Proof of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line

Preliminary: the members of $\mathcal{I}$ and of $\mathcal{R}$ are Euclidean open. Let $p,q$ be real and let $x\in(p,q)$, so that $p<x$ and $x<q$. Put $\delta=\min(x-p,q-x)$, a positive real number. Every real $y$ with $x-\delta<y<x+\delta$ satisfies $y>x-(x-p)=p$ and $y<x+(q-x)=q$, hence lies in $(p,q)$; by Euclidean Open Box Criterion in $\mathbb{R}^n$, applied in $\mathbb{R}^1$, the interval $(p,q)$ is Euclidean open, the criterion being satisfied vacuously when the interval is empty. Likewise, if $a$ is real and $a<x$, then with $\delta=x-a$ every real $y$ with $x-\delta<y<x+\delta$ satisfies $y>a$, so every ray is Euclidean open. Since $\mathcal{B}(\mathbb{R})$ contains every Euclidean open subset of $\mathbb{R}$, we get $\mathcal{I}\subseteq\mathcal{B}(\mathbb{R})$ and $\mathcal{R}\subseteq\mathcal{B}(\mathbb{R})$.

Claim 1. Let $U$ be Euclidean open. If $U$ is empty, take $I_k=(0,0)$ for every $k$: this interval belongs to $\mathcal{I}$ because $0$ is rational, and it is empty because no real $x$ satisfies both $0<x$ and $x<0$, so the union of the $I_k$ is $U$.

Suppose now that $U$ is nonempty, and let $D$ be the set of those $\theta\in\mathbb{Q}^2$ for which the open interval $(\theta_1,\theta_2)$ is contained in $U$.

First, every point of $U$ lies in an interval indexed by $D$. Indeed, let $x\in U$. By Euclidean Open Box Criterion in $\mathbb{R}^n$, applied in $\mathbb{R}^1$, there is a real $\delta>0$ such that every real $y$ with $x-\delta<y<x+\delta$ lies in $U$. By claim 1 of The Rational Numbers are Dense in the Real Numbers there are rationals $p$ and $q$ with $x-\delta<p<x$ and $x<q<x+\delta$. Every $y$ with $p<y<q$ then satisfies $x-\delta<y<x+\delta$ and so lies in $U$; hence $(p,q)\subseteq U$, the pair $(p,q)$ belongs to $D$, and $x\in(p,q)$.

In particular $D$ is nonempty. By claim 3 of The Integers and the Rational Numbers are Countable the set $\mathbb{Q}^2$ is countable, so its subset $D$ is countable by claim 3 of Basic Properties of Countable Sets. Being countable and nonempty, $D$ is the set of terms of a sequence $(\theta^k)_{k\in\mathbb{N}}$ in $D$. Put $I_k=(\theta^k_1,\theta^k_2)$; each $I_k$ belongs to $\mathcal{I}$ and is contained in $U$, so $\bigcup_{k}I_k\subseteq U$. Conversely each $x\in U$ lies in an interval $(\theta_1,\theta_2)$ with $\theta\in D$ by the previous paragraph, and $\theta$ is a term $\theta^k$ of the sequence, so $x\in I_k$. Hence $U=\bigcup_{k\in\mathbb{N}}I_k$.

Claim 2. By Generated Sigma-Algebra a generated $\sigma$-algebra is the smallest one containing the generating family. From $\mathcal{I}\subseteq\mathcal{B}(\mathbb{R})$ we get $\sigma(\mathcal{I})\subseteq\mathcal{B}(\mathbb{R})$. Conversely, by claim 1 every Euclidean open subset of $\mathbb{R}$ is a union of a sequence of members of $\mathcal{I}$ and hence lies in $\sigma(\mathcal{I})$, which is closed under countable unions; as $\mathcal{B}(\mathbb{R})$ is the smallest $\sigma$-algebra containing the Euclidean open sets, $\mathcal{B}(\mathbb{R})\subseteq\sigma(\mathcal{I})$. Therefore $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{I})$.

Similarly $\mathcal{R}\subseteq\mathcal{B}(\mathbb{R})$ gives $\sigma(\mathcal{R})\subseteq\mathcal{B}(\mathbb{R})$. For the reverse inclusion let $(h_k)_{k\in\mathbb{N}}$ be a sequence of positive real numbers with limit $0$, which exists by Existence of a Sequence of Positive Real Numbers with Limit Zero, and fix a real number $q$. Then

$$\bigcap_{k\in\mathbb{N}}\{x\in\mathbb{R}:q-h_k<x\}=\{x\in\mathbb{R}:q\le x\}.$$

Indeed, if $q\le x$ then $q-h_k<q\le x$ for every $k$, because $h_k>0$; and if $x<q$ then $q-x>0$, so, the sequence having limit $0$ and positive terms, there is $k$ with $h_k<q-x$, that is $x<q-h_k$, and $x$ fails to lie in the $k$-th set. Each set in the intersection belongs to $\mathcal{R}$, and a $\sigma$-algebra is closed under countable intersections by Sigma-Algebra and Measurable Space, so $\{x\in\mathbb{R}:q\le x\}$ lies in $\sigma(\mathcal{R})$ and hence so does its complement $\{x\in\mathbb{R}:x<q\}$. For any reals $p,q$ we therefore have

$$(p,q)=\{x\in\mathbb{R}:p<x\}\cap\{x\in\mathbb{R}:x<q\}\in\sigma(\mathcal{R}).$$

In particular $\mathcal{I}\subseteq\sigma(\mathcal{R})$, so $\mathcal{B}(\mathbb{R})=\sigma(\mathcal{I})\subseteq\sigma(\mathcal{R})$, and the two $\sigma$-algebras coincide.

Claim 3. Suppose $f$ is measurable. Each ray is Borel by the preliminary step, and $\{x\in X:f(x)>a\}$ is the preimage under $f$ of the ray $\{t\in\mathbb{R}:a<t\}$, so it lies in $\mathcal{F}$. Conversely, suppose $\{x\in X:f(x)>a\}\in\mathcal{F}$ for every real $a$. Then the preimage under $f$ of every member of $\mathcal{R}$ lies in $\mathcal{F}$, and $\mathcal{B}(\mathbb{R})$ is generated by $\mathcal{R}$ by claim 2, so claim 2 of Generator Criterion for Measurability makes $f$ measurable with respect to $\mathcal{F}$ and $\mathcal{B}(\mathbb{R})$.