I was reading about use of ultrafilters in general topology and found a document by Dror Bar-Natan. It contained this problem, so I embarked on solving it. In my opinion, it's rather interesting.
Let $\mathcal{F}$ be a non--principle ultrafilter on $\mathbb{N}$. Define $A = \left\{ \sum_{n \in F} 2^{-n} | F \in \mathcal{F} \right\}$. Here $\lambda$ denotes Lebesgue measure on [0,1].
The problem is to identify if $A$ is Lebesgue measurable.
My attempt:
(1)As $\mathcal{F}$ is a filter it must be true that $\forall F,G \in \mathcal{F} \; . \; F \cap G \neq 0$, so it must be the case that $\forall x \in A \; . \; 1 - x \in A^\complement$, at least, then $x$ is not dyadic . Also, as ultrafilters are maximal,actually , $A^\complement = (1 - A) \setminus Z_2$, where $Z_2$ is countable set of dyadic non-zero numbers in $[0,1]$. The function $1-x$ preserves Lebesgue measure and $\lambda(Z_2) = 0$, so, if $A$ were measurable, then $\lambda(A) = \lambda(A^\complement)$. As
$$ 1 = \lambda[0,1] = \lambda(A \cup A^\complement) = \lambda(A) + \lambda(A^\complement) =2\lambda(A), $$
it follows, that $\lambda(A) = 1/2$ . From now on, assume that $A$ is measurable.
(2) being non-principal means for $\mathcal{F}$ that $\forall n \in \mathbb{N} . \exists F \in \mathcal{F} : n \not \in F$. This implies that both $A$ and $A^\complement$ are everywhere dense in $[0,1]$. Moreover, if two numbers are different in a finite number of digits in their binary expansion, they must be either in $A$ or $A^\complement$ together. Whence, repeated reasoning from (1) with function $b + a - x $ shows that for any dyadic half-intervals $(a,b]$ ( or $[a,b),[a,b],(a,b)$ ) we have $\lambda(A \cap (a,b]) = \lambda(A^\complement \cap (a,b]) = \frac{1}{2}\lambda (a,b]$.
(3) Define auxiliary measures $\mu$ and $\mu'$ by $\mu(X) = \lambda(A \cap X)$ and $\mu'(X) = \lambda(A^\complement \cap X)$. Note, that dyadic intervals as above generate Borel $\sigma$-algebra on $[0,1]$ . So by (2) $\mu = \mu' = \frac{1}{2} \lambda$ . However, by construction $\mu(A) = 1/2$ and $\mu'(A) = 0$, a contradiction! Thus, $A$ is not measurable$\square$
If I'm wrong here you are welcome to bash me.
If my result is correct, then it must be impossible to construct such non-principal ultrafilters without some weaker form of choice, as existence of non-measurable sets is equivalent to it.