# (Product) Lebesgue measure on infinite dimensional spaces?

I am trying to understand measure construction procedures on infinite-dimensional spaces. Why is it not possible in general to construct Lebesgue measure on $\mathbb{R}^\mathbb{N}$ or $\mathbb{R}^\mathbb{R}$?

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– Samuel Nov 22 '12 at 3:41
Thanks. This is a very nice link! – Learner Nov 22 '12 at 3:44
@Samuel: Note that the proof in the link needs modification, since it is written for Banach spaces, but the idea is the same: any open set contains infinitely many translates of a smaller open set. – Nate Eldredge Nov 22 '12 at 3:55

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@NateEldredge How would you modify this proof if we drop the Banach assumption? – Thomas E. Nov 22 '12 at 7:56

## 3 Answers

I will prove it for $\mathbb R^{\mathbb N}$. A proof for all Banach spaces is given at http://en.wikipedia.org/wiki/There_is_no_infinite-dimensional_Lebesgue_measure#Proof_of_the_theorem .

Consider the cube $(-1,1]^{\mathbb N}$. It can be partitioned into infinitely many translated copies of the cube $(0,1]^{\mathbb N}$, so if we want them all to have the same volume, and the cube $(-1,1]^{\mathbb N}$ to have finite volume, then the volume of each must be $0$. Now, we can cover the entire space $\mathbb R^{\mathbb N}$ with countably many copies of the cube $(0,1]^{\mathbb N}$, so the entire space $\mathbb R^{\mathbb N}$ must have measure $0$, and thus all subsets must also have measure $0$. Thus the only finite translation-invariant complete measure on $\mathbb R^{\mathbb N}$ is the trivial measure $0$.

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First constructions of Lebesgue measure" on~$\mathbb{R}^{\infty}$ can be found in papers:

[1] Baker R., Lebesgue measure" on~$\mathbb{R}^{\infty}$, \textit{Proc. Amer. Math. Soc.}, vol. 113, no. 4, 1991, pp.1023--1029.

[2] Baker R., Lebesgue measure" on $\mathbb{R}^{ \infty}$. II. \textit{Proc. Amer. Math. Soc.} vol. 132, no. 9, 2003, pp. 2577--2591.

Some generalizations of Baker constructions can be found in the following articles:

[3] G.Pantsulaia , On ordinary and Standard Lebesgue Measures on $R^{\infty}$, \textit{Bull. Polish Acad. Sci.} 73(3) (2009), 209-222.

[4] G.Pantsulaia , On a standard product of an arbitrary family of -finite Borel measures with domain in Polish spaces, \textit{Theory Stoch. Process,} vol. 16(32), 2010, no 1, p.84-93.

[5] G.Pantsulaia , On ordinary and standard products of infinite family of $\sigma$ -finite measures and some of their applications. \textit{Acta Math. Sin. (Engl. Ser.)} 27 (2011), no. 3, 477--496

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Thanks for the references! – Learner Dec 29 '12 at 6:21

I am not quite sure what you mean. But the construction of Lebesgue measure on $X$ I learnt is by using the Riesz representation theorem to identify Lebesgue measure as one continuous linear functional over $C(X)$. But this requires the $X$ to be locally compact, and if $X$ is a vector space, the only locally compact ones are $\mathbb{R}^n$.

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The following old article of Oxtoby [Oxtoby J.C., Invariant measures in groups which are not locally compact, Trans. AMS., 60 (1946), 215--237] contains general constructions of left-invariant quasi-finite Borel measures on Polish groups that are not locally compact. This article contains answers to all questions stated above. – Gogi  Pantsulaia Dec 30 '12 at 6:30
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