TITLE: Infinite-dimensional translation-invariant measure
QUESTION [5 upvotes]: Why is there no translation-invariant measure on an infinite-dimensional Euclidean space? Is there a reasonably short, insightful proof? 
I am interested in an infinite-dimensional space with a definite inner product but not necessarily complete in the corresponding topology. Thus it need not be a Hilbert space.

REPLY [14 votes]: The statement in the question "Why is there no translation-invariant measure on an infinite-dimensional Euclidean space?" is not correct.  
(i) A counting measure defined in infinite-dimensional Euclidean space is  an example of such measure which  is translation-invariant.
(ii)There does not exist a translation-invariant Borel measure  in an infinite-dimensional Euclidean space $\ell_2$ which gets the value 1 on the unit ball.  Indeed, assume the contrary  and let $\mu$ be such a measure. let $(e_k)_{k \in N}$ be a standard basis with $||e_k||=1$ for $k \in N$. Let $B_k$ be an open ball with center at $\frac{e_k}{2}$ and with radius $r$ less than $\frac{\sqrt{2}}{4}$. Then $(B_k)_{k \in N}$ is a family of pairwise disjoint open balls with radius $r$. On the one hand, $\mu$ measure of $B_k$ must be zero because in other case the $\mu$ measure of the unit ball will be $+\infty$. On the other hand, since $\ell_2$ is separable, $\ell_2$ can be covered by countably many translations of $B_1$ which together with an invariance of $\mu$ implies that $\mu$ measure of $\ell_2$ is zero. This is a contradiction and assertion (ii) is proved.    
(iii) There exists  a translation-invariant measure  on an infinite-dimensional Euclidean space $\ell_2$ which gets the value 1 on the parallelepiped $P$ defined by $P=\{x : x \in \ell_2 ~\&~ |<x,e_k>|\le \frac{1}{2^k}\}$.
Let $\lambda$ be infinite-dimensional  Lebesgue measure in $R^{\infty}$ (see, Baker R., ``Lebesgue measure" on~$\mathbb{R}^{\infty}$,Proc. Amer. Math. Soc., vol. 113, no. 4, 1991,
pp.1023--1029).  We set 
$$
(\forall X)(X \in {\cal{B}}(\ell_2) \rightarrow \mu(X)=\lambda(T(X)))
$$ 
where ${\cal{B}}(\ell_2)$  denotes the $\sigma$-algebra of  Borel subsets of $\ell_2$
and the mapping  $T : \ell_2 \to R^{\infty} $ is defined by: $T(\sum_{k \in N}a_ke_k)=(2^{k-1}a_k)_{k \in N}$.
Then  $\mu$ satisfies all conditions participated in (iii).
P.S. There exist many interesting translation-invariant non-sigma finite Borel measures in infinite-dimensional separable Banach spaces(see, for example,  G.Pantsulaia , On generators of shy sets on  Polish topological vector spaces, New  York   J.  Math.,14 ( 2008) ,  235 – 261)