Lebesgue measurability of a set $A = \{\sigma_i \in B\ /\ {2^i}, B \in F\}$ for a a non-principal ultrafilter $F$ 
Problem Let $F$ be a non-principal ultrafilter on natural number
  set $\mathbb N$. Determine if the set $A = \{\sigma_i \in  B\ /\ {2^i}, B \in
 F\}$ is Lebesgue measurable and if it is measurable, determine its
  Lebesgue measure.

This is a problem in the lecture download from internet, and now I guess it is not measurable, but my friends think it may be measurable ,and then the measure is $\frac12$.
I don't know how to prove if it is measurable or not.
 A: HINT: In line with my comment, I’m going to use $\mathscr{F}$ for the ultrafilter and assume that $A$ is supposed to be
$$A=\left\{\sum_{n\in F}\frac1{2^n}:F\in\mathscr{F}\right\}\;.$$
This makes it likely that $\Bbb N$ here is $\Bbb Z^+$, so that $A\subseteq[0,1]$, and I will assume as much.


*

*Show that since $\mathscr{F}$ is non-principal and therefore contains the cofinite filter, every dyadic rational is in $A$. 


Let $D$ be the set of dyadic rationals.


*

*Show that for each $x\in[0,1]\setminus D$, $x\in A$ iff $1-x\notin A$.  

*Conclude that if $A$ is measurable, then $\mu(A)=\frac12$.  

*Adapt this argument to show that $\mu\left(A\cap\left[0,\frac12\right]\right)=\frac14$. Remember that if $F\in\mathscr{F}$, and $F'$ differs from $F$ by a finite set, then $F'\in\mathscr{F}$.  

*Further generalize it to show that if $p,q\in D\cap[0,1]$, with $p<q$, then $\mu(A\cap[p,q])=\frac12(q-p)$.


Then use the Lebesgue density theorem or this question and answer to get a contradiction.
