There is no infinite-dimensional Lebesgue measure

In mathematics, there is a folklore claim that there is no analogue of Lebesgue measure on an infinite-dimensional Banach space. The theorem this refers to states that there is no translationally invariant measure on a separable Banach space - because if any ball has nonzero non-infinite volume, a slightly smaller ball has zero volume, and countable many such smaller balls cover the space. The folklore statement, however, is entirely false. The countable product of Lebesgue measure is translationally invariant and gives the intuitive notion of volume as the infinite product of lengths, only the domain on which this product measure is defined must necessarily be non-separable, and the measure itself is not sigma finite.

There are other kinds of measures with support entirely on separable Banach spaces: the abstract Wiener space construction gives the analog of products of Gaussian measures, which are not translationally invariant. Alternatively, one may consider Lebesgue measure on finite-dimensional subspaces of the larger space and consider so-called prevalent and shy sets.

The Hilbert cube carries the product Lebesgue measure, and the compact topological group given by the Tychonoff product of infinitely many copies of the circle group is infinite-dimensional, and carries a Haar measure that is translation-invariant. These two spaces can be mapped onto each other in a measure preserving way by unwrapping the circles into intervals. The infinite product of the additive real numbers has the analogous product Haar measure, which is precisely the infinite dimensional analog of Lebesgue measure.

Motivation

It can be shown that Lebesgue measure $\lambda$ on Euclidean space $\Reals^n$ is locally finite, strictly positive and translation-invariant, explicitly:
* every point $x$ in $\Reals^n$ has an open neighbourhood $N_x$ with finite measure $\lambda(N_x) < + \infty;$
* every non-empty open subset $U$ of $\Reals^n$ has positive measure $\lambda(U) > 0;$ and
* if $A$ is any Lebesgue-measurable subset of $\Reals^n,$ $T_n : \Reals^n \to \Reals^n,$ $T_h(x) = x + h,$ denotes the translation map, and $(T_h)_*(\lambda)$ denotes the push forward, then $(T_h)_*(\lambda)(A) = \lambda(A).$

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.

Statement of the theorem

Let $(X, \, \cdot\, )$ be an infinite-dimensional, separable Banach space. Then the only locally finite and translation-invariant Borel measure $\mu$ on $X$ is the trivial measure, with $\mu(A) = 0$ for every measurable set $A.$ Equivalently, every translation-invariant measure that is not identically zero assigns infinite measure to all open subsets of $X.$

Proof of the theorem

Let $X$ be an infinite-dimensional, separable Banach space equipped with a locally finite, translation-invariant measure $\mu.$
Like every separable metric space, $X$ is a Lindelöf space, which means that every open cover of $X$ has a countable subcover.
To prove that $\mu$ is the trivial measure, it is sufficient (and necessary) to show that $\mu(X) = 0.$ To prove this, it is enough to show that there exists some non-empty open set $N$ of measure zero because then $\$ will be an open cover of $X$ by sets of measure $\mu(x + N) = \mu(N) = 0$ (by translation-invariance); after picking any countable subcover of $X$ by these measure zero sets, $\mu(X) = 0$ will follow from the σ-subadditivity of $\mu.$

Using local finiteness, suppose that, for some $r > 0,$ the open ball $B(r)$ of radius $r$ has finite $\mu$-measure. Since $X$ is infinite-dimensional, by Riesz's lemma there is an infinite sequence of pairwise disjoint open balls $B_n(r/4),$ $n \in \N,$ of radius $r/4,$ with all the smaller balls $B_n(r/4)$ contained within the larger ball $B(r).$ 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 $\mu$-measure zero.

See also

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References

* (See section 1: Introduction)
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{{Functional analysis

Articles containing proofs
Banach spaces
Measure theory
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