There is no infinite-dimensional Lebesgue measure

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In mathematics, it is a theorem that there is no analogue of Lebesgue measure on an infinite-dimensional space. This fact forces mathematicians studying measure theory on infinite dimensional spaces to use other kinds of measures: often, the abstract Wiener space construction is used.

[edit] Motivation

It can be shown that Lebesgue measure $λ n$ on Euclidean space $\mathbb{R}^{n}$ is locally finite, strictly positive and translation-invariant. Explicitly:

every point $x \in \mathbb{R}^{n}$ has an open neighbourhood $N x$ with finite measure $\lambda^{n} (N_{x}) < + \infty$ ;
every non-empty open set $U \subseteq \mathbb{R}^{n}$ has positive measure $λ n (U) > 0$ ;
if $A \subseteq \mathbb{R}^{n}$ is any Lebesgue measurable subset, $T_{h} : \mathbb{R}^{n} \to \mathbb{R}^{n} : x \mapsto x + h$ is the translation map, and $(T h) * (λ n)$ denotes the push forward, then $(T h) * (λ n)(A) = λ n (A)$ .

Geometrically speaking, these three properties make Lebesgue measure very nice to work with. When we consider an infinite dimensional space such as an L^p space or the space of continuous paths in Euclidean space, it would be nice to have a similarly nice measure to work with. Unfortunately, this is not possible.

[edit] Statement of the theorem

Suppose that $H$ is a Hilbert space with $\dim H = + \infty$ . Then the only locally finite translation-invariant measure on $H$ is the trivial measure $\mu \equiv 0$ .

(Many authors assume that $H$ is separable. This assumption simplifies the proof considerably, since it provides a countable basis for $H$ . However, the result holds even without the separability assumption.)

[edit] Proof of the theorem

Suppose that $H$ is separable (see the remark above). Suppose that $μ$ is a locally finite and translation-invariant measure on $H$ ; suppose also that $μ$ is not the trivial measure. Since $μ$ is locally finite, there exists an open set $U$ with $\mu (U) < + \infty$ . Since $U$ is open, there exists an open ball

$B_{r_{0}} (x) \subseteq U$ for some $x \in H$ ,

r 0 > 0.

Clearly, this ball also has finite measure. By translation,

$\mu \left( B_{r} (y) \right) < + \infty$ for every $y \in H$ and $r \leq r_{0}.$

Fix any radius $0 < r < r 0$ and form the open cover

$H = \bigcup_{y \in H} B_{r} (y).$

Since $H$ is separable, this open cover admits a countable subcover:

$H = \bigcup_{j = 1}^{\infty} B_{r} (y_{j})$ for some sequence $(y_{j})_{j = 1}^{\infty} \subset H.$

Since $μ$ is not the trivial measure, $μ(H) > 0$ , so $μ(B r (y j)) > 0$ for some $j$ , and so $μ(B r (y)) > 0$ for all $y \in H$ and $r > 0$ . Now set $c := \mu (B_{r_{0} / 30} (y))$ for any (i.e. all) $y \in H$ . Observe that if $(e_{i})_{i = 1}^{\infty} \subset H$ is a countable orthonormal basis for $H$ , then

$B_{r_{0} / 30} (e_{i} / 2) \subseteq B_{r_{0}} (0)$ for all

i .

By Pythagoras' theorem, these balls $B_{r_{0} / 30} (e_{i} / 2)$ are pairwise (and hence collectively) disjoint. Now

$\mu \left( B_{r_{0}} (0) \right) \geq \sum_{j = 1}^{\infty} c = + \infty$ unless $\mu \equiv 0.$

But $\mu \not\equiv 0$ implies that $\mu \left( B_{r_{0}} (0) \right) < + \infty$ by local finiteness. This is a contradiction, and completes the proof.