Theorem 2.6 — the Invariant Extension Theorem

Statement

Let $G$ be a group (an arbitrary group: no finiteness, countability, topology or discreteness assumption beyond being a group). If $G$ is amenable in the sense below, then $G$ has the invariant extension property in the sense below.

Throughout, $[0,\infty]$ denotes the extended non-negative reals, with $a + \infty = \infty + a = \infty$ for every $a$.

Vocabulary

Finitely additive measure. For a set $Y$, a function $m : \mathcal P(Y) \to [0,\infty]$, defined on every subset of $Y$, is called finitely additive here when

$$m(\varnothing) = 0 \qquad\text{and}\qquad m(S \cup T) = m(S) + m(T) \ \text{ for all } S, T \subseteq Y \text{ with } S \cap T = \varnothing .$$

No normalisation, finiteness or countable additivity is required; the value $\infty$ is allowed.

Translate of a subset. If $G$ acts on a set $Y$ (on the left) and $g \in G$, $S \subseteq Y$, then $gS = \{\, g\cdot y : y \in S \,\}$ is the image of $S$ under $y \mapsto g\cdot y$. For $Y = G$ acting on itself, $g\cdot y = gy$ is left multiplication, so $gS = \{ gy : y \in S\}$ is the left translate.

Invariant. A function $m : \mathcal P(Y) \to [0,\infty]$ is $G$-invariant when $m(gS) = m(S)$ for every $g \in G$ and every subset $S \subseteq Y$.

Hypothesis: $G$ is amenable

There exists a function $m : \mathcal P(G) \to [0,\infty]$, defined on all subsets of $G$, such that

1. $m$ is finitely additive (as above);
2. $m(G) = 1$;
3. $m(gS) = m(S)$ for every $g \in G$ and every $S \subseteq G$, where $gS = \{gs : s \in S\}$ is the left translate.

Conclusion: $G$ has the invariant extension property

For every set $X$, every left action of $G$ on $X$, every family $\mathcal R$ of subsets of $X$, and every two functions $\mu, \nu : \mathcal P(X) \to [0,\infty]$, the following holds. If

• (a) $\mathcal R$ is closed under translation: $S \in \mathcal R \Rightarrow gS \in \mathcal R$ for all $g \in G$;
• (b) $\mu$ is invariant on $\mathcal R$ only: $\mu(gS) = \mu(S)$ for all $g \in G$ and all $S \in \mathcal R$;
• (c) $\nu$ agrees with $\mu$ on $\mathcal R$: $\nu(S) = \mu(S)$ for all $S \in \mathcal R$;
• (d) $\nu$ is finitely additive on all of $\mathcal P(X)$,

then there exists $\bar\mu : \mathcal P(X) \to [0,\infty]$ such that

• $\bar\mu$ is finitely additive on all of $\mathcal P(X)$;
• $\bar\mu(S) = \mu(S)$ for every $S \in \mathcal R$;
• $\bar\mu$ is $G$-invariant on all subsets of $X$: $\bar\mu(gS) = \bar\mu(S)$ for every $g \in G$, $S \subseteq X$.

Here "left action" means a map $G \times X \to X$ with $1\cdot x = x$ and $(gh)\cdot x = g\cdot(h\cdot x)$.

Notes on the quantifiers and edge cases

• Size of $X$. $X$ ranges over sets in every universe at least as large as the one containing $G$; the statement is asserted separately for each such choice (it has a free universe level for $X$ in addition to the one for $G$). In ordinary mathematical terms: $X$ is an arbitrary set.
• Nothing is assumed about $\mathcal R$ beyond (a). It need not contain $\varnothing$ or $X$, and need not be closed under unions, intersections or complements. It may be empty.
• Nothing is assumed about $\mu$ beyond (b) and (c). $\mu$ is an arbitrary function on all subsets of $X$; outside $\mathcal R$ its values play no role in the hypotheses or the conclusion. Its only link to additivity is through $\nu$: on $\mathcal R$ it coincides with some finitely additive $\nu$. The function $\nu$ itself is not assumed invariant.
• Values $\infty$ are allowed for $\mu$, $\nu$ and $\bar\mu$; nothing requires $\bar\mu(X)$ to be finite or equal to $1$, and $\bar\mu$ is not required to relate to $\nu$ outside $\mathcal R$.
• Empty $\mathcal R$. Then (a)–(c) are vacuous, and the conclusion asks only for some $G$-invariant finitely additive function on $\mathcal P(X)$, with no other constraint; the identically zero function is such a function.
• Empty $X$. Then $\mathcal P(X) = \{\varnothing\}$, finite additivity forces value $0$ at $\varnothing$, and $\mathcal R$ is either empty or $\{\varnothing\}$.
• Hypothesis on $G$. In the amenability hypothesis, finite additivity together with $m(G) = 1$ forces every value $m(S)$ to lie in $[0,1]$, since $m(S) + m(G\setminus S) = 1$.