# (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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en.wikipedia.org/wiki/… –  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
@NateEldredge How would you modify this proof if we drop the Banach assumption? –  Thomas E. Nov 22 '12 at 7:56

I will prove it for $\mathbb R^{\mathbb N}$. (A proof for all Banach spaces is given here.)

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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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

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

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

[2] Baker R., "Lebesgue measure" on $\mathbb{R}^{ \infty}$. II. 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 $\mathbb R^{\infty}$, 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, 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, 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

Feynman has done something like this when he defines path integral. It is infinite dimensional measure.

$\int_{\mathbb{R}^\mathbb{N}}d^{\infty}x \mathcal{Dx}e^{i\mathcal{S}[x]}$

Is such beautiful formula but many mathematician have problem justifying? Answer: Is not infinite dimensional measure.

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