A countable norming family determines Borel laws

For every $x\in B$ and $r>0$, the given norming identity gives the identity $B(x,r)={\bigcup }_{\begin{array}{c}q\in \mathbb{Q}\\ 0\le q<r\end{array}}{\bigcap }_{j\ge 1}\{y:\mathrm{\mid }{\mathrm{\ell }}_j(y-x)\mathrm{\mid }\le q\}\mathrm{.}$ If $\mathrm{\parallel }y-x\mathrm{\parallel }<r$, use [F2] to choose rational $q$ between that norm and $r$; conversely, membership in one intersection implies $\mathrm{\parallel }y-x\mathrm{\parallel }\le q<r$. Thus every open ball belongs to $\sigma ({\mathrm{\ell }}_j:j\ge 1)$. By [F3], fix a countable dense set $D\subseteq B$. The balls $B(d,s)$ with $d\in D$ and positive rational $s$ form a countable family by [F2]. For any open $U$ and $y\in U$, take $a>0$ with $B(y,a)\subseteq U$, choose $d\in D$ with $\mathrm{\parallel }d-y\mathrm{\parallel }<a\mathrm{/}4$, and choose rational $s$ with $a\mathrm{/}4<s<a\mathrm{/}2$ by [F2]. Then $y\in B(d,s)\subseteq B(y,a)\subseteq U$. Thus this family is a base, and every open set is the union of its members contained in that set, a countable union. All those balls are in $\sigma ({\mathrm{\ell }}_j)$ by the first part, so every open set is in that sigma-algebra. Conversely each continuous linear ${\mathrm{\ell }}_j$ is Borel measurable, giving the reverse inclusion; hence it is the Borel sigma-algebra.

The finite-coordinate cylinder sets form a pi-system generating $\sigma ({\mathrm{\ell }}_j:j\ge 1)$. The assumed equality of finite-dimensional pushforwards makes $\mu$ and $\nu$ agree on each such cylinder. The cylinder pi-system contains $B$, and both probabilities assign it mass one; the constant exhaustion by $B$ meets [F1], which gives equality on the generated Borel sigma-algebra.