Borell–Sudakov–Tsirelson expansion for a Banach Gaussian law

First let $F=T^{-1}(M)$ be a closed finite-coordinate cylinder, with $T=({\mathrm{\ell }}_1,\dots ,{\mathrm{\ell }}_k):B\to {\mathbb{R}}^k$ continuous and $M$ closed. Let $Q=\mathrm{span}\{[{\mathrm{\ell }}_i]:1\le i\le k\}\subset H$, choose an orthonormal basis $e_1,\dots ,e_m$, and choose representatives ${\phi }_j\in B^{\ast }$ for $e_j$. Set $G={\sum }_{j=1}^m{\phi }_j(W)e_j$ and $S=T\circ i{\mathrm{\mid }}_Q$. The vector of coefficients is standard Gaussian on $Q$ because its covariance matrix is the identity. For each $i$, $[{\mathrm{\ell }}_i]={\sum }_j⟨[{\mathrm{\ell }}_i],e_j{⟩}_H{e}_j$; by the covariance quotient and [F2], the residual ${\mathrm{\ell }}_i(W)-{\mathrm{\ell }}_i(iG)$ has variance $\mathrm{\parallel }[{\mathrm{\ell }}_i]-{\sum }_j⟨[{\mathrm{\ell }}_i],e_j{⟩}_H{e}_j{\mathrm{\parallel }}_H^2=0$. Hence $T(W)=S(G)$ almost surely. The map $S$ is injective: if $S(q)=0$, then $⟨q,[{\mathrm{\ell }}_i]{⟩}_H=0$ for every $i$, and these classes span $Q$, so $q=0$.

Now let $A\subseteq B$ be Borel with $0<\mu (A)<1$, and fix $\beta <{\mathrm{\Phi }}^{-1}(\mu (A))$. By [F4] choose compact $K\subseteq A$ with $\mu (K)>\mathrm{\Phi }(\beta )$. For each $x\mathrm{\notin }K$ and $y\in K$, [F7] gives ${\mathrm{\ell }}_y\in B^{\ast }$ with ${\mathrm{\ell }}_y(x)\mathrm{\ne }{\mathrm{\ell }}_y(y)$. The open sets $U_y=\{z:\mathrm{\mid }{\mathrm{\ell }}_y(z)-{\mathrm{\ell }}_y(y)\mathrm{\mid }<\mathrm{\mid }{\mathrm{\ell }}_y(x)-{\mathrm{\ell }}_y(y)\mathrm{\mid }\mathrm{/}3\}$ cover $K$; compactness gives a finite subcover $U_{y_1},\dots ,U_{y_q}$. For every $z\in K$, some $i$ has $z\in U_{y_i}$, which forces ${\mathrm{\ell }}_{y_i}(z)\mathrm{\ne }{\mathrm{\ell }}_{y_i}(x)$. Thus the finite coordinate map $T_x=({\mathrm{\ell }}_{y_1},\dots ,{\mathrm{\ell }}_{y_q})$ satisfies $T_x(x)\mathrm{\notin }{T}_x(K)$. The cylinder $F_x=T_x^{-1}(T_x(K))$ is closed, contains $K$, and excludes $x$, since $T_x(K)$ is compact. The open complements $B\setminus F_x$ cover $B\setminus K$. By [F5] this cover has a countable subcover: for every basis element meeting $B\setminus K$ and contained in a cover member, countable choice selects one such member; every point is covered by one selected member. If the subcover is finite, repeat one member to index it by $j\ge 1$.

If $m=0$, then $T(W)=0$ almost surely, so $\mu (F)$ is zero or one; the endpoint convention, $0\in H_1$, and inner regularity [F4] give the claimed inequality. If $m>0$, define $E=S^{-1}(M)\subset Q$. It is closed and $\mu (F)={\gamma }_Q(E)$. For $r\ge 0$, $E_r=E+r(Q\cap H_1)$ is closed because its second summand is compact. Thus [F4] identifies its inner and ordinary Gaussian measures, and [F1] gives ${\gamma }_Q(E_r)\ge \mathrm{\Phi }({\mathrm{\Phi }}^{-1}({\gamma }_Q(E))+r)$. Let $C_r=T^{-1}(S(E_r))$, a closed finite-coordinate cylinder because $S$ is a linear isomorphism from finite-dimensional $Q$ onto the closed subspace $S(Q)$; injectivity of $S$ and $T(W)=S(G)$ give $\mu (C_r)={\gamma }_Q(E_r)$. If $y\in C_r$, write $T(y)=S(u+v)$ with $u\in E$ and $v\in r(Q\cap H_1)$. Then $T(y-i(v))=S(u)\in M$, so $y-i(v)\in F$ and $y\in F+r{H}_1$. Hence $C_r\subseteq F+r{H}_1$, and inner regularity [F4] gives ${\mu }_{\ast }(F+r{H}_1)\ge \mu (C_r)\ge \mathrm{\Phi }({\mathrm{\Phi }}^{-1}(\mu (F))+r)$.

Let $G_n={\bigcap }_{j\le n}{F}_{x_j}$ for the countable subcover in step 1.2. Each $G_n$ is a closed finite-coordinate cylinder because it is a finite intersection of such cylinders; the sequence decreases, ${\bigcap }_n{G}_n=K$, and $\mu (G_n)\ge \mu (K)>\mathrm{\Phi }(\beta )$. Applying step 2.1 yields ${\mu }_{\ast }(G_n+r{H}_1)\ge \mathrm{\Phi }(\beta +r)$. The set $G_n+r{H}_1$ is closed: if $a_j+h_j$ converges with $a_j\in G_n$ and $h_j\in r{H}_1$, compactness [F3] gives a convergent subsequence of $h_j$, after which closedness of $G_n$ gives a limit representation. Thus [F4] gives equality of its inner and ordinary measures.

The closed sets $G_n+r{H}_1$ decrease and their intersection is $K+r{H}_1$. Indeed, if $x=a_n+h_n$ with $a_n\in G_n$ and $h_n\in r{H}_1$ for every $n$, compactness [F3] gives a subsequence $h_{n_j}\to h\in r{H}_1$; then $a_{n_j}\to x-h$, and for each fixed $N$ the tail belongs to the closed set $G_N$, so $x-h\in K$. The reverse inclusion is immediate. Continuity from above [F6] gives $\mu (K+r{H}_1)\ge \mathrm{\Phi }(\beta +r)$. Since $K\subseteq A$, ${\mu }_{\ast }(A+r{H}_1)\ge \mathrm{\Phi }(\beta +r)$. Letting $\beta \uparrow {\mathrm{\Phi }}^{-1}(\mu (A))$ proves the interior-probability case. If $\mu (A)=0$ the right side is zero; if $\mu (A)=1$, inner regularity [F4] and $0\in H_1$ give ${\mu }_{\ast }(A+r{H}_1)=1$.