Nonexistence of infinite-dimensional Lebesgue measure

Remarks

By clause 1 and clause 2, some open ball $B(0, r)$ with $r > 0$ has finite measure. By Riesz's lemma, there are infinitely many points $x_1, x_2, \ldots$ with $\norm{x_i} = 3r/4$ and $\norm{x_i - x_j} \ge r/2$ for $i \ne j$. The balls $B(x_i, r/4)$ are disjoint, lie in $B(0, r)$, and all have the same measure by translation invariance; so this measure is $0$. By separability, $X$ is a countable union of balls of radius $r/4$, and so $\mu(X) = 0$.

The result is often stated in an equivalent way: a translation-invariant Borel measure that is not identically zero gives infinite measure to every nonempty open set. In finite dimension the conclusion fails, because Lebesgue measure on $\R^n$ is translation-invariant and locally finite. Gaussian measures on infinite-dimensional spaces avoid the theorem by giving up translation invariance.

Going deeper

For a student

The theorem explains why integration over spaces of functions, as in probability theory and quantum field theory, uses other measures. The standard choices are Gaussian measures, such as Wiener measure on paths. Their behaviour under translations is described by the Cameron–Martin theorem: only translations along a small subspace keep the measure class.