# Infinite-dimensional Lebesgue measure

Short description: Mathematical folklore

In functional analysis and measure theory, one may seek to extend the Lebesgue measure on $\displaystyle{ \mathbb{R}^n }$ to an analogous measure on an infinite dimensional separable Banach space. Without departing substantially from the essential properties of the Lebesgue measure, this is impossible. Indeed, any translation invariant Borel measure on an infinite dimensional separable Banach space is either infinite on all open sets, or the zero measure.

On the other hand, numerous examples of such measures exist once one allows non-separable Banach spaces, or allows the measure to not be translation invariant. In the same vein, one may extend the Lebesgue measure to various subsets of infinite-dimensional space, like the Hilbert cube.

## Motivation

It can be shown that the Lebesgue measure $\displaystyle{ \lambda }$ on Euclidean space $\displaystyle{ \Reals^n }$ is locally finite, strictly positive, and translation-invariant. That is:

• every point $\displaystyle{ x }$ in $\displaystyle{ \Reals^n }$ has an open neighbourhood $\displaystyle{ N_x }$ with finite measure $\displaystyle{ \lambda(N_x) \lt + \infty; }$
• every non-empty open subset $\displaystyle{ U }$ of $\displaystyle{ \Reals^n }$ has positive a measure $\displaystyle{ \lambda(U) \gt 0; }$ and
• if $\displaystyle{ A }$ is any Lebesgue-measurable subset of $\displaystyle{ \Reals^n, }$ $\displaystyle{ T_n : \Reals^n \to \Reals^n, }$ $\displaystyle{ T_h(x) = x + h, }$ denotes the translation map, and $\displaystyle{ (T_h)_*(\lambda) }$ denotes the push forward, then $\displaystyle{ (T_h)_*(\lambda)(A) = \lambda(A). }$

Geometrically speaking, these three properties make the Lebesgue measure very useful. When we consider an infinite-dimensional space such as an $\displaystyle{ L^p }$ space or the space of continuous paths in Euclidean space, it would be clean to have a similar measurement to work with; however, this is not possible.

## Statement of the theorem

On a non locally compact Polish group $\displaystyle{ G }$, there cannot exist a σ-finite, left-invariant Borel measure.[1]

## Separable Banach spaces

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

### Proof of the Theorem

Let $\displaystyle{ X }$ be an infinite-dimension, separable Banach space equipped with a locally finite translation-invariant measurement $\displaystyle{ \mu. }$

Like every separable metric space, $\displaystyle{ X }$ is a Lindelöf space, which means that every open cover of $\displaystyle{ X }$ has a countable subcover.

To prove that $\displaystyle{ \mu }$ is the trivial measurement, it is sufficient and necessary to show that $\displaystyle{ \mu(X) = 0. }$ To prove this show that there exist some non-empty open sets of $\displaystyle{ N }$ that measure zero because then $\displaystyle{ \{x + N: x \in X\} }$ will be an open cover of $\displaystyle{ X }$ by sets of the measurement $\displaystyle{ \mu(x + N) = \mu(N) = 0 }$ (by translation-invariance); after picking any countable subcover of $\displaystyle{ X }$ by these measurement zero sets, $\displaystyle{ \mu(X) = 0 }$ will follow from the σ-subadditivity of $\displaystyle{ \mu. }$

Using local finiteness, suppose that for some $\displaystyle{ r \gt 0, }$ the open ball $\displaystyle{ B(r) }$ of radius $\displaystyle{ r }$ has a finite $\displaystyle{ \mu }$-measurement. Since $\displaystyle{ X }$ is defined as being infinite-dimensional by Riesz's lemma there is an infinite sequence of pairwise disjoint open balls $\displaystyle{ B_n(r/4), }$ $\displaystyle{ n \in \N, }$ of radius $\displaystyle{ r/4, }$ with all the smaller balls $\displaystyle{ B_n(r/4) }$ contained within the larger ball $\displaystyle{ B(r). }$ By translation-invariance, all of the smaller balls have the same measurement; since the sum of these measurements is finite, the smaller balls must all have $\displaystyle{ \mu }$-measurement of zero.

## Nontrivial measures

Here we describe examples where a notion of an infinite-dimensional Lebesgue measure exists, once one loosens the conditions of the above theorem.

There are other kinds of measures that support entirely separable Banach spaces such as the abstract Wiener space construction, which gives the analog of products of Gaussian measures. Alternatively, one may consider a Lebesgue measurement of finite-dimensional subspaces on the larger space and consider the so-called prevalent and shy sets.[2]

The Hilbert cube carries the product Lebesgue measure[3] and the compact topological group given by the Tychonoff product of an infinite number of copies of the circle group which 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 the Lebesgue measure.[citation needed]