(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 $\mathbb{R}^\mathbb{R}$?
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 $\mathbb{R}^n$.
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.