Weakly measurable need not be strongly measurable

Verify that the target is a Banach space. Finite-dimensional Cauchy--Schwarz gives the triangle inequality after taking the supremum over finite $F$; homogeneity and definiteness are immediate, so the displayed formula is a norm. If $(y_n)$ is Cauchy in this norm, then every coordinate sequence $(y_n(t))$ is Cauchy. Let $y(t)={\lim }_n{y}_n(t)$. Given $\epsilon >0$, choose $N$ such that $\parallel y_m-y_N\parallel <\epsilon$ for $m\ge N$. For every finite $F$, passage to the scalar limit gives ${\sum }_{t\in F}\mathrm{\mid }y(t)-y_N(t){\mathrm{\mid }}^2\le {\epsilon }^2$. Taking the supremum shows $y-y_N\in H$ with norm at most $\epsilon$. Hence $y\in H$ and $y_n\to y$, so [L3] makes $H$ Banach. This also proves directly that $\parallel e_s-e_t\parallel =\sqrt{2}$ whenever $s\mathrm{\ne }t$.

Every scalar evaluation of the range has countable support. Fix $\phi \in H^{\ast }$ and put $a_t=\phi (e_t)$. For a finite $F\subseteq I$, apply $\phi$ to ${\sum }_{t\in F}\overline{a_t}{e}_t$ (with conjugation trivial over $\mathbb{R}$) to obtain

Prove weak measurability. The trace measure space on $I$ is complete: any subset of a trace-null set is an ambient subset of a Lebesgue-null set and hence is Lebesgue measurable by [L1]. For the fixed $\phi$, the scalar function $\phi \circ f:t\mapsto a_t$ vanishes off the countable null support from step 2.1. The inverse image of an open scalar set is either a subset of that support or the complement of such a subset, according as the open set omits or contains zero. Completeness makes every such inverse image measurable. Since $\phi$ was arbitrary, $f$ is weakly measurable.

Rule out an essentially separable range. Suppose there were a null $N\subseteq I$ and a separable closed subspace $Y\subseteq H$ containing every $e_t$ for $t\in I\setminus N$. The set $I\setminus N$ is uncountable: if it were countable, [L1] would make both it and $N$ null, contrary to $\mu (I)=1$. Let $D$ be an at most countable dense subset of $Y$. For each $t\in I\setminus N$, assign the first member of a fixed enumeration of $D$ lying within $\sqrt{2}\mathrm{/}3$ of $e_t$. The assignment is injective, because one point of $D$ cannot lie within that radius of two vectors at distance $\sqrt{2}$. This would make $I\setminus N$ countable, a contradiction. Thus the range is not essentially separably valued.

Conclude failure of strong measurability and audit boundaries. [A1, L2, step 1.1, step 3.1, step 4.1] The trace measure is complete, $H$ is Banach, and step 3.1 proves weak measurability, but step 4.1 disproves the other necessary condition in [L2]. Hence $f$ is not strongly measurable. Every coordinate vector has norm one; the zero functional has empty support and gives the constant zero scalar map; the real and complex cases are both covered by the finite coefficient calculation. The positive-measure interval, rather than a singleton or a null domain, is essential to the failed conclusion. AC is used through [L1], [L2], and the displayed simultaneous countability arguments only.