What are the differences between the Lebesgue measure on the Hilbert cube $[0,1]^\mathbb{N}$ and the standard Lebesgue measure on $[0,1]^n$? There are no Lebesgue measure on infinite dimensional Banach space. However, there is a Lebesgue measure on the Hilbert cube $[0,1]^\mathbb{N}$. What are the differences between this measure and the finite-dimensional Lebesgue measure on $[0,1]^n$? 
 A: Here is a difference: consider the set $D_t=\{x\in S\mid m(x)\ \text{exists},\ \frac12-t\leqslant m(x)\leqslant\frac12+t\}$ where $S=[0,1]^n$ or $[0,1]^\mathbb N$, $m(x)=\frac1n\sum\limits_{k=1}^nx_k$ if $x$ is in $[0,1]^n$, and $m(x)=\lim\limits_{n\to\infty}\frac1n\sum\limits_{k=1}^nx_k$ if $x$ is in $[0,1]^\mathbb N$ and the limit exists. Then, $\mathrm{Leb}(D_t)\to0$ when $t\to0$ if $S=[0,1]^n$, while $\mathrm{Leb}(D_t)=1$ for every positive $t$ if $S=[0,1]^\mathbb N$.
A: Corollary 0.8 in Folland's Intro to Partial Differential Equations is that the volume of the unit ball in $\mathbb{R}^n$ is given by
$$\omega_n = \frac{2\pi^{n/2}}{\Gamma(n/2)},$$
which tends to 0 as $n\to\infty$. This implies that the volume of any $\mathbb{N}$-sphere will be zero, which probably has some weird implications, e.g. you cannot cover any set of positive measure with countably many $\mathbb{N}$-spheres.
A: Lebesgue measure in $[0,1]^{N}$ is the product of infinitely many copies of linear Lebesgue measures in  $[0,1]$ unlike the Lebesgue measure in $[0,1]^n$ which is the product of  $n$ copies of linear Lebesgue measures in  $[0,1]$.     
