Half-Line Tests for Real-Valued Measurability

Why Test Upper Rays?

A real-valued function can be tested for measurability by examining a simple family of subsets of the real line: the open upper rays \((a,\infty)\), one for each real number \(a\). The point is not that these are the only Borel sets, but that they generate all Borel sets through countable sigma-algebra operations. Once this generation is established, checking the preimages of the rays is enough to check the preimages of every Borel set.

Throughout, let \((X,\mathcal{M},\mu)\) be an arbitrary measure space: no finiteness, sigma-finiteness, or completeness assumption is imposed. A function \(f:X\to\mathbb{R}\) is real-valued, meaning it is finite at every point. It is \(\mathcal{M}/\mathcal{B}_{\mathbb{R}}\)-measurable if \(f^{-1}(B)=\{x\in X:f(x)\in B\}\) belongs to \(\mathcal{M}\) for every \(B\in\mathcal{B}_{\mathbb{R}}\). These are set equalities everywhere, not assertions up to a null set. The criterion below concerns measurability only; it does not assert that \(f\) is integrable.

Why Upper Rays Generate the Borel Sets

Write \(\sigma(\mathcal{E})\) for the smallest sigma-algebra containing a family of sets \(\mathcal{E}\). Let \(\mathcal{H}=\{(a,\infty):a\in\mathbb{R}\}\). We first verify that \(\sigma(\mathcal{H})=\mathcal{B}_{\mathbb{R}}\). This generation fact is what allows one family of tests to control preimages of every Borel set.

Proof. Each \((a,\infty)\) is open, hence Borel, so \(\sigma(\mathcal{H})\subseteq\mathcal{B}_{\mathbb{R}}\). For the reverse inclusion, fix \(b\in\mathbb{R}\). Since a sigma-algebra is closed under countable intersections and complements, it contains \[ \mathbb{R}\setminus\bigcap_{n=1}^{\infty}(b-1/n,\infty)=(-\infty,b). \] Indeed, a real number \(x\) belongs to every ray in that intersection exactly when \(x\ge b\): if \(x\ge b\), then \(x>b-1/n\) for every \(n\); if \(x<b\), choose \(n\) with \(1/n<b-x\), so \(x\le b-1/n\) and \(x\) is not in that ray. Thus \((a,b)=(a,\infty)\cap(-\infty,b)\) belongs to \(\sigma(\mathcal{H})\).

Every open set \(U\subseteq\mathbb{R}\) is the union of the rational-endpoint intervals \((p,q)\) that are contained in \(U\). To see coverage, for each \(x\in U\), openness gives an interval around \(x\) contained in \(U\); density of the rationals gives \(p,q\in\mathbb{Q}\) with \(p<x<q\) and \((p,q)\subseteq U\). There are only countably many rational pairs, so this union is countable. The empty open set is the empty union. Each such interval belongs to \(\sigma(\mathcal{H})\), hence every open set does. Since the Borel sigma-algebra is generated by the open sets, \(\mathcal{B}_{\mathbb{R}}\subseteq\sigma(\mathcal{H})\). Both inclusions give the claim. \(\square\)

The Half-Line Criterion

Proof. Suppose first that \(f\) is \(\mathcal{M}/\mathcal{B}_{\mathbb{R}}\)-measurable. For every \(a\in\mathbb{R}\), the ray \((a,\infty)\) is Borel, so its preimage \(f^{-1}((a,\infty))=\{x:f(x)>a\}\) belongs to \(\mathcal{M}\).

Conversely, suppose \(\{x:f(x)>a\}\in\mathcal{M}\) for every real \(a\). Define \[ \mathcal{D}=\{B\subseteq\mathbb{R}: f^{-1}(B)\in\mathcal{M}\}. \] This is a sigma-algebra: \(f^{-1}(\mathbb{R})=X\), \(f^{-1}(B^c)=(f^{-1}(B))^c\), and \(f^{-1}(\bigcup_{j=1}^{\infty}B_j)=\bigcup_{j=1}^{\infty}f^{-1}(B_j)\). By the assumption, \(\mathcal{D}\) contains every ray in \(\mathcal{H}\). Therefore it contains \(\sigma(\mathcal{H})=\mathcal{B}_{\mathbb{R}}\), by the proposition. Thus the preimage of every Borel set is in \(\mathcal{M}\), which is precisely the stated measurability of \(f\). \(\square\)

The strict inequality is part of the test: the sets tested are \(\{f>a\}\), not \(\{f\ge a\}\). The theorem asks for every real threshold, and the resulting sets must be in the given sigma-algebra \(\mathcal{M}\). It neither assumes that all subsets of \(X\) are measurable nor changes the definition of measurability.

Applying the Test

A Common Endpoint Mistake

A frequent error is to replace the strict upper ray \((a,\infty)\) with the closed upper ray \([a,\infty)\) without changing the set identity. For the identity function \(f(x)=x\) on \((\mathbb{R},\mathcal{L},m)\), the wrong calculation \(\{x:f(x)\ge0\}=\{x:f(x)>0\}\) would identify \([0,\infty)\) with \((0,\infty)\). This is false: \(0\) belongs to the first set and not the second.

The correct relation, useful when a closed upper level set is needed, is \[ \{x:f(x)\ge b\}=\bigcap_{n=1}^{\infty}\{x:f(x)>b-1/n\}. \] If \(f(x)\ge b\), then \(f(x)>b-1/n\) for every \(n\). If \(f(x)<b\), choose \(n\) with \(1/n<b-f(x)\); then \(b-1/n>f(x)\), so \(x\) is absent from that particular set in the intersection. This proves the equality pointwise. In the identity-function example with \(b=0\), the intersection is \(\bigcap_{n\ge1}(-1/n,\infty)=[0,\infty)\), not \((0,\infty)\). This countable-intersection formula also shows why the strict-ray tests retain enough information to recover boundary points.

The criterion matters because it reduces measurability to a structured family of threshold events. In applications, each such event can often be described by inequalities or intervals, as for the square and floor functions. The generation argument is essential: it is what turns those threshold checks into the full requirement of Borel measurability, without any assumption that the measure space is complete or finite.