Main result (p. 431) — an amenable discrete measured equivalence relation is, up to a null set, generated by one non-singular transformation

What the statement asserts

Data and standing assumptions

The statement is universally quantified over the following data.

• A space $X$. $X$ is an arbitrary set (of any size, possibly empty) carrying a σ-algebra, whose members are called the measurable sets of $X$.
• $X$ is a standard Borel space. There exists some topology on $X$ that is Polish (second countable and completely metrizable) and whose Borel σ-algebra is exactly the given σ-algebra of $X$.
• A measure $\mu$ on $X$ (countably additive, with values in $[0,\infty]$).
• $\mu$ is σ-finite. There is a sequence $S_0, S_1, S_2, \dots$ of subsets of $X$ with $\mu(S_i) < \infty$ for every $i$ and $\bigcup_i S_i = X$.
• A subset $R \subseteq X \times X$. It is arbitrary until the two hypotheses below are imposed.

Conventions used throughout:

• $X \times X$ carries the product σ-algebra (generated by the sets $B \times X$ and $X \times B$ with $B$ measurable), and $\mathbb{R}$ carries its Borel σ-algebra.
• $\mu$ is also applied to sets that are not known to be measurable. For an arbitrary $s \subseteq X$, $\mu(s)$ means the outer measure:

$$\mu(s) = \inf\{\mu(t) : t \supseteq s,\ t \text{ measurable}\}.$$

• "For $\mu$-almost every $x$, $Q(x)$" means that the set $\{x : Q(x) \text{ fails}\}$ has outer measure $0$. For functions $F, G : X \to \mathbb{R}$, "$F = G$ a.e." means $\mu(\{x : F(x) \neq G(x)\}) = 0$.

Hypothesis 1: $R$ is a discrete measured equivalence relation

All four of the following hold.

1. $R$ is a measurable subset of $X \times X$.
2. The relation $x \sim y \iff (x,y) \in R$ is an equivalence relation on $X$: $(x,x) \in R$ for every $x$; $(x,y) \in R$ implies $(y,x) \in R$; and $(x,y) \in R$ together with $(y,z) \in R$ implies $(x,z) \in R$.
3. For every $x \in X$, the set $\{y \in X : (x,y) \in R\}$ is countable. Here countable means finite or countably infinite.
4. For every measurable $A \subseteq X$ with $\mu(A) = 0$, the set

$$[A]_R = \{x \in X : \text{there is } y \in A \text{ with } (x,y) \in R\}$$

has $\mu([A]_R) = 0$. This set is not assumed to be measurable, so $\mu$ here is the outer measure.

Hypothesis 2: amenability of $R$ with respect to $\mu$

Admissible functions. Call a function $f : X \times X \to \mathbb{R}$ admissible if both of the following hold:

• $f$ is measurable on all of $X \times X$.
• $f$ is bounded on $R$: there is a real $C$ with $|f(p)| \le C$ for every $p \in R$.

No bound is required at points of $(X \times X) \setminus R$.

Partial transformations. A partial transformation of $R$ is a triple $\varphi = (D, E, e)$ consisting of:

• measurable sets $D, E \subseteq X$;
• a bijection $e : D \to E$ such that $e$ and $e^{-1}$ are both measurable. Here $D$ and $E$ carry their subspace σ-algebras, whose members are the sets $B \cap D$, respectively $B \cap E$, for measurable $B \subseteq X$;
• the condition $(a, e(a)) \in R$ for every $a \in D$.

No condition relating $e$ to $\mu$ is imposed. The empty partial transformation $D = E = \varnothing$ is allowed.

A partial transformation $\varphi$ acts on functions as follows. For $f : X \times X \to \mathbb{R}$, define $\varphi \cdot f : X \times X \to \mathbb{R}$ by

$$(\varphi\cdot f)(x,z) = \begin{cases} f\big(e^{-1}(x),\, z\big) & \text{if } x \in E,\\ 0 & \text{if } x \notin E.\end{cases}$$

For $F : X \to \mathbb{R}$, define $\varphi \cdot F : X \to \mathbb{R}$ by

$$(\varphi\cdot F)(y) = \begin{cases} F\big(e^{-1}(y)\big) & \text{if } y \in E,\\ 0 & \text{if } y \notin E.\end{cases}$$

The hypothesis. There exists a map $P$ that assigns to every function $f : X \times X \to \mathbb{R}$ a function $Pf : X \to \mathbb{R}$, satisfying all seven of the following conditions.

1. A.e. measurability. For every admissible $f$, there is a measurable $g : X \to \mathbb{R}$ with $Pf = g$ a.e.
2. Dependence on $f$ only off a null set of first coordinates. Let $f, g$ be admissible. Suppose

$$\mu\Big(\pi_1\big(\{p \in R : f(p) \neq g(p)\}\big)\Big) = 0,$$

where $\pi_1(x,y) = x$ is the first-coordinate projection and $\mu$ is the outer measure. Then $Pf = Pg$ a.e. 3. Additivity. For admissible $f, g$: $P(f+g) = Pf + Pg$ a.e. Sums are pointwise. 4. Homogeneity. For every $c \in \mathbb{R}$ and admissible $f$: $P(cf) = c \cdot Pf$ a.e. Scalar multiples are pointwise. 5. Positivity. For admissible $f$ with $f(p) \ge 0$ for every $p \in R$, we have $Pf(x) \ge 0$ for $\mu$-almost every $x$. 6. Normalization. $P(\mathbf{1}) = \mathbf{1}$ a.e. Here the argument is the constant function $1$ on $X \times X$, and the right-hand side is the constant function $1$ on $X$. 7. Invariance. For every partial transformation $\varphi$ of $R$ and every admissible $f$:

$$P(\varphi\cdot f) = \varphi\cdot(Pf) \quad \text{a.e.}$$

Every conclusion about $P$ holds only $\mu$-almost everywhere. The values $Pf$ for non-admissible $f$ are unconstrained, except that the normalization condition concerns the constant function $1$, which is admissible.

Conclusion

There exist a map $T : X \to X$ and a set $N \subseteq X$, chosen in that order (so $N$ may depend on $T$), such that all of the following hold.

• $T$ is a measurable automorphism. $T$ is a bijection of $X$ onto itself, and both $T$ and $T^{-1}$ are measurable.
• $T$ is nonsingular. $T$ is measurable, and for every measurable $A \subseteq X$,

$$\mu\big(T^{-1}(A)\big) = 0 \iff \mu(A) = 0.$$

This condition is stated for preimages under $T$.

• $N$ is a null set. $N$ is measurable and $\mu(N) = 0$.
• $T$ generates $R$ outside $N$. For all $x, y \in X$ with $x \notin N$ and $y \notin N$,

$$(x,y) \in R \iff \text{there exists } n \in \mathbb{Z} \text{ with } T^n(x) = y.$$

Here $T^0$ is the identity, $T^n$ for $n > 0$ is the $n$-fold composite $T \circ \cdots \circ T$, and $T^{-n} = (T^{-1})^n$ for $n > 0$.

Edge cases and what the quantifiers include

• Pairs touching $N$. The final equivalence constrains only pairs with both $x \notin N$ and $y \notin N$. Nothing is asserted about pairs where $x \in N$ or $y \in N$. In particular:
  • $(x, T(x)) \in R$ is asserted only when both $x$ and $T(x)$ lie outside $N$;
  • $N$ is not asserted to be invariant under $T$ or to be a union of $R$-classes.
• $n = 0$ is allowed. The integer $n$ may be $0$. With $n = 0$ the right-hand side reads $x = y$.
• No measure preservation. $T$ is required only to be nonsingular as above. It need not preserve $\mu$.
• The zero measure. If $\mu = 0$, condition 4 of Hypothesis 1 holds automatically, and so does Hypothesis 2: every a.e. condition is then vacuous, so any $P$, for example $P \equiv 0$, qualifies. Also $N$ is then allowed to be all of $X$, and in that case the final equivalence quantifies over no pairs.
• Empty $X$. $X$ may be empty. Then every quantifier over points is vacuous.
• Finite classes. Countable classes in Hypothesis 1 include finite ones.