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Threshold characterisations of real-valued and extended-real-valued measurability

Statement

Let (X,A) be a measurable space and let f:X→R‾. The following are equivalent:

  1. f is measurable;
  2. {x:f(x)>a}∈A for every real a;
  3. {x:f(x)≥a}∈A for every real a;
  4. {x:f(x)<a}∈A for every real a;
  5. {x:f(x)≤a}∈A for every real a.

Moreover, in any one of conditions 2 through 5 it is enough to test only rational thresholds a∈Q.

Facts & Assumptions

Given: A measurable space (X,A) and a function f:X→R‾.

[L1]

The Borel sigma-algebra on R‾ is generated by the rays (a,+∞] with a∈R. (The Borel sigma-algebra on the extended real line)

[L2]

A generating family on the codomain suffices to test measurability. (A generating family on the codomain suffices to test measurability)

[L3]

Between any two distinct real numbers there lies a rational number. (The rationals embed densely in the reals)

[L4]

The rationals admit a countable enumeration, without choice. (Q is countably infinite)

[L5]

A sigma-algebra contains X and the empty set and is closed under complements, countable unions and countable intersections. (Sigma-algebras Sigma-algebras are closed under countable intersections, differences, symmetric differences, and set limits)

Proof

technique · direct
1.1L1L2

For every real a, the set {f>a} is the preimage f−1((a,+∞]). By [L1] these rays generate the codomain Borel sigma-algebra, so [L2] gives the equivalence of conditions 1 and 2.

1.2L1L5givenalgebra

For every real a, total order on the extended real line gives {f≤a}=X∖{f>a},{f<a}=X∖{f≥a}. Complement closure [L5] therefore makes conditions 2 and 5 equivalent and conditions 3 and 4 equivalent. No rational-threshold hypothesis is needed here.

1.3L1L3L4L5algebra

For every real a, {f>a}=⋃q∈Q, q>a{f>q}. One inclusion follows from f(x)>q>a. Conversely, a finite f(x)>a admits a<q<f(x) by [L3]; if f(x)=+∞, take any rational in (a,a+1), and f(x)=−∞ belongs to neither side. The union is countable by [L4], so measurability of all rational strict upper threshold sets implies their measurability for all real thresholds by [L5]. The converse is immediate by restriction. This proves the rational reduction for condition 2.

2.1L1L3L4L5step 1.1step 1.2algebra

For every real a, {f≥a}=⋂q∈Q, q<a{f>q},{f>a}=⋃q∈Q, q>a{f≥q}. For a finite value v=f(x), the first identity follows because v≥a implies v>q for every q<a, whereas v<a admits v<q<a by [L3] and fails that term. For the second, v≥q>a implies v>a, and v>a admits a<q<v by [L3]. At v=+∞ both sides of both identities contain x; at v=−∞ neither does. The rational index sets are nonempty: apply [L3] in (a−1,a) and (a,a+1). By [L4], these operations are countable: fix an enumeration of Q, use X for excluded indices in intersections and the empty set for excluded indices in unions. Hence [L5] makes conditions 2 and 3 equivalent. Together with steps 1.1 and 1.2, all five conditions are equivalent.

3.1L1L3L4L5step 2.1algebra

The other rational reductions follow from the identities {f≥a}=⋂q∈Q, q<a{f≥q},{f<a}=⋃q∈Q, q<a{f<q},{f≤a}=⋂q∈Q, q>a{f≤q}. For the first, v≥a meets every lower rational threshold, while a finite v<a fails a threshold v<q<a. For the second, v<q<a implies v<a, and a finite v<a admits such a q. For the third, v≤a meets every upper rational threshold, whereas a finite v>a fails a threshold a<q<v. All separating rationals are supplied by [L3]. At v=+∞ the first identity has membership on both sides, and the second and third have membership on neither; at v=−∞ the first has membership on neither and the second and third on both. The relevant index sets are nonempty by the rational intervals used in step 2.1. All operations are countable by [L4], using the same padded enumeration as in step 2.1. Thus [L5] promotes each respective rational-only hypothesis to its full-real condition 3, 4 or 5. Conversely each full-real condition contains its rational restriction.

4.1L1L3L4L5step 1.1step 1.2step 2.1step 1.3step 3.1∎

Infinity fibres are also explicitly measurable under any of these equivalent conditions: {f=+∞}=⋂q∈Q{f>q},{f=−∞}=⋂q∈Q{f<q}. The indicated infinity satisfies all the comparisons. A finite value v fails the first intersection at a rational in (v,v+1) and fails the second at a rational in (v−1,v) by [L3]; the opposite infinity also fails. The countable intersections are measurable by [L4] and [L5]. At infinite thresholds, {f>+∞}={f<−∞}=∅ and {f≥−∞}={f≤+∞}=X; the remaining strict rays {f>−∞} and {f<+∞} are the complements of the corresponding infinity fibres, and the remaining non-strict rays are those fibres themselves. These arguments include real-valued functions as the case where both infinity fibres are empty. Steps 1.1 through 3.1 prove the stated equivalences and every rational-threshold reduction without restricting the extended-real values.

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