There is no infinite dimensional Lebesgue measure

Imagine that there was a countably additive and translation invariant measure, {m}, on a Hilbert Space {\mathcal{H}} such that every ball {B(v, \varepsilon)} has positive measure, finite measure. Because {\mathcal{H}} is infinite dimensional, there exists {\{e_{i} | i = 1, \cdots, \infty\}} such that {\langle e_{i}, e_{j}\rangle = \delta_{ij}}. Note that the points {e_{i}} and {e_{j}} are far apart: {\|e_{i} - e_{j}\|^2 = \|e_{i}\|^2 + \|e_{j}\|^2 = 2}, if {i \neq j}. Now, the ball, {B(0, 2)}, contains the countable collection of disjoint balls: {\{B(e_{i}, \frac{1}{2}) | i = 1,\cdots, \infty\}} (you can see where this is going). Thus, because all balls, {B(e_{i}, \frac{1}{2})} have the same measure (by translation invariance), and {m} is countable additive, it follows that {\infty = \sum_{i=1}^\infty m(B(e_{i}, \frac{1}{2})) < B(0, 2) < \infty}. Thus, we’ve reached a contradiction.

In some cases, it’s desirable to have an analogue of a measure on a Hilbert space {\mathcal{H}}. The most common application is to make statements such as “Property X is true for a.e. function.” One way to do make these statements is through the concept of Prevalence. Another is the definition of a Gaussian like measure called the Abstract Wiener Space.

Note that I wrote this article before realizing that Wikipedia has a similar article.