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 λⁿ on Euclidean space Rⁿ is locally finite, strictly positive and translation-invariant. Explicitly:

every point x in Rⁿ has an open neighbourhood N_x with finite measure λⁿ(N_x) < +∞;
every non-empty open subset U of Rⁿ has positive measure λⁿ(U) > 0;
if A is any Lebesgue measurable subset of Rⁿ, T_h : Rⁿ → Rⁿ, T_h(x) = x + h, denotes the translation map, and (T_h)_∗(λⁿ) denotes the push forward, then (T_h)_∗(λⁿ)(A) = λⁿ(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) = +∞. Then the only locally finite and translation-invariant measure on H is the trivial measure μ, with μ(A) = 0 for every measurable set A.

(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

Let H be an infinite-dimensional separable Hilbert space and suppose μ is a translation-invariant, locally finite measure on H with μ(H) > 0 (so μ is not the trivial measure). For any point x in H and any real r > 0, the notation B(x,r) denotes the open ball around x of radius r.

The first step is to show that for every positive r and every x in H, μ(B(x,r)) > 0. Suppose not, and fix x and r with μ(B(x,r)) = 0. Then, because μ is translation-invariant, μ(B(y,r)) = 0 for every point y in H. Now

$H = \bigcup_{y \in H} B(y,r),$

and so, because H is separable, there is a sequence (y_i)_i∈N of points of H such that

$H = \bigcup_{i \in \mathbb{N}} B(y_i,r).$

Thus,

$\mu (H) \leq \sum_{i \in \mathbb{N}} \mu(B(y_i,r)) = 0,$

contradicting the assumption that μ(H) > 0. This proves the first step.

Since μ is locally finite, there is an open set U of finite measure. Since μ is assumed to be translation-invariant, we may assume that U contains the 0 element of H. Thus, there is some positive real number r₀ such that μ(B(0,r₀)) is finite. Thus, in light of the first step, for any positive r < r₀ and any point x in H, 0 < μ(B(x,r)) < +∞.

Let (e_i)_i∈N be a countable, orthonormal basis for H. Then, by the triangle inequality, for each e_i,

$B \left(\frac{r_0}{2} e_i, \frac{r_0}{4} \right) \subseteq B(\mathbf{0}, r_0)$ .

Now, Pythagoras' theorem holds in any Hilbert space, and thus

$d \left(\frac{r_0}{2} e_i ,\frac{r_0}{2} e_j \right) = \frac{r_0\sqrt{2}}{2}$

for all distinct i and j. Because

$\frac{r_0\sqrt{2}}{2} > 2 \frac{r_0}{4},$

the balls B(r₀e_i/2, r₀/4) and B(r₀e_j/2, r₀/4) must be disjoint for all distinct i and j. Thus

$\mu (B(\mathbf{0}, r_0)) \geq \sum_{i \in \mathbb{N}} \mu \left( B \left( \frac{r_0}{2}e_i, \frac{r_0}{4} \right) \right)$ .

Because μ(B(r₀e_i/2, r₀/4)) has the same nonzero value independent of i, the right-hand side of this inequality is +∞, and so μ(B(0, r₀)) is infinite, contradicting the choice of r₀.