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On $\mathbb{R}^n$, we of course have the usual Lebesgue meausre. In many ways, separable, infinite-dimesional Hilbert space is the most natural generalization of $\mathbb{R}^n$ to infinite-dimensions, so it is natural to ask, does there exist a Lebesgue-like measure on separable, infinite dimensional Hilbert space? For the sake of concreteness, is there a natural, Lebesgue-like measure on $\ell ^2$? (For this purposes of this question, I don't believe it should make a difference whether we are working over $\mathbb{R}$ or $\mathbb{C}$.)

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What do you mean by "natural, Lebesgue-like"? Translation-invariant Borel measure? If so, then there's a theorem of Mackey and Weil telling you that if such a measure exists on a separable completely metrizable group then the group must be locally compact. So if you want such a measure to exist on a Hilbert space, it must be finite-dimensional. –  t.b. Oct 26 '11 at 1:53
    
@t.b. Yes, at the very least it should a translation-invariant Borel measure. This is essentially a solution then. Can you point me to reference I can find this theorem in? –  Jonathan Gleason Oct 26 '11 at 1:57
    
Maybe this thread here and this thread on MO contain further pointers of interest. Edit: I didn't see your comment before posting this one. I'll look for a reference. –  t.b. Oct 26 '11 at 1:57
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Lebesgue measure is not only translation-invariant, it satisfies $\mu(\lambda A) = \lambda^n \mu(A)$ for real $\lambda \ge 0$, and I don't see a reasonable way to generalize this to an infinite-dimensional space. –  Qiaochu Yuan Oct 26 '11 at 2:08
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Okay, it's always a bit hard to find a reference for these folklore results. The only essentially self-contained one I know of is Theorem 5.41 in Varadarajan's Geometry of quantum theory, Springer, 1973. –  t.b. Oct 26 '11 at 2:38

4 Answers 4

up vote 14 down vote accepted

As clarified, you are looking for translation-invariant Borel measures. Here are two: the zero measure, and counting measure. Obviously those are not going to satisfy you, but you can't really do better.

Theorem. A translation-invariant Borel measure on an infinite-dimensional separable Banach space is either the zero measure, or assigns infinite measure to every open set.

You can find a proof on Wikipedia, or in Theorem 1.1 of these notes I wrote.

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Is the parenthetical in Exercise 1.2 meant to be easily answerable or just thought provoking? :-) Finding a counterexample may not be entirely straight forward. –  Willie Wong Jul 6 '12 at 15:45
    
@WillieWong: It was more a question for myself. I don't know the answer, but I wish I did. –  Nate Eldredge Jul 6 '12 at 16:51
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dunno if you've seen this. The answer to your question is apparently independent of ZF + DC! –  Willie Wong Jul 7 '12 at 15:43
    
Theorem presented by Nate Eldredge admits the following extension: {\bf Theorem.} Any Borel measure on an infinite-dimensional separable Banach space, which is invariant with respect to everywhere dense vector subspace, is either the zero measure, or assigns infinite measure to every open set. –  Gogi Pantsulaia Mar 20 at 17:39

The following facts are valid:

Fact 1. There exists a sigma-finite Borel measure in $\ell_2$ which is invariant under the group of all eventually zero sequences and takes the value $1$ on the Hilbert cube $\prod_{k \in N}[0,\frac{1}{k}]$.

The proof of Fact 1 can be found in [Kharazishvili A.B., On invariant measures in the Hilbert space.Bull. Acad. Sci.Georgian SSR, 114(1) (1984),41--48 (in Russian)].

Fact 2. There exists a translation-invariant Borel measure $\mu$ in $\ell_2$ which takes the value $1$ on the Hilbert cube $\prod_{k \in N}[0,\frac{1}{k}]$.

The proof of Fact 2 can be obtained by Baker measure $\lambda$ [Baker R., ``Lebesgue measure" on $\mathbb{R}^{ \infty}$. II. \textit{Proc. Amer. Math. Soc.} vol. 132, no. 9, 2003, pp. 2577--2591] as follows:

Let $T:\ell_2 \to \mathbb{R}^{ \infty}$ be defined by $T((x_k)_{k \in N})=(k x_k)_{k \in N}$ for $(x_k)_{k \in N} \in \ell_2$. For each Borel subset $X \subseteq \ell_2$ we set $\mu(X)=\lambda(T(X))$. Then $\mu$ satisfies all conditions of Fact 2.

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Welcome to Math.SE! Thank you for your contribution. –  user53153 Jan 2 '13 at 20:57

Some authors take local compactness of the space to be part of the definition of Borel measure, so that leaves out infinite-dimensional Hilbert spaces right away. I think the Mackey-Weil result is talking about $\sigma$-finite measures. If you don't require that, you might consider $r$-dimensional Hausdorff measure for any nonnegative real $r$. These are translation-invariant measures, and all Borel sets are measurable.

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I have the following questions: Question1. Does there exists an r-dimensional Hausdorf measure which takes the value 1 on Hilbert cube for some non-negative $r$? Question 2. Please, indicate a Borel set $F_r \subseteq l_2$ on which $r$-dimensional Hausdorf measure takes the vaslue 1? –  Gogi Pantsulaia Jan 3 '13 at 13:48
    
I have the following questions: Question1. Does there exists an r-dimensional Hausdorf measure which takes the value 1 on Hilbert cube for some non-negative $r$? Question 2. Please, indicate a Borel set $F_r \subseteq l_2$ on which $r$-dimensional Hausdorf measure takes the value 1? –  Gogi Pantsulaia Jan 3 '13 at 13:57
    
1) No, the $r$-dimensional Hausdorff measure of any set of dimension greater than $r$ is infinite. The Hilbert cube has $d$-dimensional subspaces for every finite $d$. 2) If $r$ is a positive integer, $F_r = \{x: 0 \le x_j \le 1 \text{ for }1 \le j \le r, x_j = 0 \text{ otherwise}\}$. –  Robert Israel Jan 4 '13 at 0:55
    
Thank you for your answers. Are you agree with me that a translation invariant Borel measure in $\ell_2$ which takes a numerical value $1$ on the Hilbert cube is not Hausdorf measure? Am I correct in my estimate? –  Gogi Pantsulaia Jan 4 '13 at 17:10
    
Yes, of course. Note, btw, that $\ell_2$ contains uncountably many disjoint translations of the Hilbert cube, so your measure is not $\sigma$-finite. –  Robert Israel Jan 4 '13 at 19:12

Below I present a new construction of translation-invariant measures in a separable Banach space $B$ with the Schauder basis $(e_k)_{k \in N}$. We say that a Borel measure $\mu$ in $B$ is generator of shy sets(equivalently, Haar null sets) if the condition $\mu(X)=0$ implies that $X$ is shy(equivalently, Haar null).

Definition 1. A universal measurable set $S$ in a separable Banach space $B$ is said to be an $n$-dimensional Preiss-Ti$\check{s}$er null set if every Lebesgue measure $\mu$ concentrated on any $n$-dimensional vector space $\Gamma$ is transverse to $S$.

We denote the class of all n-dimensional Preiss-Ti$\check{s}$er null sets in $B$ by $$\mathcal{P~T~N}(B, n).$$

Let $(\Gamma_i)_{i \in I}$ be a family of all $n$-dimensional vector spaces and let $\mu_i$ be an $n$-dimensional Lebesgue measure concentrated on $\Gamma_i$ for $i \in I.$

Let $\Gamma_i^{\perp}$ be a linear complement of the vector space $\Gamma_i$ for $i \in I$. We put $$ (\forall X)(X \in \mathcal{B}(B) \rightarrow G_{P~\&~T}^{(n)}(X)=\sum_{i \in I }\sum_{g \in \Gamma_i^{\perp}}\mu_i(X-g \cap \Gamma_i)). $$

Theorem 1. A functional $G_{P~\&~T}^{(n)}$ is a translation-invariant quasi-finite generator of shy sets in $B$ such that $$\mathcal{P~T~N}(B, n)=\mathcal{N}(\overline{G_{P~\&~T}^{(n)}}).$$

Remark 1. The proof of Theorem 1 can be found in [G.Pantsulaia , On generators of shy sets on Polish topological vector spaces, New York J. Math.,14 ( 2008) , 235 – 261].

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