Non-principal Ultrafilter on Measurable Subsets of $\mathbb R$

Question

Let $\Omega$ be the set of all measurable subsets of $\mathbb R$ ordered by inclusion. Is it possible to construct a non-principal ultrafilter on $\Omega$?

Answer 1

Take a maximal filter generated by all complements of null sets in the collection of all Lebesgue measurable sets, by Zorn.

Answer 2

I think that it is possible. Let $U$ be a non-principal ultrafilter on $\mathcal{P}(\mathbb{N})$. Define $W := \{ A \in \Omega \mid \exists X \in U,\ X\subset A\}$. Then, $W$ is an ultrafilter on $\Omega$: If $A,B \in W$, then there are $X_A , X_B \in U$ such that $X_A \subset A$ and $X_B \subset B$. Thus $X_A \cap X_B \subset A \cap B$. If $A \in \Omega \setminus W$, then $A \cap \mathbb{N} \notin U$ so that $\mathbb{N}\setminus A \in U$. Since $\mathbb{N} \setminus A \subset \mathbb{R} \setminus A $, we have $\mathbb{R} \setminus A \in W$.

Answer 3

$\Omega$ contains the cofinite topology $\tau$ on $\mathbb R$, which $\tau$ can then be extended to a ultrafilter $U$ on $\mathbb R$ by the ultrafilter lemma. $U$ is non-principal as for any $a \in \mathbb R$, $I \setminus a \in \tau$. I believe $U \cap \Omega$ is then the desired non-principal ultrafilter.