Proof of A Function Nondecreasing on a Borel Subset of the Real Line and Vanishing Outside It is Borel

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: $T(x)\le T(x')$ for all $x,x'\in E$ with $x\le x'$, and $T(x)=0$ for every $x\in\mathbb{R}\setminus E$.

For a real number $c$ put

Step 1 ($A_{c}$ is Borel). Fix $c$ and distinguish three cases.

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

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

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

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

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

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

Step 3 (Measurability). By Step 2 the set $S_{c}=\{x\in\mathbb{R}:c<T(x)\}$ belongs to $\mathcal{B}(\mathbb{R})$ for every real number $c$. Claim 3 of Rational Intervals and Rays 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 $T$ is measurable.