Proof of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets

Throughout, the generator criterion is claim 2 of Generator Criterion for Measurability: a map into a measurable space whose $\sigma$-algebra is generated by a family is measurable as soon as the preimage of every member of that family is measurable.

Claim 1. Fix $j\in\{1,\dots,m\}$ and a real number $a$. The set $(a,\infty)$ is Euclidean open in $\mathbb{R}$, since for $x>a$ every $y$ with $x-\delta<y<x+\delta$ satisfies $y>a$ when $\delta=x-a$; hence $(a,\infty)$ belongs to $\mathcal{B}(\mathbb{R})$, and so does $\mathbb{R}$. Therefore the set $A_1\times\dots\times A_m$ with $A_j=(a,\infty)$ and $A_i=\mathbb{R}$ for $i\ne j$ is a Borel rectangle in the sense of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$, and every Borel rectangle lies in $\mathcal{B}_m$ because those rectangles generate $\mathcal{B}_m$. This set is exactly $\{x\in\mathbb{R}^m:\pi_j(x)>a\}$, so claim 3 of Rational Intervals and Rays Generate the Borel Sigma-Algebra of the Real Line makes $\pi_j$ measurable.

Claim 2. Suppose first that every component $f^j$ is measurable, and let $A_1\times\dots\times A_m$ be a Borel rectangle. A point $x\in X$ satisfies $f(x)\in A_1\times\dots\times A_m$ exactly when $f^j(x)\in A_j$ for every $j$, so

$$f^{-1}(A_1\times\dots\times A_m)=\bigcap_{j=1}^{m}(f^j)^{-1}(A_j),$$

a finite intersection of members of $\mathcal{F}$ and hence a member of $\mathcal{F}$ by Sigma-Algebra and Measurable Space. Since the Borel rectangles generate $\mathcal{B}_m$, the generator criterion makes $f$ measurable.

Conversely suppose $f$ is measurable and let $B\in\mathcal{B}(\mathbb{R})$ and $j\in\{1,\dots,m\}$. Since $f^j=\pi_j\circ f$ we have $(f^j)^{-1}(B)=f^{-1}\bigl(\pi_j^{-1}(B)\bigr)$, and $\pi_j^{-1}(B)\in\mathcal{B}_m$ by claim 1; hence $(f^j)^{-1}(B)\in\mathcal{F}$.

Claim 3(a). Let $g:\mathbb{R}^n\to\mathbb{R}$ be sequentially continuous. Apply Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable with the measurable space $(\mathbb{R}^n,\mathcal{B}_n)$, with $E=\mathbb{R}^n$, with the function $g$ on $E$, and with the map $\mathbb{R}^n\to E$ taken to be the identity: its components are the projections $\pi_1,\dots,\pi_n$, which are measurable with respect to $\mathcal{B}_n$ and $\mathcal{B}(\mathbb{R})$ by claim 1, and the hypothesis of sequential continuity on $E$ in that lemma is exactly the one assumed here. The conclusion is that $g$, being the composition of $g$ with the identity, is measurable with respect to $\mathcal{B}_n$ and $\mathcal{B}(\mathbb{R})$.

Now let $g:\mathbb{R}^n\to\mathbb{R}$ be continuous from $(\mathbb{R}^n,d)$ to $(\mathbb{R},d)$, and note that $d$ on $\mathbb{R}$ is the absolute-value metric by The Euclidean Distance on the Real Line is the Absolute Value Metric. Let $(x^k)_{k\in\mathbb{N}}$ be a sequence in $\mathbb{R}^n$ with $d(x^k,x)\to0$ and let $\varepsilon>0$. Continuity at $x$ supplies $\delta>0$ such that $|g(y)-g(x)|<\varepsilon$ whenever $d(y,x)<\delta$, and convergence supplies $K$ with $d(x^k,x)<\delta$ for all $k\ge K$; hence $|g(x^k)-g(x)|<\varepsilon$ for all $k\ge K$. So $g(x^k)\to g(x)$, $g$ is sequentially continuous, and the previous paragraph applies.

Claim 3(b). Each component $\Phi^j$ is measurable with respect to $\mathcal{B}_n$ and $\mathcal{B}(\mathbb{R})$ by claim 3(a), so $\Phi$ is measurable with respect to $\mathcal{B}_n$ and $\mathcal{B}_m$ by claim 2 applied with $(X,\mathcal{F})=(\mathbb{R}^n,\mathcal{B}_n)$.

Claim 4. Let $C\subseteq\mathbb{R}^m$ be closed. If $C$ is empty then $C\in\mathcal{B}_m$, since a $\sigma$-algebra contains the empty set. Suppose $C$ is nonempty and let $g(x)$ denote the distance from $x$ to $C$ in the metric space $(\mathbb{R}^m,d)$. Each value $g(x)$ is the infimum of a set of nonnegative real numbers, for which $0$ is a lower bound, so $g(x)\ge0$ for every $x$. By The Distance to a Set is Nonexpansive we have $|g(x)-g(y)|\le d(x,y)$ for all $x,y$, so $d(x^k,x)\to0$ forces $g(x^k)\to g(x)$; thus $g$ is sequentially continuous and claim 3(a), applied with $n=m$, makes $g$ measurable with respect to $\mathcal{B}_m$ and $\mathcal{B}(\mathbb{R})$.

By Vanishing of the Distance to a Set Characterizes the Closure, $g(x)=0$ if and only if $x$ lies in the closure of $C$, and that closure is $C$ itself because $C$ is closed, by The Closure is the Smallest Closed Superset. Since $g$ is nonnegative, the complement of $C$ is therefore $\{x\in\mathbb{R}^m:g(x)>0\}$, which belongs to $\mathcal{B}_m$ by measurability of $g$; hence $C\in\mathcal{B}_m$, a $\sigma$-algebra being closed under complements.

Finally let $U\subseteq\mathbb{R}^m$ be Euclidean open. Its complement $C=\mathbb{R}^m\setminus U$ is closed, because the complement of $C$ is $U$, which is open. By the preceding paragraph $C\in\mathcal{B}_m$, and therefore $U=\mathbb{R}^m\setminus C\in\mathcal{B}_m$.

Claim 5. By claim 4 every Euclidean open subset of $\mathbb{R}^m$ lies in $\mathcal{B}_m$, and $\mathcal{B}(\mathbb{R}^m)$ is the smallest $\sigma$-algebra containing those sets, so $\mathcal{B}(\mathbb{R}^m)\subseteq\mathcal{B}_m$.

For the reverse inclusion fix $j\in\{1,\dots,m\}$ and let $A\subseteq\mathbb{R}$ be Euclidean open. Then $\pi_j^{-1}(A)$ is Euclidean open in $\mathbb{R}^m$: if $x\in\pi_j^{-1}(A)$ then $x_j\in A$, so Euclidean Open Box Criterion in $\mathbb{R}^n$ applied in $\mathbb{R}^1$ supplies $\delta>0$ such that every real $t$ with $x_j-\delta<t<x_j+\delta$ lies in $A$; every $y\in\mathbb{R}^m$ with $x_i-\delta<y_i<x_i+\delta$ for all $i$ then has $y_j\in A$, that is $y\in\pi_j^{-1}(A)$, and the same criterion applied in $\mathbb{R}^m$ gives openness of $\pi_j^{-1}(A)$. Hence $\pi_j^{-1}(A)\in\mathcal{B}(\mathbb{R}^m)$ for every open $A\subseteq\mathbb{R}$, and since $\mathcal{B}(\mathbb{R})$ is generated by the open subsets of $\mathbb{R}$, the generator criterion makes $\pi_j$ measurable with respect to $\mathcal{B}(\mathbb{R}^m)$ and $\mathcal{B}(\mathbb{R})$. Consequently, for every Borel rectangle,

$$A_1\times\dots\times A_m=\bigcap_{j=1}^{m}\pi_j^{-1}(A_j)\in\mathcal{B}(\mathbb{R}^m),$$

using closure of a $\sigma$-algebra under finite intersections. The Borel rectangles generate $\mathcal{B}_m$, so $\mathcal{B}_m\subseteq\mathcal{B}(\mathbb{R}^m)$, and the two $\sigma$-algebras coincide.