In mathematics, an infinite-dimensional Lebesgue measure is a measure defined on infinite-dimensional normed vector spaces, such as Banach spaces, which resembles the Lebesgue measure used in finite-dimensional spaces.

However, the traditional Lebesgue measure cannot be straightforwardly extended to all infinite-dimensional spaces due to a key limitation: any translation-invariant Borel measure on an infinite-dimensional separable Banach space must be either infinite for all sets or zero for all sets. Despite this, certain forms of infinite-dimensional Lebesgue-like measures can exist in specific contexts. These include non-separable spaces like the Hilbert cube, or scenarios where some typical properties of finite-dimensional Lebesgue measures are modified or omitted.

The Lebesgue measure ${\displaystyle \lambda }$ on the Euclidean space ${\displaystyle \mathbb {R} ^{n}}$ is locally finite, strictly positive, and translation-invariant. That is:

Motivated by their geometrical significance, constructing measures satisfying the above set properties for infinite-dimensional spaces such as the ${\displaystyle L^{p}}$ spaces or path spaces is still an open and active area of research.

Let ${\displaystyle X}$ be an infinite-dimensional, separable Banach space. Then, the only locally finite and translation invariant Borel measure ${\displaystyle \mu }$ on ${\displaystyle X}$ is a trivial measure. Equivalently, there is no locally finite, strictly positive, and translation invariant measure on ${\displaystyle X}$.

More generally: on a non locally compact Polish group ${\displaystyle G}$, there cannot exist a σ-finite and left-invariant Borel measure.

This theorem implies that on an infinite dimensional separable Banach space (which cannot be locally compact) a measure that perfectly matches the properties of a finite dimensional Lebesgue measure does not exist.

Let ${\displaystyle X}$ be an infinite-dimensional, separable Banach space equipped with a locally finite translation-invariant measure ${\displaystyle \mu }$. To prove that ${\displaystyle \mu }$ is the trivial measure, it is sufficient and necessary to show that ${\displaystyle \mu (X)=0.}$