In this post, I would like to share an idea that finally leads to a breakthrough in my research work. My main reference book ^[1] is the two volumes monograph Measure Theory by V.I.Bogachev. Any cited propositions in the following are meant to be found in this book.

Problem setting

Let $X$ be a space equipped with various topologies. For example, $X$ might be a space of functions or functionals, where different metrics exist. Fix one possible topology and a Borel measure $\mu$ with respect to it.

Recall that a Borel measure $\mu$ on X is called a Radon measure if for $B \in \mathcal{B}(X)$ and $\varepsilon > 0$, there exists a compact set $K_\varepsilon \subset B$ such that $|\mu|(B\setminus K_{\varepsilon})<\varepsilon$. It is well-known that, stated as Theorem 7.1.7 in the book, any finite Borel measure on a complete and separable space is Radon. Hence, there is usually no trouble to obtain a non-trivial compact set for a given Borel measure $\mu$.

Here is the main problem today. Given a Borel measure $\mu$ and Borel set $B \subset X$ with respect to a given topology $(X, \tau_1)$, is it possible to obtain a non-trivial compact subset of $B$ with respect to another given topology $(X, \tau_2)$? Regardless possible topological assumptions on $(X, \tau_1)$ or $(X, \tau_2)$, there are two things to resolve if we want to apply the previous Radon measure trick:

1. $\mu$ must be a Borel measure with respect to the topology $(X, \tau_2)$;
2. $B$ must be a Borel set of $(X, \tau_2)$.

We introduce the concept of Souslin spaces to solve this problem by giving mild assumptions that imply the identity between the two Borel $\sigma$-algebras generated by $(X, \tau_1)$ and $(X, \tau_2)$, which will denoted by $\sigma_1$ and $\sigma_2$. In other words, we shall use Souslin spaces to discuss that under what conditions we will have $\sigma_1 := \sigma(\tau_1)$ equals to $\sigma_2 := \sigma(\tau_2)$.

Souslin spaces ^[2]

The name Souslin might be written as Suslin, which is the case on Wikipedia. While Souslin was a research student of Luzin, at the age of 22 or 23 ^[3], he found an error in an argument of Lebesgue ^[4], who believed he had proved that for any Borel set in $\mathbb{R}^{2}$, the projection onto the real axis was also a Borel set. It is clear that this discovery inspires nowadays definition of Souslin space.

Definitions

A set in a Hausdorff space is called Souslin if it is the image of a complete separable metric space under a continuous mapping. A Souslin space is a Hausdorff space that is a Souslin set. The empty set is Souslin as well.

Souslin sets are also called analytic sets. The complement of a Souslin set in a Souslin space is called co-Souslin or coanalytic.

This definition looks ordinary and less powerful as expected. But as happened frequently in the domain of measure theory, we also have an unexpected equivalent definition for it. It is stated in Theorem 6.6.8 from the book that every Souslin set in a Hausdorff space can be obtained from closed sets by means of the $\mathcal{A}$-operation. The rough idea is that $\mathcal{A}$-operation includes at least countable unions and intersections. We shall show this equivalence in the next section.

Properties

Let's sketch the proofs of some interesting properties of the Souslin space. To simplify our arguments, we introduce the concept of Polish spaces.

A topological space homeomorphic to a complete separable metric space is called Polish Space. The empty set is also included in the class of Polish spaces.

The representation of a Polish space as the continous image from the Baire space $\mathbb{N}^\mathbb{N}$^[5], which is endowed with product topology of the discrete topology on $\mathbb{N}$, is the starting point of classical descriptive set theory. We can always find the spirit of its proof in the theory of Souslin space. Let us define the $\mathcal{A}$-operation to reveal this idea.

Let $X$ be a nonempty set and let $\mathcal{E}$ be some class of its subsets. We say that we are given a Souslin scheme (or a table of sets) $\{A_{n_0, \ldots , n_k}\}$ with values in $\mathcal{E}$ if, to every finite sequence $(n_0, \ldots , n_k)$ of natural numbers, there corresponds a set $A_{n_0,\ldots,n_k} \in \mathcal{E}$. The $\mathcal{A}$-operation (or the Souslin operation) over the class $\mathcal{E}$ is a mapping that to every Souslin scheme $\{A_{n_0,\ldots,n_k}\}$ with values in $\mathcal{E}$ associates the set

$$A = \bigcup_{ (n_i) \in \mathbb{N}^\mathbb{N} } \bigcap_{ k=0 }^ \infty A_{n_0,\ldots,n_k}.$$

The sets of this form are called $\mathcal{E}$-Souslin or $\mathcal{E}$-analytic.

One can use the $\mathcal{A}$-operation to construct the following continuous surjection.

Every nonempty complete separable metric space is the image of $\mathbb{N}^\mathbb{N}$ under a continuous mapping.

Let us equip $\mathbb{N}^\mathbb{N}$ with the metric

$$\varrho(\alpha,\beta)=\sum_{j=0}^{\infty}2^{-j-1}\frac{|n_{j}-m_{j}|} {|n_{j}-m_{j}|+1},\quad\alpha=(n_{j}),\,\beta=(m_{j}),$$

which gives rise to the topology of the Baire space. We represent the given space X in the form $X = \bigcup_{j = 0}^\infty E(j)$ where the sets $E(j)$ are closed (not necessarily disjoint) of diameter less than $2^{-2}$. By induction, for every $k$, we find a closed set $E(n_0, \ldots , n_k)$ of diameter less than $2^{−k−2}$ with

$$E(n_{0},\ldots,n_{k}) = \bigcup_{j=0}^{\infty}E(n_{0},\ldots,n_{k},j).$$

For every $\alpha = (n_i) \in \mathbb{N}^\mathbb{N}$, the closed sets $E(n_0, . . . , n_k)$ are decreasing and have diameters less than $2^{−k−2}$. Hence, they shrink to a single point denoted by $f(\alpha)$. Note that $f(\mathbb{N}^\mathbb{N}) = X$. Indeed, every point $x$ belongs to some set $E(n_0)$, then to $E(n_0, n_1)$ and so on, which yields an element $\alpha$ such that $f(\alpha) = x$. In addition, $f$ is locally Lipschitzian. If $\varrho(\alpha, \beta) < \frac{1}{4}$, then there exists $k$ with $2^{−k−2} ≤ \varrho(\alpha, \beta) < 2^{−k−1}$. Then $\alpha_i = \beta_i$ if $i ≤ k$. Hence $f(\alpha)$ and $f(\beta)$ belong to $E(n_0, \ldots , n_k)$, whence we obtain $\rho( f(\alpha), f(\beta) ) < 2^{−k−2} ≤ \varrho(\alpha, \beta)$, where $\rho$ be the metric in X. Thus, $f$ is continuous.

Let $\mathcal{S}(X)$ denote the class of all Souslin sets in a topological space X, then a. $\mathcal{S}(X)$ includes all open and closed subsets of $X$; b. $\mathcal{S}(X)$ is closed with respect to the $\mathcal{A}$-operation; c. Any element of $\mathcal{S}(X)$ can be written as the $\mathcal{A}$-operation of closed subsets of $X$.

Pleas click to expand any of the following details if you are interested in.

Proof for (a):

We remark that any open or closed subsets of a Polish space is Polish. In fact, to construct a metric on the open subset $X \setminus Y \subset X$ with $Y$ a closed subset of a given Polish metric space $(X, \varrho)$, we simply define

$$\varrho_{0}(x,y)=\varrho(x,y)+\frac{|\mathrm{dist}(x,X\setminus Y)-\mathrm{dist}(y,X\setminus Y)|} {|\mathrm{dist}(x,X\setminus Y)-\mathrm{dist}(y,X\setminus Y)|+1},$$

which turns out to be complete.

Proof for (b):

One can construct infinite product of Polish spaces to show that $\mathcal{S}(X)$ is closed with respect to countable union and intersection. For the general case, we first lift a given Souslin scheme of $\mathcal{S}(X)$ to a monotone disjoint Souslin scheme of $\mathcal{S}(X \times \mathbb{N}^\mathbb{N})$. A scheme $\{ A_{n_0, \ldots, n_k} \}$ is monotone if $A_{n_0,\ldots,n_k} \subset A_{n_0,\ldots,n_{k-1}}$, and it is disjoint if $A_\alpha \cap A_\beta = \emptyset$ for $\alpha \neq \beta \in \mathbb{N}^\mathbb{N}$. We can trivialize the $\mathcal{A}$-operation for monotone disjoint Souslin scheme $\{A_{n_0,\ldots,n_k}\}$ to countable unions and intersections since for this scheme we have

$$\bigcup_{ (n_i) \in \mathbb{N}^\mathbb{N} } \bigcap_{k=0}^\infty A_{n_0,\ldots,n_k} = \bigcap_{k=0}^\infty \bigcup_{ n_i \in \mathbb{N}} A_{n_0,\ldots,n_k}.$$

For our proof, first notice that we can transform any scheme to be monotone without changing the result of $\mathcal{A}$-operation by re-defining

$$A_{n_0,\ldots, n_k} := A_{n_0} \cap A_{n_0, n_1} \cap \ldots \cap A_{n_0, \ldots, n_k} .$$

To render a scheme $A$ disjoint, we define $A^\prime_{n_0,\ldots,n_k} := A_{n_0,\ldots,n_k} \times C_{n_0, \ldots, n_k},$ where $C_{n_0, \ldots, n_k}$ is the set of sequences in $\mathbb{N}^\mathbb{N}$ whose first $k+1$ element is $(n_0, \ldots, n_k)$. Then the $\mathcal{A}$-operation gives us a Souslin set in the space $X \times \mathbb{N}^\mathbb{N}$. Finally, one can get the original $\mathcal{A}$-operation result by projection.

Proof for (c):

Let a set $A$ be the image of the space $\mathbb{N}^\mathbb{N}$ under a continuous mapping $f$. For every finite sequence $n_0, \ldots , n_k$, we denote by $F_{n_0, \ldots ,n_k}$ the closure of $f(C_{n_0,\ldots,n_k})$, where we define $C_{n_0,\ldots,n_k}$ as in the proof of (b):

$$C_{n_{0},\ldots,n_{k}}= \left\{(m_{i}) \in \mathbb{N}^{\mathbb{N}} :\, (m_{0},\ldots,m_{k})=(n_{0},\ldots,n_{k}) \right\}.$$

One can then show that $A = \bigcup_{(n_i) \in \mathbb{N}^\mathbb{N}} \bigcap_{k=0}^\infty F_{n_0,\ldots,n_k}$ by noticing that $f((n_i)) = \bigcap_{k=0}^\infty F_{n_0,\ldots,n_k}$.

We thus finish the proof.

Radon measure and Souslin space

In practice, it is necessary to know the following result, stated as Theorem 7.4.3 in the book.

If $X$ is a Souslin space, then every Borel measure $\mu$ on $X$ is Radon.

Theorems by Souslin or Lusin

After getting a bit acquaintance of Souslin spaces, we now discuss several fondamental and useful results in this theory.

(Lusin's Separation Theorem) Let $X$ be a Hausdorff space and let $A,B \subset X$ be two disjoint Souslin sets. Then there is a Borel set $C \subset X$ separating $A$ from $B$.

We can express $A$ and $B$ as the continuous images of $\mathbb{N}^\mathbb{N}$ under $f$ and $g$. If we cannot separate $A$ and $B$ with Borel sets, it must be that we cannot separate two Souslin subsets, say $f(C_{n_0}) \subset A$ and $g(C_{m_0}) \subset B$, with Borel sets, where $C_i \subset \mathbb{N}^\mathbb{N}$ is the set of sequences starting with $i \in \mathbb{N}$. Continue this argument applying to the two chosen subsets in each step, we will finally get a contradiction that we cannot even separate two points in a Hausdorff space.

Suppose that the complement of a Souslin set $A$ in a Hausdorff space $X$ is Souslin. Then $A$ is a Borel set.

Let $(X, \tau)$ be a Souslin space. If $\tau^\prime \subset \tau$ is another topology on $X$, then $(X, \tau)$ and $(X, \tau^\prime)$ have the same Borel set.

Let $f: (X, \tau) \rightarrow (X, \tau^\prime)$ be the continuous identity map that sends $x \in X$ to $x$. For $B$ a Borel set in $(X, \tau)$, $B = f(B)$ is a Souslin set in $(X, \tau^\prime)$ since $f$ is continuous and $B$, being Souslin in $(X, \tau)$, is the continuous image of a Polish space. For the same reason, $X \setminus B$ is a Souslin set for $(X, \tau^\prime)$, which implies that $B$ is Borel set for $(X, \tau^\prime)$.

Exemplar usage in my research

Let $(E,d)$ be a Polish space, then the Wasserstein-$2$ space $\mathcal{W}_2(E)$ is also a Polish space. Fix a reference measure $\mu$ on $(E, d)$. Define the set $\mathbb{B}: = \mathcal{W}_2(E) \cap P_{\operatorname{ac}}(E)$ of absolutely continuous probability measures in $\mathcal{W}_2(E)$ with respect to $\mu$. Via the identification $f \cdot \mu \in \mathbb{B} \leftrightarrow f$, we consider $\mathbb{B}$ as a subset of the unit sphere of $L^1(\mu)$.

We can consider $\mathbb{B}$ with different topologies. We denote by $(\mathbb{B}, \tau_w)$ the topological space with respect to the weak convergence of probability measures, denote by $(\mathbb{B}, \tau_W)$ the topological subspace of $(\mathcal{W}_2(M), W)$, denote by $(\mathbb{B}, \tau)$ the weak topological space induced by the dual space $L^\infty(\mu)$ of $L^1(\mu)$ and denote by $(\mathbb{B}, \tau_L)$ the topological subspace of the Lebesgue space $L^1(\mu)$. From these definitions, we always have $\tau_w \subset \tau \subset \tau_L$ and $\tau_w \subset \tau_W$. Now it is a good exercise to show that

The four topological spaces $(\mathbb{B}, \tau_w)$, $(\mathbb{B}, \tau_W)$, $(\mathbb{B}, \tau)$ and $(\mathbb{B}, \tau_L)$ have the same Borel sets.

In particular, if $\mathbb{P} \in \mathcal{W}_2(\mathcal{W}_2(E))$ gives mass to the set $\mathbb{B}$, then it gives mass to a compact subset of $(\mathbb{B}, \tau_L)$.

---

1. I share an PDF copy of this book on Google Drive in case you might need. ↩︎

2. One can consult section 6.6 and section 6.7 in the book for a detailed understanding of this concept. ↩︎

3. Souslin died at the age of 24, comparing to the fact that Galois died at the age of 20 and Abel died at the age of 26. ↩︎

4. It is in the proof of Théorème XVIII in his paper Sur les fonctions représentables analytiquement, p.193, Journal de mathématiques pures et appliquées 6e série, tome 1 (1905), p. 139-216. ↩︎

5. The Baire space is homeomorphic to topological subspace of irrationals on (0,1) using the continued fraction. ↩︎