Measurable sets and Valuation Theory Consider the 2-adic valuation on rationals and then extend it to a valuation on the real numbers. Lets call this extension $\phi$. Let $A$ be the set of all points $(x,y)$ in the plane such that $\phi(x) < 1$ and $\phi(y)<1$. Is it true that $A$ is Lebesgue non-measurable ?
 A: edit, January 13 
Write $|\cdot|$ for the extended $2$-adic absolute value.  I am assuming you mean $A = \{ (x,y) : |x|<1, |y|<1\}$, but since you say ``valuation'' maybe that is not right.  Anyway, your set $A$ is simply related to my set $A$, perhaps by taking complements or multiplying by a constant.
Note that $A$ is a group under addition, since $|-x| = |x|$ and $|x+y| \le \max\{|x|,|y|\}$.
Assume (for purposes of contradiction) that $A$ is measurable.
If $A$ has positive measure we get a contradiction:  indeed,
the set $A - A$ contains a neighborhood of zero (for the usual topology).
But $A-A = A$ and $A+A=A$, so $A$ is the whole plane, which is false.
Now consider sets $A_n = 2^{-n}A = \{(2^{-n}x,2^{-n}y): (x,y) \in A\}$.  These sets are also groups under addition.  The map $(x,y) \mapsto 2^{-n}(x,y)$ is an affine bijection, so all sets $A_n$ are Lebesgue measurable.  But also note that multiplication is continuous with respect to $|\cdot|$, and $|2^n| \to 0$, and $A$ is a neighborhood of zero for the $|\cdot|$ topology.  So for any $(x,y) \in \mathbb R^2$ there is $n$ so that $2^n(x,y) \in A$, and that means $(x,y) \in A_n$.  Thus
$$
\bigcup_{n=1}^\infty A_n = \mathbb R^2 .
$$
A union of measurable sets.  So some $A_n$ has positive measure.  Get a contradiction as before.
