A trivial invariant by translations measure

Question

In a course about Gaussian measures in infinite dimensional space, they assert that a set function which is invariant under translation is infinite over open sets of a hilbert space and here is the main proposition:

Proposition:

Let $X$ be a separable Hilbert space. If $\mu: \mathcal{B}(X)\to [0,+\infty[$ is a $\sigma$-additive set function s.t.:

1. $\mu(x+B)=\mu (B)$ for every $x\in X,\ B\in \mathcal{B}(X) $,
2. $\mu(B(0,r))>0$ for every $r>0$, then $\mu(A)=+\infty$ for every open set $A$.

My problem is I didn't get quit the proof:

Assume thet $\mu$ satisfies $i)$ and $ii)$, and let $(e_n)$ be an orthonormal basis in $X$. For any $n\in \mathbb N$ consider the balls $B_n$ with center $2re_n$ and radius $r>0$; they are pairwise disjoint and by assumption they have the same measure, say $\mu(B_n)$ for all $n\in \mathbb N$. disjoint and by assumption they have the same measure, say $\mu (B_n)=m>0$ for all $n\in \mathbb N$.

Then, \begin{equation} \cup_{n=1}^\infty B_n\subset B(0,3r) \Rightarrow \mu(B(0,3r))\geq \sum_{n=1}^\infty \mu(B_n)=\sum_{n=1}^\infty m=+\infty \end{equation} hence $\mu(A)=+{\infty}$ for every open set $A$.

My questions are :

• Why the balls $B_n$ are disjoint ?
• How did they get the inclusion of the union of $B_n$ in $B(0,3r)$?

Thank you for your time.

Answer 1

For both questions, the answer boils down to the triangle inequality. It is more straight forward for the second: if $x\in B(2re_i,r)$, then $d(x,0)\leq d(x,2re_i)+d(2re_i,0)<r+2r=3r$.

For the first question, we first need to calculate the distance between the centers of two circles. Since $d(2re_i,2re_j)=2rd(e_i,e_j)=2r\sqrt{2}$, a point in our space must be at least $2r\sqrt{2}/2=r\sqrt{2}>r$ away from one of $2re_i$ or $2re_j$ (again, by the triangle inequality) and therefore no point can be in the intersection of two of these balls.

We note from the calculation for the first question that we have not chosen the smallest possible distance to place our balls at, and one could replace $2re_i$ with $r\sqrt{2}e_i$ without changing the argument. However, one does not need things to be tight for the argument to work.