\}_\alpha$ of disjoint elements of $\mathcal{F}$ such that any element of $\mathcal{F}$ is a union of a subcollection of $\{E_\alpha\}_\alpha$. It's not too hard to show that any sigma-algebra for a finite or countably infinite $\Omega$ have this property. [Details at the end].

Suppose $\mathcal{F} = \sigma(\{E_\alpha\}_\alpha)$ for some collection of disjoint subsets $E_\alpha$ of $\Omega$. Define $\mu : \mathcal{P}(\Omega) \to [0,\infty]$ via $\mu(A) = \#\{\alpha: A \cap E_\alpha \not = \emptyset\}$. It's easy to see that $\mu$ is indeed an outer measure. We first show any set $F \in \mathcal{F}$ is $\mu$-measurable. Take $F \in \mathcal{F}$ and write $F = \cup_\beta E_{\beta}$ for some subcollection $\{\beta\}$ of $\{\alpha\}$. Take any $A \subseteq \Omega$. Then $\mu(A \cap F) = \#\{\alpha : A \cap F \cap E_\alpha\} = \#\{\beta : A \cap E_\beta \not = \emptyset\}$. And similarly, $\mu(A \cap F^c) = \#\{\alpha \not \in \{\beta\} : A \cap E_\alpha \not = \emptyset\}$. So, $\mu(A \cap F) + \mu(A \cap F^c) = \mu(A)$, as desired. Now take $F \not \in \mathcal{F}$. Then there is some $\alpha$ and $x,y \in E_\alpha$ such that $x \in F$ but $y \not \in F$. Letting $A = \{x,y\}$, we see that $\mu(A) = 1$ while $\mu(A \cap F)+\mu(A \cap F^c) = 1+1 = 2$. Therefore, each $F \not \in \mathcal{F}$ is not $\mu$-measurable.

I'll quickly show that any finite or countably infinite set $\Omega$ has the property described above. For each $x \in \Omega$, let $F_x = \cap \{F \in \mathcal{F} : F \ni x\}$. The crucial part is that $F_x \in \mathcal{F}$ if $\Omega$ is finite or countably infinite. If $\Omega$ is finite, this is obvious. If $\Omega$ is countably infinite, then although the intersection might be taken over an uncountable set, it's still easy to see that $F_x \in \mathcal{F}$, since we only need to intersect a countable number of $F$'s containing $x$ [think about this].

We show that for $x \not = y \in \Omega$ either $F_x = F_y$ or $F_x \cap F_y = \emptyset$. Let $F := F_x \cap F_y$ and suppose $F \not = \emptyset$. If $x \in F$, then by minimality, we must have $F = F_x$, which implies $F_x \subseteq F_y$. Then, unless $F_x = F_y$, we must have $y \not \in F_x$, in which case $F_y \setminus F_x$ is a smaller element of $F$ containing $y$, a contradiction. So, $x \not \in F$. But then $F_x \setminus F_y$ is an element of $\mathcal{F}$ containing $x$ and is a subset of $F_x$; so it must be $F_x$. But this means $F_x \cap F_y = \emptyset$, a contradiction.

Our collection $\{E_\alpha\}_\alpha$ is just $\{F_x : x \in \Omega\}$ (we remove duplicates). Each pair of sets in this collection are disjoint, and for any $A \in \mathcal{F}$, $A = \cup_{x \in A} F_x$.

I'm not sure whether any sigma-algebra has this disjoint collection property. I'll think more about it.

ADDED: I think a necessary and sufficient condition on $(\Omega,\mathcal{F})$ for the existence of such a $\mu$ is that for each $x \in \Omega$, the intersection of all elements of $\mathcal{F}$ containing $x$ lies in $\mathcal{F}$. We already showed this condition is sufficient. To see that it is necessary, it suffices to show that for each $x \in \Omega$ the set $\cap_{\substack{F \in \mathcal{F} \\ F \ni x}} F$ is $\mu$-measurable for any outer measure $\mu$ making each element of $\mathcal{F}$ $\mu$-measurable. The proof of this I think should be along the same lines as the proof that the set of all $\mu$-measurable sets form a sigma-algebra.