(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 $\mathbb{R}^\mathbb{R}$?
 A: 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 $0$.
A: 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 $\mathbb{R}^n$.
A: 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.
A: 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.
