There is no infinite-dimensional Lebesgue measure
From Wikipedia, the free encyclopedia
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. Alternatively, one may consider Lebesgue measure on finite-dimensional subspaces of the larger space and consider so-called prevalent and shy sets.
Contents |
[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 Nx with finite measure λn(Nx) < +∞;
- every non-empty open subset U of Rn has positive measure λn(U) > 0;
- if A is any Lebesgue-measurable subset of Rn, Th : Rn → Rn, Th(x) = x + h, denotes the translation map, and (Th)∗(λn) denotes the push forward, then (Th)∗(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 Lp 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
Let (X, || ||) be an infinite-dimensional, separable Banach space. Then the only locally finite and translation-invariant Borel measure μ on X is the trivial measure, with μ(A) = 0 for every measurable set A. Equivalently, every translation-invariant measure that is not identically zero assigns infinite measure to all open subsets of X.
(Many authors assume that X is separable. This assumption simplifies the proof considerably, since it provides a countable basis for X, and if X is a Hilbert space then the basis can even be chosen to be orthonormal. However, if X is not separable, one is still left with the undesirable property that some open sets have zero measure, so μ is not strictly positive even if it is not the trivial measure.)
[edit] Proof of the theorem
Let X be an infinite-dimensional, separable Banach space equipped with a locally finite, translation-invariant measure μ. Using local finiteness, suppose that, for some δ > 0, the open ball B(δ) of radius δ has finite μ-measure. Since X is infinite-dimensional, there is an infinite sequence of pairwise disjoint open balls Bn(δ/4), n ∈ N, of radius δ/4, with all the smaller balls Bn(δ/4) contained within the larger ball B(δ). By translation-invariance, all of the smaller balls have the same measure; since the sum of these measures is finite, the smaller balls must all have μ-measure zero. Now, since X is separable, it can be covered by a second countable collection of balls of radius δ/4; since each such ball has μ-measure zero, so must the whole space X, and so μ is the trivial measure.
[edit] References
- Hunt, Brian R. and Sauer, Tim and Yorke, James A. (1992). "Prevalence: a translation-invariant "almost every" on infinite-dimensional spaces". Bull. Amer. Math. Soc. (N.S.) 27: 217–238. doi:. (See section 1: Introduction)
