The Effros Borel structure

Written by Claude Opus 5.5 (Anthropic), September 2026. Self-checked by the writing AI. Original text: CC0 1.0.

A direct integral of von Neumann algebras is assembled from a family $\gamma\mapsto M(\gamma)$ of von Neumann algebras that depends measurably on a parameter $\gamma$. To make sense of this, one needs a Borel structure on a set of von Neumann algebras, and workable tests for measurability. This lesson supplies both.

We give a Borel structure, the Effros Borel structure, to three sets: the weak*-closed subspaces of the dual of a separable Banach space, the closed subspaces of a separable Hilbert space, and the von Neumann algebras acting on a fixed separable Hilbert space. We show that these Borel spaces are standard, that their elements can be chosen in a Borel way, and that adjoints, commutants, intersections, joins and the set of factors are Borel. We then embed every measurable field of Hilbert spaces isometrically into one fixed space, with ranges that depend measurably on the point. With this we show that commutants, centres, intersections and joins of measurably generated families of von Neumann algebras are again measurably generated. These are the measurability facts that direct integrals of von Neumann algebras need.

Sections 1–5 need only basic functional analysis and metric topology. Sections 6, 7 and 10 use measurable fields of Hilbert spaces and of operators, as developed in Measurable fields of Hilbert spaces and their direct integrals, and a few facts from Decomposable operators and the diagonal algebra. Sections 8–10 use the predual of $B(\mathcal K)$ and its $\sigma$-weak topology. The facts used without proof are stated in full below. No measure is used anywhere: every result holds over an arbitrary measurable space.

Effros introduced the Borel space of von Neumann algebras on a separable Hilbert space in 1965 [Effros 1965]. Basic references are [Effros 1965] and [Takesaki I].

Conventions

• $\mathbb K$ is $\mathbb R$ or $\mathbb C$, and Banach spaces are over $\mathbb K$. Hilbert spaces are complex, inner products are linear in the first variable, and the zero space is allowed. "Separable" includes finite-dimensional.

• For a Banach space $E$, $E^*$ is its dual with the weak* topology, $E^*_1$ is the closed unit ball of $E^*$, and $d(x,A)=\inf\{\|x-a\|:a\in A\}$. For $S\subseteq E$, $S^\perp=\{f\in E^*:f=0\text{ on }S\}$. For $T\subseteq E^*$, $T_\perp=\{x\in E:f(x)=0\text{ for all }f\in T\}$.

• $(\Gamma,\Sigma)$ is a measurable space, and measurable means $\Sigma$-measurable. A map into a topological space is measurable if the preimage of every Borel set lies in $\Sigma$. No standardness of $\Gamma$ is needed, and no measure is needed either.

• A measurable field of Hilbert spaces over $(\Gamma,\Sigma)$ is a family $(H(\gamma))_{\gamma\in\Gamma}$ of Hilbert spaces together with a linear space $\mathfrak M$ of sections (maps $\xi$ with $\xi(\gamma)\in H(\gamma)$ for every $\gamma$), called measurable sections, such that:

  • (F1) $\gamma\mapsto\|\xi(\gamma)\|$ is measurable for every $\xi\in\mathfrak M$;
  • (F2) a section $\eta$ belongs to $\mathfrak M$ whenever $\gamma\mapsto\langle\eta(\gamma),\xi(\gamma)\rangle$ is measurable for every $\xi\in\mathfrak M$;
  • (F3) some sequence in $\mathfrak M$, called fundamental, has values whose linear span is dense in $H(\gamma)$ for every $\gamma$.

  A family of bounded operators $x(\gamma):H(\gamma)\to K(\gamma)$ between two measurable fields is a measurable field of bounded operators if $\gamma\mapsto x(\gamma)\xi(\gamma)$ is a measurable section for every measurable section $\xi$.

• The lesson Measurable fields of Hilbert spaces and their direct integrals develops these fields over a measure space $(\Gamma,\Sigma,\mu)$, but nothing there before the direct integral itself uses $\mu$. So we use those results over $(\Gamma,\Sigma)$; formally this is the case $\mu=0$. The same applies to what we use from Decomposable operators and the diagonal algebra: the matrix-unit fields, the matrix-entry criterion for operator fields on a constant field, and two of its example fields.

• For a Hilbert space $\mathcal K$, $B(\mathcal K)_1$ is the closed unit ball of $B(\mathcal K)$, and $B(\mathcal K)_*$ is the space of $\sigma$-weakly continuous linear functionals on $B(\mathcal K)$.

• We call a $*$-subalgebra $M\subseteq B(\mathcal K)$ with $M''=M$ a von Neumann algebra on $\mathcal K$, and a factor if moreover $M\cap M'=\mathbb C1$. For a set $S$ with $S^*=S$, $S''$ is the von Neumann algebra generated by $S$. The von Neumann algebra generated by an arbitrary set $S$ is $(S\cup S^*)''$, and $M\vee N=(M\cup N)''$. On the zero space, $B(0)=\{0\}=\mathbb C1$ is the only von Neumann algebra, and it is a factor by this definition. Every commutant is $\sigma$-weakly closed, because $x\mapsto xa-ax$ is $\sigma$-weakly continuous for each $a$; so every von Neumann algebra is $\sigma$-weakly closed. The bicommutant theorem is not needed.

Background used without proof

Let $E$ be a Banach space and $\mathcal K$ a Hilbert space.

• (D1) (Hahn–Banach theorem) Let $q$ be a seminorm on a vector space $W$ over $\mathbb K$, $V\subseteq W$ a subspace and $g:V\to\mathbb K$ linear with $|g|\leq q$ on $V$. Then $g$ extends to a linear functional on $W$ with $|g|\leq q$. Over $\mathbb R$ it is enough that $g\leq q$, and the extension satisfies $g\leq q$. This is Theorems 2.1 and 2.2 of Hahn–Banach, Baire and the basic theorems on Banach spaces.

• (D2) (Annihilators) For $S\subseteq E$, $(S^\perp)_\perp$ is the norm-closed linear span of $S$. For $T\subseteq E^*$, $(T_\perp)^\perp$ is the weak*-closed linear span of $T$. This is Theorem 5.1 of Weak topologies: Tychonoff, Banach–Alaoglu, Mazur, bipolars, Krein–Milman and Eberlein–Šmulian, applied to the linear spans.

• (D3) (Banach–Alaoglu theorem) $E^*_1$ is weak*-compact. This is Theorem 3.1 of Weak topologies: Tychonoff, Banach–Alaoglu, Mazur, bipolars, Krein–Milman and Eberlein–Šmulian.

• (P1) Every $\psi\in B(\mathcal K)_*$ has the form $\psi(x)=\sum_n\langle x\xi_n,\eta_n\rangle$ with $\sum\|\xi_n\|^2<\infty$ and $\sum\|\eta_n\|^2<\infty$.

• (P2) $B(\mathcal K)_*$ is a Banach space whose dual is $B(\mathcal K)$, and the weak* topology on $B(\mathcal K)$ is the $\sigma$-weak topology.

• (P3) On $B(\mathcal K)_1$ the $\sigma$-weak and weak operator topologies coincide.

  For (P1)–(P3) see [Blackadar, Sections I.8.5 and I.8.6]. The proofs below use only (P1) and (P2); (P3) links the $\sigma$-weak topology with the weak operator topology used for operator fields.

• (R) (Reduced algebras, used only in Exercise 5) If $M$ is a von Neumann algebra on $\mathcal K$ and $p\in M$ is a projection, then $pMp$, acting on $p\mathcal K$, is a von Neumann algebra. This is proved in The double commutation theorem.

Elementary facts from Hilbert space and metric space theory are used without comment: the Riesz representation of bounded functionals on a Hilbert space, Parseval's identity, the polarization identity, the compactness of $[0,1]^{\mathbb N}$, and the equivalence of all norms on a finite-dimensional space.

1. Borel spaces and Polish spaces

Definitions 1.1.

• A Borel space consists of a set $X$ and a $\sigma$-algebra $\mathcal B$ of subsets of $X$, whose members are called Borel sets. A subset $Y\subseteq X$ carries the relative Borel structure $\{B\cap Y:B\in\mathcal B\}$.
• A map $f:X\to Y$ of Borel spaces is Borel if $f^{-1}(B)$ is Borel for every Borel $B\subseteq Y$. A Borel isomorphism is a bijection $f$ such that $f$ and $f^{-1}$ are Borel.
• $\sigma(\mathcal S)$ is the $\sigma$-algebra generated by a family $\mathcal S$ of sets. For maps $f_i:X\to Y_i$ into Borel spaces, $\sigma(f_i:i\in I)$ is the smallest $\sigma$-algebra that makes every $f_i$ Borel. A family of Borel sets is generating if it generates $\mathcal B$, and separating if for any two distinct points some member contains one of them and not the other. $X$ is countably generated or countably separated if it has a countable family of that kind. A stronger condition asks for a single countable family that both generates and separates. Proposition 3.2(5) supplies such a family, so its conclusion holds in either sense.
• The Borel structure of a topological space is the $\sigma$-algebra generated by its open sets.
• A topological space is Polish when its topology comes from a complete metric and it has a countable dense subset. A Borel space is standard when some Polish space, with its Borel sets, is Borel isomorphic to it.
• A measure on $(X,\mathcal B)$ is a countably additive map $\mu:\mathcal B\to[0,\infty]$. It is standard if $X\setminus N$, with its relative Borel structure, is standard for some Borel set $N$ with $\mu(N)=0$.

Example 1.2. A Borel map need not carry Borel sets to Borel sets. The identity map from $\mathbb R$ with its Borel sets to $\mathbb R$ with the $\sigma$-algebra of countable and co-countable sets is Borel, since every set of the second kind is Borel. It carries the Borel set $[0,1]$ to a set that is neither countable nor co-countable.

Lemma 1.3.

1. (Testing on generators) Suppose $\mathcal B=\sigma(f_i:i\in I)$. A map $g:\Gamma\to X$ is measurable if and only if every $f_i\circ g$ is measurable.
2. (Countable bases) If a topological space has a countable base $\mathcal U$, every open set is a countable union of members of $\mathcal U$, so the Borel sets are $\sigma(\mathcal U)$. The Borel structure of a product of countably many second countable spaces is generated by the coordinate maps. In particular, for a countable set $D$, the Borel structure of $\mathbb R^D$ is generated by the coordinates.
3. (A stock of Polish spaces) Closed subsets and countable products of Polish spaces are Polish, and so is the disjoint union of two Polish spaces, each piece open in the union. The space $\mathbb N^{\mathbb N}$ is Polish. A separable metrizable space, and hence every standard Borel space, has at most $2^{\aleph_0}$ points.
4. (Pieces) Let $\Gamma$ be the union of countably many sets $\Gamma_d\in\Sigma$. A map $g$ on $\Gamma$ whose restriction to every $\Gamma_d$ is measurable for the relative $\sigma$-algebra is measurable.

Proof. (1) The sets $B\subseteq X$ with $g^{-1}(B)\in\Sigma$ form a $\sigma$-algebra. If every $f_i\circ g$ is measurable, it contains every $f_i^{-1}(C)$ with $C$ Borel, so it contains $\mathcal B$. The converse is clear.

(2) Let $V$ be open. For each $x\in V$ pick $U_x\in\mathcal U$ with $x\in U_x\subseteq V$. Only countably many distinct sets $U_x$ occur, and their union is $V$. For a product $\prod_kX_k$ with countable bases $\mathcal U_k$, the finite intersections of sets $\pi_k^{-1}(U)$, $U\in\mathcal U_k$, form a countable base. Each of them lies in $\sigma(\pi_k:k)$, and the coordinates are continuous, so this $\sigma$-algebra is the Borel one.

(3) A closed subset of a complete metric space is complete, and a subset of a separable metric space is separable. For Polish spaces $X_k$, choose complete compatible metrics $d_k\leq1$ (replace $d_k$ by $\min(d_k,1)$). Then $\rho(x,y)=\sum_k2^{-k}d_k(x_k,y_k)$ is a metric for the product topology. A $\rho$-Cauchy sequence is Cauchy in every coordinate, so it converges in every coordinate, and then in $\rho$, because the tails of the series are uniformly small. The points that agree with a fixed point outside finitely many coordinates, and lie in fixed countable dense sets inside them, form a countable dense set. For two Polish spaces, use their metrics bounded by $1$ inside each piece and distance $1$ across. $\mathbb N^{\mathbb N}$ is a countable product of copies of the complete separable discrete space $\mathbb N$. Finally, if $Q$ is a countable dense set in a separable metrizable space, sending each point to a sequence in $Q$ that converges to it is injective, and there are at most $|Q^{\mathbb N}|\leq2^{\aleph_0}$ such sequences.

(4) For Borel $B$, $g^{-1}(B)=\bigcup_d(g|_{\Gamma_d})^{-1}(B)$, and each term is a subset of $\Gamma_d$ lying in the relative $\sigma$-algebra, hence in $\Sigma$. $\square$

2. Polish subspaces and Borel subsets

Theorem 2.1 (Polish subspaces). Let $X$ be a Polish space.

1. Every $G_\delta$ subset of $X$ is Polish in the relative topology.
2. Conversely, a subset of $X$ that is Polish in the relative topology is a $G_\delta$ subset of $X$.
3. Every Polish space is homeomorphic to some $G_\delta$ subset of $[0,1]^{\mathbb N}$, and every such subset is Polish.

Proof. Fix a complete compatible metric $d$ on $X$.

(1) Let $Y=\bigcap_nG_n$ with $G_n$ open. Drop every $G_n$ equal to $X$; if none is left, $Y=X$. For $y\in Y$ put $h_n(y)=1/d(y,X\setminus G_n)$, a positive continuous function on $G_n$, and $$d_Y(y,y')=d(y,y')+\sum_n2^{-n}\min\big(1,|h_n(y)-h_n(y')|\big).$$ This is a metric on $Y$, and it defines the relative topology: $d_Y\geq d$, and if $d(y_k,y)\to0$ inside $Y$, then $h_n(y_k)\to h_n(y)$ for each $n$, so the series tends to $0$ term by term under the bounds $2^{-n}$. Let $(y_k)$ be $d_Y$-Cauchy. It is $d$-Cauchy, so it converges in $X$ to some $x$. For each $n$, $(h_n(y_k))_k$ is Cauchy, hence bounded by some $c_n$. Then $d(y_k,X\setminus G_n)\geq1/c_n$ for all $k$, so $d(x,X\setminus G_n)\geq1/c_n>0$ and $x\in G_n$. Hence $x\in Y$ and $d_Y(y_k,x)\to0$. Finally $Y$ is separable, being a subset of a separable metric space.

(2) Let $\rho$ be a complete metric on $Y$ compatible with its relative topology. For $n\geq1$, let $Y_n$ be the set of points $x$ of the closure $\overline Y$ that have an open neighbourhood $U$ in $X$ with $\rho$-diameter of $U\cap Y$ at most $1/n$. Each $Y_n$ is relatively open in $\overline Y$, because a witness $U$ for $x$ is a witness for every point of $U\cap\overline Y$. Also $Y\subseteq Y_n$: for $y\in Y$, the ball $\{\rho(\cdot,y)<1/(3n)\}$ is relatively open in $Y$, so it equals $U\cap Y$ for some open $U\ni y$ in $X$. Conversely, let $x\in\bigcap_nY_n$, with witnesses $U_n$. The open sets $W_n=U_1\cap\cdots\cap U_n\cap\{d(\cdot,x)<1/n\}$ contain $x\in\overline Y$, so we can pick $y_n\in W_n\cap Y$. Then $y_n\to x$ in $X$, and $\rho(y_m,y_n)\leq1/n$ for $m\geq n$, since both lie in $U_n\cap Y$. So $(y_n)$ converges in $(Y,\rho)$ to some $y\in Y$, hence to $y$ in $X$, and $y=x$. Thus $Y=\bigcap_nY_n$. Write $Y_n=G_n\cap\overline Y$ with $G_n$ open in $X$. Since $\overline Y=\bigcap_m\{d(\cdot,\overline Y)<1/m\}$ is a $G_\delta$, so is $Y$.

(3) $[0,1]^{\mathbb N}$ is a compact metrizable space, hence Polish, so its $G_\delta$ subsets are Polish by (1). Conversely, let $X$ be Polish, with compatible metric $d$ and dense sequence $(a_n)$, and put $\varphi(x)=(\min(1,d(a_n,x)))_n$. The map $\varphi$ is continuous. It is injective: for $x\neq y$ put $\delta=\min(1,d(x,y))$ and choose $a_n$ with $d(a_n,x)<\delta/3$; then the $n$-th coordinate of $\varphi(x)$ is below $\delta/3$ and that of $\varphi(y)$ is at least $2\delta/3$. Its inverse is continuous on $\varphi(X)$: if $\varphi(x_k)\to\varphi(x)$ and $0<\varepsilon<1/2$, choose $a_n$ with $d(a_n,x)<\varepsilon$; then eventually $d(a_n,x_k)<\varepsilon$, so $d(x_k,x)<2\varepsilon$. So $\varphi(X)$ is homeomorphic to $X$, hence Polish, hence a $G_\delta$ subset of $[0,1]^{\mathbb N}$ by (2). $\square$

Theorem 2.2 (Borel subsets of standard spaces). Let $(X,\tau)$ be a Polish space and $B\subseteq X$ a Borel set. There is a Polish topology $\tau'\supseteq\tau$ on $X$, with the same Borel sets as $\tau$, in which $B$ is open and closed. Consequently $B$, with its relative Borel structure, is a standard Borel space. The same holds for a Borel subset of any standard Borel space.

The proof uses only part (1) of Theorem 2.1.

Proof. Call a topology $\tau'$ on $X$ admissible if it is Polish, contains $\tau$, and has the same Borel sets as $\tau$.

(a) Open sets. Let $U$ be $\tau$-open, and let $\tau_U$ be the topology generated by $\tau$ together with the set $X\setminus U$. Its open sets are the unions of sets $V$ and $V\cap(X\setminus U)$ with $V\in\tau$. So $(X,\tau_U)$ is the disjoint union of the open subspace $U$ and the closed subspace $X\setminus U$ of $(X,\tau)$, each open in $\tau_U$. Both pieces are Polish: $U$ by Theorem 2.1(1), since an open set is a $G_\delta$, and $X\setminus U$ because closed subsets of Polish spaces are Polish (Lemma 1.3(3)). By the same lemma, their disjoint union is Polish. The new open sets are $\tau$-Borel. So $\tau_U$ is admissible, and $U$ is $\tau_U$-clopen.

(b) Countable joins. Let $\tau_1,\tau_2,\ldots$ be admissible, and let $\tau_\infty$ be the topology generated by their union. The diagonal map $x\mapsto(x,x,\ldots)$ from $(X,\tau_\infty)$ to the Polish space $\prod_n(X,\tau_n)$ is a homeomorphism onto the diagonal $\Delta$, since the preimage of $\pi_n^{-1}(V)$ is $V$ for $V\in\tau_n$. The diagonal is closed: if $y\notin\Delta$, then $y_m\neq y_n$ for some $m,n$; since $\tau$ is Hausdorff, there are disjoint $\tau$-open sets $A\ni y_m$ and $C\ni y_n$, and $\pi_m^{-1}(A)\cap\pi_n^{-1}(C)$ is a neighbourhood of $y$ that misses $\Delta$. So $\tau_\infty$ is Polish, because a closed subset of a Polish space is Polish. It has a countable base consisting of finite intersections of members of countable bases of the $\tau_n$. These are $\tau$-Borel, and a countable base generates the Borel sets (Lemma 1.3(2)). So $\tau_\infty$ has the same Borel sets as $\tau$, and it is admissible.

(c) All Borel sets. Let $\mathcal A$ be the family of sets that are clopen for some admissible topology. It contains $\tau$ by (a), and it is closed under complements. If $B_n\in\mathcal A$ is clopen for an admissible $\tau_n$, then $\bigcup_nB_n$ is open for the $\tau_\infty$ of (b); applying (a) to $(X,\tau_\infty)$ makes it clopen for a topology that is admissible. So $\mathcal A$ is a $\sigma$-algebra containing $\tau$, and it contains every Borel set.

Now let $B$ be clopen for an admissible $\tau'$. Then $B$ is $\tau'$-closed, hence Polish in the relative $\tau'$-topology (Lemma 1.3(3)). The Borel sets of a subspace are the traces of the Borel sets of the whole space: the traces form a $\sigma$-algebra on $B$ containing the relatively open sets, and the sets $C$ whose trace is Borel in $B$ form a $\sigma$-algebra containing the open sets. The $\tau'$-Borel sets are the $\tau$-Borel sets, so the relative Borel structure of $B$ is the Borel structure of a Polish space. For a standard Borel space, transport $B$ to a Polish space by a Borel isomorphism. $\square$

3. The Effros Borel structure

Throughout, $E$ is a separable Banach space, and $\mathfrak W(E^*)$ is the set of weak*-closed linear subspaces of $E^*$. For $F\in\mathfrak W(E^*)$ put $F_1=F\cap E^*_1$ and $$p_F(x)=\sup\{|f(x)|:f\in F_1\},\qquad x\in E. \tag{3.1}$$ This is the norm of $x$ as a functional on $F$.

Definition 3.1. The Effros Borel structure on $\mathfrak W(E^*)$ is $\sigma(F\mapsto p_F(x):x\in E)$.

Proposition 3.2. Let $F\in\mathfrak W(E^*)$, and let $D\subseteq E$ be dense.

1. $p_F$ is a seminorm with $p_F\leq\|\cdot\|$. Hence $|p_F(x)-p_F(y)|\leq\|x-y\|$. Its kernel is $F_\perp$.
2. (Distance formula) $p_F(x)=d(x,F_\perp)$ for every $x\in E$.
3. (Domination) A linear functional $f$ on $E$, not assumed continuous, lies in $F_1$ if and only if $|f(x)|\leq p_F(x)$ for every $x\in E$. For $\mathbb K=\mathbb R$ this is the same as $f\leq p_F$.
4. (Subspaces of $E$) $F\mapsto F_\perp$ is a bijection of $\mathfrak W(E^*)$ onto the set $\mathcal S(E)$ of closed linear subspaces of $E$, with inverse $N\mapsto N^\perp$. It carries $p_F$ to $d(\cdot,F_\perp)$.
5. $F$ is determined by the values $p_F(x)$, $x\in D$, and the Effros structure equals $\sigma(F\mapsto p_F(x):x\in D)$. If $D$ is countable, the sets $\{F:p_F(x)<r\}$, $x\in D$, $r\in\mathbb Q$, are countably many Borel sets that generate and separate. So $\mathfrak W(E^*)$ is countably generated and countably separated.
6. If $\Phi,\Psi:X\to\mathfrak W(E^*)$ are Borel maps on a Borel space $X$, the set $\{x:\Phi(x)=\Psi(x)\}$ is Borel. In particular, points of $\mathfrak W(E^*)$ are Borel sets, and so is the fixed-point set of any Borel map $\mathfrak W(E^*)\to\mathfrak W(E^*)$.
7. (The unit ball) $E^*_1$ with the weak* topology is a compact metrizable space. For countable dense $D$, its Borel structure is $\sigma(f\mapsto f(x):x\in D)$. A map $a:\Gamma\to E^*_1$ is measurable if and only if $\gamma\mapsto a(\gamma)(x)$ is measurable for every $x\in D$, or equivalently for every $x\in E$.
8. (Norm Borel structure) Let $(\ell_j)$ be a sequence in $E^*_1$ with $\|x\|=\sup_j|\ell_j(x)|$ for all $x$. Then the Borel structure of the norm topology of $E$ is $\sigma(\ell_j:j\geq1)$. So a map $y:\Gamma\to E$ is measurable if and only if every $\ell_j\circ y$ is.

Proof. (1) $p_F$ is a supremum of the seminorms $|f(\cdot)|$ with $\|f\|\leq1$. The Lipschitz bound follows from $p_F(x)\leq p_F(y)+p_F(x-y)$ and the same with $x,y$ exchanged. $p_F(x)=0$ means $f(x)=0$ for $f\in F_1$, hence for all $f\in F$ by scaling; so the kernel is $F_\perp$.

(2) For $f\in F_1$ and $y\in F_\perp$, $|f(x)|=|f(x-y)|\leq\|x-y\|$, so $p_F(x)\leq d(x,F_\perp)$. For the reverse, let $x\neq0$ (the case $x=0$ is trivial). $q=d(\cdot,F_\perp)$ is a seminorm, and the functional $\lambda x\mapsto\lambda q(x)$ on $\mathbb Kx$ satisfies $|\lambda q(x)|=q(\lambda x)$. By (D1) it extends to a linear $g$ on $E$ with $|g|\leq q\leq\|\cdot\|$. Then $g\in E^*_1$, and $g$ vanishes on $F_\perp$, so $g\in(F_\perp)^\perp=F$ by (D2). Hence $p_F(x)\geq|g(x)|=q(x)$.

(3) Elements of $F_1$ satisfy the bound by (3.1). Conversely, if $|f|\leq p_F\leq\|\cdot\|$, then $f\in E^*_1$ and $f$ vanishes on $\ker p_F=F_\perp$, so $f\in(F_\perp)^\perp=F$ by (D2). For $\mathbb K=\mathbb R$, $f\leq p_F$ also gives $-f(x)=f(-x)\leq p_F(x)$.

(4) $F_\perp$ is a closed subspace, and $(F_\perp)^\perp=F$ by (D2). For $N\in\mathcal S(E)$, $N^\perp$ is weak*-closed and $(N^\perp)_\perp=N$ by (D2). The last claim is (2).

(5) $p_F$ is continuous by (1), so its values on $D$ determine it, and $p_F$ determines $F=(\ker p_F)^\perp$ by (1) and (4). For $x\in E$ choose $x_j\in D$ with $x_j\to x$. Then $p_F(x)=\lim_jp_F(x_j)$ for every $F$, so $F\mapsto p_F(x)$ is measurable for $\sigma(F\mapsto p_F(y):y\in D)$. The sets $\{F:p_F(x)<r\}$ generate this $\sigma$-algebra, and they separate points, because $F$ is determined by $p_F$ on $D$.

(6) Take a countable dense $D$. By (5), $\Phi(x)=\Psi(x)$ exactly when $p_{\Phi(x)}(y)=p_{\Psi(x)}(y)$ for all $y\in D$. Each of these countably many conditions defines a Borel set. For points, take $\Phi$ the identity and $\Psi$ constant; for fixed points, take $\Psi$ the identity.

(7) Let $D=\{x_1,x_2,\ldots\}$. On $E^*_1$ the weak* topology is the weakest topology making the maps $f\mapsto f(x_k)$ continuous: if these values converge along a net $(f_i)$ to those of $f$, then $|f_i(x)-f(x)|\leq|f_i(x_k)-f(x_k)|+2\|x-x_k\|$, so $f_i(x)\to f(x)$ for all $x$. The metric $\sum_k2^{-k}\min(1,|f(x_k)-g(x_k)|)$ defines this topology, and $E^*_1$ is compact by (D3). The topology has a countable base of sets defined by finitely many conditions $|f(x_k)-c|<r$, with $c$ Gaussian rational and $r$ rational. Since a countable base generates the Borel sets (Lemma 1.3(2)), the Borel sets are $\sigma(f\mapsto f(x_k):k)$. The measurability test follows by testing on generators (Lemma 1.3(1)); values at points outside $D$ are limits of values on $D$.

(8) Each $\ell_j$ is continuous. Conversely, an open ball $\{x:\|x-x_0\|<r\}$ equals $\bigcup_m\bigcap_j\{x:|\ell_j(x)-\ell_j(x_0)|\leq r-1/m\}$, which lies in $\sigma(\ell_j:j)$. $E$ is separable, so every open set is a countable union of balls centred at points of a countable dense set. The last clause follows by testing on generators (Lemma 1.3(1)). $\square$

4. A Polish topology and standardness

We now show that the Effros Borel structure is standard, by exhibiting a Polish topology that generates it.

Theorem 4.1. Let $\tau_E$ be the weakest topology on $\mathfrak W(E^*)$ for which every function $F\mapsto p_F(x)$, $x\in E$, is continuous. Then $(\mathfrak W(E^*),\tau_E)$ is a Polish space, and its Borel sets are exactly the Effros Borel sets. In particular, the Effros Borel structure is standard.

Proof. Let $D\subseteq E$ be a countable dense subset that is a vector space over $\mathbb Q$ (over $\mathbb Q+i\mathbb Q$ if $\mathbb K=\mathbb C$), for instance the rational span of a dense sequence.

The topology. $\tau_E$ is also the weakest topology making $F\mapsto p_F(x)$ continuous for $x\in D$: for $x\in E$ and $x_j\in D$ with $x_j\to x$, the function $F\mapsto p_F(x)$ is the uniform limit of the functions $F\mapsto p_F(x_j)$, because every $p_F$ is $1$-Lipschitz (Proposition 3.2(1)). The map $\Theta(F)=(p_F(x))_{x\in D}\in\mathbb R^D$ is injective, because $F$ is determined by the values of $p_F$ on $D$ (Proposition 3.2(5)). So $\Theta$ is a homeomorphism of $(\mathfrak W(E^*),\tau_E)$ onto its image $\Theta(\mathfrak W(E^*))$, with the product topology on $\mathbb R^D$.

Seminorms. Let $S\subseteq\mathbb R^D$ be the set of $c$ with $c(x)\geq0$, $|c(x)-c(y)|\leq\|x-y\|$, $c(x+y)\leq c(x)+c(y)$ and $c(qx)=|q|c(x)$ for all $x,y\in D$ and all rational (Gaussian rational) $q$. Each condition involves finitely many coordinates, so $S$ is closed. The elements of $S$ are exactly the restrictions to $D$ of the seminorms $p$ on $E$ with $p\leq\|\cdot\|$. Such a restriction satisfies the conditions. Conversely, $c\in S$ is $1$-Lipschitz on $D$, so it extends uniquely to a $1$-Lipschitz function $p$ on $E$; by continuity $p$ is subadditive, $p(\lambda x)=|\lambda|p(x)$ for all scalars $\lambda$, and $p\leq\|\cdot\|$, since $p(0)=0$.

The condition (W). Consider, for $c\in S$:

(W) for every $x\in D$ and every rational $\varepsilon>0$ there is $y\in D$ with $c(y)<\varepsilon$ and $\|x-y\|<c(x)+\varepsilon$.

For fixed $x,\varepsilon,y$, the set $\{c:c(y)<\varepsilon,\ c(x)>\|x-y\|-\varepsilon\}$ is open in $\mathbb R^D$. So the set of $c\in S$ that satisfy (W) is a $G_\delta$ subset of $\mathbb R^D$.

Claim: for $c=p|_D\in S$, (W) holds if and only if $p(x)=d(x,\ker p)$ for all $x\in E$. First, $p(x)=p(x-z)\leq\|x-z\|$ for $z\in\ker p$, so $p\leq d(\cdot,\ker p)$ always. If $p=d(\cdot,\ker p)$, let $x\in D$ and $\varepsilon>0$. Pick $z\in\ker p$ with $\|x-z\|<p(x)+\varepsilon/2$ and $y\in D$ with $\|y-z\|<\varepsilon/2$. Then $c(y)=p(y)\leq p(z)+\|y-z\|<\varepsilon$ and $\|x-y\|<p(x)+\varepsilon$. Conversely, assume (W), and let $x\in D$ and $\varepsilon>0$ be rational. Applying (W) first to $x$ and $\varepsilon/2$, and then to each $y_j$ and $2^{-j-1}\varepsilon$, choose $y_1,y_2,\ldots\in D$ with $$p(y_1)<\varepsilon/2,\quad\|x-y_1\|<p(x)+\varepsilon/2,\qquad p(y_{j+1})<2^{-j-1}\varepsilon,\quad\|y_j-y_{j+1}\|<p(y_j)+2^{-j-1}\varepsilon<2^{1-j}\varepsilon .$$ The sequence $(y_j)$ is Cauchy, so it converges to some $z\in E$. Then $p(z)=\lim p(y_j)=0$, and $\|x-z\|<p(x)+\varepsilon/2+\sum_{j\geq1}2^{1-j}\varepsilon=p(x)+5\varepsilon/2$. Thus $d(x,\ker p)\leq p(x)$ for $x\in D$, hence for all $x\in E$, because both sides are continuous.

The image. If $F\in\mathfrak W(E^*)$, then $\Theta(F)\in S$, and $p_F=d(\cdot,F_\perp)=d(\cdot,\ker p_F)$ by the distance formula and the description of the kernel in Proposition 3.2(1)–(2); so $\Theta(F)$ satisfies (W). Conversely, let $c=p|_D\in S$ satisfy (W). Put $N=\ker p$, a closed subspace, and $F=N^\perp\in\mathfrak W(E^*)$. Then $F_\perp=N$ by Proposition 3.2(4), and $p_F=d(\cdot,N)=p$ by the distance formula and the claim. So $c=\Theta(F)$. Hence the image of $\Theta$ is the $G_\delta$ set above.

Conclusion. The space $\mathbb R^D$ is Polish, as a countable product of copies of $\mathbb R$ (Lemma 1.3(3)). The image is a $G_\delta$ subset of it, hence Polish by Theorem 2.1(1), and so is $(\mathfrak W(E^*),\tau_E)$. Its Borel sets are the preimages under $\Theta$ of the relative Borel sets of the image, which are generated by the coordinates (Lemma 1.3(2)). So they form $\sigma(F\mapsto p_F(x):x\in D)$, which is the Effros structure by Proposition 3.2(5). $\square$

The claim used only the metric of $E$ and its completeness. The same argument therefore handles closed sets in any complete separable metric space.

Corollary 4.2 (spaces of closed sets). Let $(X,d)$ be a complete separable metric space, and $\mathcal C_0(X)$ the set of nonempty closed subsets of $X$. The weakest topology on $\mathcal C_0(X)$ that makes every function $A\mapsto d(x,A)$, $x\in X$, continuous is Polish. Its Borel sets form $\sigma(A\mapsto d(x,A):x\in X)$, and this $\sigma$-algebra is also generated by the sets $\{A:A\cap U\neq\emptyset\}$, $U$ open. In particular, either description gives a standard Borel structure on $\mathcal C_0(X)$.

This topology is also called the Wijsman topology.

Proof. We may assume $X\neq\emptyset$, since otherwise $\mathcal C_0(X)$ is empty. Let $D$ be a countable dense subset of $X$, and $S'\subseteq\mathbb R^D$ the closed set of $c\geq0$ with $|c(x)-c(y)|\leq d(x,y)$ for $x,y\in D$; each $c\in S'$ extends to a $1$-Lipschitz function $f\geq0$ on $X$. Let (W') be (W) with $d(x,y)$ in place of $\|x-y\|$. As before, the set of $c\in S'$ satisfying (W') is a $G_\delta$. If $f=d(\cdot,A)$ with $A\in\mathcal C_0(X)$, then (W') holds: pick $a\in A$ with $d(x,a)<f(x)+\varepsilon/2$ and $y\in D$ with $d(y,a)<\varepsilon/2$. Conversely, if (W') holds, the iteration in the proof of Theorem 4.1 produces, for $x\in D$ and rational $\varepsilon>0$, a limit point $z$ with $f(z)=0$ and $d(x,z)<f(x)+5\varepsilon/2$; it converges because $X$ is complete. So $A=f^{-1}(0)$ is nonempty and closed, and $d(\cdot,A)\leq f$ on $D$, hence everywhere. Also $f(x)\leq f(a)+d(x,a)=d(x,a)$ for $a\in A$, so $f=d(\cdot,A)$. A closed set is the zero set of its distance function, so $A\mapsto(d(x,A))_{x\in D}$ is injective, and it is a homeomorphism onto the $G_\delta$ set just described, because $|d(x,A)-d(x',A)|\leq d(x,x')$. The rest is as in the proof of Theorem 4.1.