TheoremBase

Monotonicity on the Borel set makes each of its superlevel sets there the intersection of the set with a ray, up to the single point given by the infimum; adjoining the complement when the level is negative gives Borel superlevel sets, and the generator criterion for the rays finishes the proof.

Proof

Throughout, each result cited is universally quantified over the data appearing in its own statement and is applied to the data named here. Assume the hypotheses of claim 1: T(x)≤T(x′)T(x)\le T(x') for all x,x′∈Ex,x'\in E with x≤x′x\le x', and T(x)=0T(x)=0 for every x∈R∖Ex\in\mathbb{R}\setminus E.

For a real number cc put

Ac={x∈E:c<T(x)}.A_{c}=\{x\in E:c<T(x)\}.

Step 1 (AcA_{c} is Borel). Fix cc and distinguish three cases.

If Ac=∅A_{c}=\varnothing, it belongs to B(R)\mathcal{B}(\mathbb{R}).

Suppose Ac≠∅A_{c}\ne\varnothing and AcA_{c} is bounded below, and let ss be its greatest lower bound. Every x∈Acx\in A_{c} satisfies s≤xs\le x, so Ac⊆E∩{x:s≤x}A_{c}\subseteq E\cap\{x:s\le x\}. Conversely let x∈Ex\in E with s<xs<x. By claim 2 of Approximation Property of the Supremum and the Infimum in R\mathbb{R} there is x′∈Acx'\in A_{c} with x′<xx'<x; since x′∈Ex'\in E and x′≤xx'\le x, monotonicity gives T(x′)≤T(x)T(x')\le T(x), and c<T(x′)c<T(x'), so c<T(x)c<T(x) by claim 2 of Elementary Order Arithmetic in an Ordered Field and x∈Acx\in A_{c}. Hence

E∩{x:s<x}⊆Ac⊆(E∩{x:s<x})∪{s},E\cap\{x:s<x\}\subseteq A_{c}\subseteq\bigl(E\cap\{x:s<x\}\bigr)\cup\{s\},

so AcA_{c} is either E∩{x:s<x}E\cap\{x:s<x\} or (E∩{x:s<x})∪{s}\bigl(E\cap\{x:s<x\}\bigr)\cup\{s\}, according as s∈Acs\in A_{c} or not. The ray {x:s<x}\{x:s<x\} is open in the real line with the absolute-value metric, hence belongs to B(R)\mathcal{B}(\mathbb{R}) by claim 1 of Borel Measurability and Bounded Integration on a Metric Space together with claim 2 of that lemma, and {s}\{s\} belongs to B(R)\mathcal{B}(\mathbb{R}) by Pairs of Euclidean Points: Coordinate Projections, Pairings, the Product Measure on a Euclidean Space, Borel Norm Functions and Finite Sets (v). In either case Ac∈B(R)A_{c}\in\mathcal{B}(\mathbb{R}), a σ\sigma-algebra being closed under finite intersections and unions.

Suppose finally that Ac≠∅A_{c}\ne\varnothing and AcA_{c} is not bounded below. Let x∈Ex\in E. Then xx is not a lower bound of AcA_{c}, so there is x′∈Acx'\in A_{c} with x′<xx'<x; as above c<T(x′)≤T(x)c<T(x')\le T(x), so x∈Acx\in A_{c}. Hence Ac=E∈B(R)A_{c}=E\in\mathcal{B}(\mathbb{R}).

Step 2 (Superlevel sets of TT). Let cc be a real number and put Sc={x∈R:c<T(x)}S_{c}=\{x\in\mathbb{R}:c<T(x)\}. For x∈R∖Ex\in\mathbb{R}\setminus E one has T(x)=0T(x)=0, so x∈Scx\in S_{c} exactly when c<0c<0. Therefore

Sc=Acif 0≤c,Sc=Ac∪(R∖E)if c<0,S_{c}=A_{c}\quad\text{if }0\le c,\qquad S_{c}=A_{c}\cup(\mathbb{R}\setminus E)\quad\text{if }c<0 ,

and in both cases Sc∈B(R)S_{c}\in\mathcal{B}(\mathbb{R}) by Step 1, since E∈B(R)E\in\mathcal{B}(\mathbb{R}).

Step 3 (Measurability). By Step 2 the set Sc={x∈R:c<T(x)}S_{c}=\{x\in\mathbb{R}:c<T(x)\} belongs to B(R)\mathcal{B}(\mathbb{R}) for every real number cc. Claim 3 of Intervals Are Borel, and Rays, Half-Open Intervals and Rational Intervals Generate the Borel Sigma-Algebra of the Real Line, the criterion that a real-valued map on a measurable space is measurable exactly when each of these superlevel sets is measurable, therefore gives that TT is measurable.

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