Consider the 2adic 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 nonmeasurable ?
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 $AA = A$ 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. 

