Proof of Sequentially Continuous Functions of Measurable Euclidean Maps are Measurable

Step 1: preimages of open sets under $g$ are relatively open in $E$. Let $V$ be an open subset of the real line (equivalently, an open subset of Euclidean space $\mathbb{R}^1$, by Euclidean Openness Agrees with Metric Openness on $\mathbb{R}^n$) and set $W=\{x\in E:\ g(x)\in V\}$. We claim that for every $x\in W$ there is $\varepsilon>0$ such that every $y\in E$ with Euclidean distance $d(y,x)<\varepsilon$ lies in $W$. Suppose not. Then for every natural number $n$ there is $y_n\in E$ with $d(y_n,x)<1/n$ and $g(y_n)\notin V$. The sequence $(y_n)$ satisfies $d(y_n,x)\to0$, so by the sequential continuity hypothesis $g(y_n)\to g(x)$ in the sense of Limit of a Sequence of Real Numbers. Since $V$ is open and $g(x)\in V$, there is $\varepsilon'>0$ with $(g(x)-\varepsilon',g(x)+\varepsilon')\subseteq V$; convergence gives $|g(y_n)-g(x)|<\varepsilon'$ for all large $n$, so $g(y_n)\in V$ for such $n$, a contradiction. This proves the claim.

Step 2: $W$ is the trace on $E$ of a countable union of rational boxes. For each $x\in W$ fix $\varepsilon_x>0$ as in Step 1. Using the density of the rational numbers, choose for each coordinate $j\in\{1,\dots,d\}$ rational numbers $p_j<x^j<q_j$ with $q_j-p_j<\varepsilon_x/\sqrt{d}$, where $\sqrt{d}$ is the nonnegative square root of $d$, and let $Q_x=(p_1,q_1)\times\dots\times(p_d,q_d)$. If $y\in Q_x$ then $|y^j-x^j|<q_j-p_j<\varepsilon_x/\sqrt{d}$ for every $j$, since $y^j$ and $x^j$ both lie in $(p_j,q_j)$; hence $d(y,x)^2=\sum_{j=1}^d(y^j-x^j)^2<d\cdot\varepsilon_x^2/d=\varepsilon_x^2$, so $d(y,x)<\varepsilon_x$. Consequently $Q_x\cap E\subseteq W$ by the choice of $\varepsilon_x$, while $x\in Q_x\cap E$; therefore

Each $Q_x$ is an open box determined by a $2d$-tuple of rational numbers, and the set of such tuples is countable, so the distinct boxes occurring above form a finite or countable family $(Q_n)_n$ and $W=\big(\bigcup_n Q_n\big)\cap E$.

Step 3: pullback along $f$. Since $f$ takes values in $E$,

Each open interval $(p^n_j,q^n_j)$ is a Borel set, so $(f^j)^{-1}((p^n_j,q^n_j))\in\mathcal{A}$ by the measurability of $f^j$. A $\sigma$-algebra is closed under finite intersections (complements and countable unions give intersections) and countable unions, so $(g\circ f)^{-1}(V)\in\mathcal{A}$ for every open $V$.

Step 4: from open sets to Borel sets. Let $\mathcal{S}=\{S\subseteq\mathbb{R}\ \text{Borel}:\ (g\circ f)^{-1}(S)\in\mathcal{A}\}$. Preimages commute with complements and countable unions, and $\mathcal{A}$ is a $\sigma$-algebra, so $\mathcal{S}$ is a $\sigma$-algebra of subsets of the real line intersected with the Borel sets; it contains all open sets by Step 3. Since the Borel $\sigma$-algebra is the $\sigma$-algebra generated by the open sets, minimality gives that $\mathcal{S}$ contains every Borel set. Hence $(g\circ f)^{-1}(S)\in\mathcal{A}$ for every Borel $S$, which is the measurability of $g\circ f$. $\blacksquare$