Convex measure

Let X be a locally convex Hausdorff vector space, and consider a probability measure μ on the Borel σ-algebra of X. Fix −∞ ≤ s ≤ 0, and define, for u, v ≥ 0 and 0 ≤ λ ≤ 1,

${\displaystyle M_{s,\lambda }(u,v)={\begin{cases}(\lambda u^{s}+(1-\lambda )v^{s})^{1/s}&{\text{if }}-\infty <s<0,\\\min(u,v)&{\text{if }}s=-\infty ,\\u^{\lambda }v^{1-\lambda }&{\text{if }}s=0.\end{cases}}}$

For subsets A and B of X, we write

${\displaystyle \lambda A+(1-\lambda )B=\{\lambda x+(1-\lambda )y\mid x\in A,y\in B\}}$

for their Minkowski sum. With this notation, the measure μ is said to be s-convex[1] if, for all Borel-measurable subsets A and B of X and all 0 ≤ λ ≤ 1,

${\displaystyle \mu (\lambda A+(1-\lambda )B)\geq M_{s,\lambda }(\mu (A),\mu (B)).}$

The special case s = 0 is the inequality

${\displaystyle \mu (\lambda A+(1-\lambda )B)\geq \mu (A)^{\lambda }\mu (B)^{1-\lambda },}$

i.e.

${\displaystyle \log \mu (\lambda A+(1-\lambda )B)\geq \lambda \log \mu (A)+(1-\lambda )\log \mu (B).}$

Thus, a measure being 0-convex is the same thing as it being a logarithmically concave measure.

The classes of s-convex measures form a nested increasing family as s decreases to −∞"

${\displaystyle s\leq t{\text{ and }}\mu {\text{ is }}t{\text{-convex}}\implies \mu {\text{ is }}s{\text{-convex}}}$

or, equivalently

${\displaystyle s\leq t\implies \{s{\text{-convex measures}}\}\supseteq \{t{\text{-convex measures}}\}.}$

Thus, the collection of −∞-convex measures is the largest such class, whereas the 0-convex measures (the logarithmically concave measures) are the smallest class.

The convexity of a measure μ on n-dimensional Euclidean space Rⁿ in the sense above is closely related to the convexity of its probability density function.[2] Indeed, μ is s-convex if and only if there is an absolutely continuous measure ν with probability density function ρ on some R^k so that μ is the push-forward on ν under a linear or affine map and ${\displaystyle e_{s,k}\circ \rho \colon \mathbb {R} ^{k}\to \mathbb {R} }$ is a convex function, where

${\displaystyle e_{s,k}(t)={\begin{cases}t^{s/(1-sk)}&{\text{if }}-\infty <s<0\\t^{-1/k}&{\text{if }}s=-\infty ,\\-\log t&{\text{if }}s=0.\end{cases}}}$

Convex measures also satisfy a zero-one law: if G is a measurable additive subgroup of the vector space X (i.e. a measurable linear subspace), then the inner measure of G under μ,

${\displaystyle \mu _{\ast }(G)=\sup\{\mu (K)\mid K\subseteq G{\text{ and }}K{\text{ is compact}}\},}$

must be 0 or 1. (In the case that μ is a Radon measure, and hence inner regular, the measure μ and its inner measure coincide, so the μ-measure of G is then 0 or 1.)[1]