# Infinite-dimensional translation-invariant measure

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.

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See also problem 18 in Halmos's Hilbert space problem book – t.b. Jul 6 '12 at 12:28
Is the asserted duplicate really a duplicate? It is about Hilbert space, whereas this one might be on $\mathbb R^\omega$ (i.e. the space of functions $\mathbb N\to\mathbb R$ with finite support). The answer to the existing question depends on the space being Banach, which $\mathbb R^\omega$ isn't. – Henning Makholm Jul 6 '12 at 12:39
@Henning: I think so. Nate's answer works just as well in any Hausdorff (separable) topological vector space as he points out in an exercise immediately following theorem 1.1 in his notes. – t.b. Jul 6 '12 at 12:49
@t.b.: Okay -- but the theorem as quoted in the answer did include an explicit requirement that it was a Banach space. I see that the question is tagged (hilbert-spaces), but that tag didn't come from the OP. – Henning Makholm Jul 6 '12 at 12:51
And the same proof works. Any ball of your inner-product space contains infinitely many disjoint balls of the same (smaller) radius which are all translates of each other. – GEdgar Jul 6 '12 at 13:36

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)

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