Measurable sets and Valuation Theory

2012-12-30T12:04:18Z

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 ?

Answer by Gerald Edgar for Measurable sets and Valuation Theory

2012-12-30T13:25:15Z

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.

Measurable sets and Valuation Theory - MathOverflow

Measurable sets and Valuation Theory

Answer by Gerald Edgar for Measurable sets and Valuation Theory