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Pettis measurability criterion for strong measurability

Statement

Assume the Axiom of Choice, let (Ω,A,μ) be a complete measure space, and let X be a real or complex Banach space. A function f:ΩX is strongly measurable if and only if both conditions hold:

  1. f is weakly measurable: xf is scalar measurable for every xX;
  2. f is essentially separably valued: there are a null set N and a separable closed subspace YX such that f(ΩN)Y.

Facts & Assumptions

[A1]

The Axiom of Choice holds (The Axiom of Choice).

[L1]

Strong measurability is a.e. pointwise norm approximation by measurable simple functions (Strongly measurable Banach-valued function).

[L2]

The dual consists of bounded scalar-valued linear functionals (The dual space X^* of a normed space and its dual norm).

[L3]

On a complete measure space, every subset of a measurable null set is measurable (Complete measure spaces), and measurability means that Borel preimages are measurable (A measurable function between measurable spaces).

[L4]

Separability means existence of an at most countable dense subset (Separability: the existence of an at most countable dense subset).

[L5]

Under AC, a dominated real linear functional extends to the whole real space (Hahn-Banach dominated extension theorem for real vector spaces).

Proof

technique · direct

Given: The assumptions and the two conditions in the Statement.

1.1

Strong measurability gives an essentially separable range. [given, L1, L4] Assume first that f is strongly measurable, witnessed by sn and N as in [L1]. The union of the finite ranges of the sn is countable. Its closed linear span Y is separable by [L4], and every f(ω) with ωN is a norm limit of points of Y. Thus f is essentially separably valued.

givenL1L4
1.2

Strong measurability gives weak measurability. [given, L1, L2, L3] For xX, [L2] gives x(sn(ω))x(f(ω)) off N. Each xsn is scalar simple and measurable. A pointwise scalar limit is measurable off N, and [L3] makes its arbitrary values on subsets of N measurable as well. Hence f is weakly measurable.

givenL1L2L3
1.3

Fix countable dense data for the reverse implication. [given, L4, choose] Conversely assume conditions 1 and 2. If Y={0}, the constant zero simple functions converge to f off N, so suppose Y{0}. By [L4] choose a sequence (ym) dense in Y and a sequence (zk) dense in its unit sphere.

givenL4choose
2.1

Construct a countable norming family. [A1, L2, L5, step 1.3] For each k, in the complex case define on the underlying real plane Czk the norm-one real functional uk(azk)=Rea; in the real case use uk(azk)=a on Rzk. Apply [L5] and [A1] to extend these simultaneously to real functionals Uk on the underlying real space of X. In the complex case put xk(x)=Uk(x)iUk(ix); in the real case put xk=Uk. Then xkX, xk=1, and xk(zk)=1. Consequently, for yY,

A1L2L5step 1.3

y=supkxk(y).

Indeed the upper bound is immediate, while a unit vector arbitrarily close to some zk makes the corresponding value arbitrarily close to 1.

3.1

Norm distances to fixed centres are measurable. [L3, step 2.1] For fixed yY, step 2.1 and weak measurability give, off N, fy=supkxk(fy). The right side is the supremum of a countable family of measurable scalar functions. With any values assigned on N, [L3] therefore makes ωf(ω)y measurable.

L3step 2.1
4.1

Build finite-valued nearest-centre approximants. [L1, step 1.3, step 3.1] For each n1 and ωN, choose the least m{1,,n} minimizing f(ω)ym; put sn(ω)=ym there and sn=0 on N. The finitely many tie-broken Voronoi cells are measurable by step 3.1, so sn is a measurable simple function. Density of (ym) gives sn(ω)f(ω)0 for every ωN.

L1step 1.3step 3.1
5.1

Steps 1.1--1.2 prove the forward implication, and step 4.1 supplies the simple approximants required by [L1] for the reverse implication. The only non-finite choice is [A1]: it supplies the Hahn--Banach extensions in step 2.1 (and hence also covers their countable simultaneous selection).

A1L1step 1.1step 1.2step 4.1

Depends on

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