Why are there no finitely additive measures on $\ell_\infty$ for which the measure of every ball is positive and finite?

Question

As the question title suggests, why are there no finitely additive measures on $\ell_\infty$ for which the measure of every ball is positive and finite? Here, we do not assume that the measure is translation invariant.

Answer 1

Arguing by contradiction, suppose such a measure, say $\nu$, does exist. The main idea is that we can place continuum many disjoint balls, or I would rather say cubes, inside a ball. Hence, there are infinitely many of them that have measure greater than some constant $\epsilon > 0$.

Let us denote by $B_r(x)$ a ball in $\ell_\infty$ with center $x = (x_i)_{i = 1}^\infty$ and radius $r > 0$. That is,$$B_r(x) = \{y \in \ell_\infty : \|y - x\|_\infty \le r\} = \left\{y = (y_i) : \sup_{i \in \mathbb{N}} |y_i - x_i| \le r\right\}.$$Denote$$Y \subset \ell_\infty, \quad Y := \{y \in \ell_\infty : y_i \in \{-2, 2\}, \text{ }i = 1,\,2, \ldots\}.$$Consider the balls $B_1(y)$, $y \in Y$. Clearly,$$B_1(y_1) \cap B_1(y_2) = \emptyset \text{ for all }y_1,\,y_2 \in Y,\,y_1 \neq y_2.$$Thus, we have a family $B_1(y)$, $y \in Y$ with continuum many disjoint balls, all of them inside $B_3(0)$. Hence, there exists an infinite, in fact with cardinality continuum, set $Y' \subset Y$, such that $\nu(B_1(y)) > \epsilon$, for all $y \in Y'$, for some $\epsilon > 0$.

This means for any $n \in \mathbb{N}$ and $y_1$, $y_2$, $\ldots$, $y_n \in Y'$, we have$$\nu(B_3(0)) > \nu\left(\bigcup_{i = 1}^n B_1(y_i)\right) > n\epsilon,$$which is a contradiction.