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The answer is that the above is equivalent to non amenability. Fix a group $(G,\*)$. Since $(G,\*)$ is non amenable if and only if every finitely generated subgroup is non amenable, we may assume that $G$ is finitely generated.

If $\mu$ and $\nu$ are finitely supported probability measures on $G$, define $$ \mu \* \nu (Z) = \sum_{x \* y \in Z} \mu (\{x\}) \nu (\{y\}) $$ Observe that $g \* \nu (E) = \nu ( g^{-1} \* E)$. If $S$ is a subset of $S$, let $P(S)$ denote all probability measures on $S$ (which are identified with probability measures on $G$ which are supported on $S$). I will identify $G$ with the point masses in $P(G)$.

If $A$ and $B$ are subsets of $G$ and $A$ is finite, we say that $B$ is $\epsilon$-Ramsey with respect to $A$ if for every $E \subseteq B$, then there is a $\nu$ in $P(B)$ such that $P(A) \* \nu \subseteq P(B)$ and $$ |\mu \* \nu (E) - \nu (E)| < \epsilon $$ for all $\mu$ in $P(A)$. Notice that in some sense $E$ is defining a partition of $P(B)$ and we are postulating the existence of a copy of $P(A)$ in $P(B)$ which is homogeneous for $E$ up to an error of $\epsilon$.

It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$ |f(\mu \* \nu) - f(\nu)| < \epsilon $$ where $f$ has been extended linearly to $P(B)$.

We say that $(G,*)$ is Ramsey if for every finite subset $A \sub</i>seteq G$ and every $\epsilon > 0$, there is a finite subset $B$ of $G$ with is $\epsilon$-Ramsey with respect to $A$. Notice that if $B$ satisfies that for every $E \subseteq B$ there is a $\nu$ in $P(B)$ such that $$ |g \* \nu (E) - \nu (E)| < \epsilon $$ for all $g$ in $A$, then $B$ is contained in a finite set which is $\epsilon$-Ramsey (we need only to replace $B$ by $A \* B \cup B$).

To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n < \infty)$ be an increasing sequence of finite sets covering $G$. Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. Otherwise there would be a $\nu$ in $P(B_n)$ such that $g \* \nu$ is in $P(B_n)$ for each $g$ in $A$ and $$ |g \* \nu (E \cap B_n) - \nu (E \cap B_n)| < \epsilon $$ Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. This order makes $T$ into an infinite finitely branching tree. By König's lemma, $T$ has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.

Now the claim is that the Ramsey property of a discrete group is equivalent to its amenability. That amenability implies the Ramsey property follows from Følner's characterization of amenability. Also observe that $G$ is amenable provided that for every $\epsilon > 0$, every finite list $E_i$ $(i < n)$ of subsets of $G$, and $g_i$ $(i < n)$ in $G$, there is a finitely supported $\mu$ such that $$ |\mu (g_i \* E_i) - \mu (E_i) | < \epsilon. $$ Set $B_{-1} = \{1_G\} \cup \{g^{-1}_i :i < n\}$ and construct a sequence $B_i$ $(i < n)$ such that $B_{i+1}$ is $\epsilon/2$-Ramsey with respect to $B_i$.

Now inductively construct $\nu_i$ $(i < n)$ by downward recursion on $i$. If $\nu_j$ $(i < j)$ has been constructed, let $\nu_i \in P(B_i)$ be such that $$ |\mu \* \nu_{i} \* \ldots \* \nu_{n-1} (E_i) - \nu_i \* \ldots * \nu_{n-1} (E_i)| < \epsilon/2 $$ for all $\mu$ in $P(B_{i-1})$. Set $\mu = \nu_0 \* \ldots \* \nu_{n-1}$. If $i < n$, then since $\nu_0 * \ldots * \nu_{i-1}$ and $g_i^{-1} * \nu_0 * \ldots * \nu_{i-1}$ are in $P(B_{i-1})$, $$ |g_i^{-1} \* \mu (E_i) - \nu_i \* \ldots \* \nu_{n-1} (E_i)| < \epsilon/2 $$ $$ |\mu (E_i) - \nu_i \* \ldots \* \nu_{n-1} (E_i)| < \epsilon/2 $$ and therefore $|\mu (g_i \* E_i) - \mu (E_i)| < \epsilon$.

The answer is that the above is equivalent to non amenability. Fix a group $(G,*)$. Since $(G,*)$ is non amenable if and only if every finitely generated subgroup is non amenable, we may assume that $G$ is finitely generated.

If $\mu$ and $\nu$ are finitely supported probability measures on $G$, define $$ \mu * \nu (Z) = \sum_{x * y \in Z} \mu (\{x\}) \nu (\{y\}) $$ Observe that $g * \nu (E) = \nu ( g^{-1} * E)$. If $S$ is a subset of $S$, let $P(S)$ denote all probability measures on $S$ (which are identified with probability measures on $G$ which are supported on $S$). I will identify $G$ with the point masses in $P(G)$.

If $A$ and $B$ are subsets of $G$ and $A$ is finite, we say that $B$ is $\epsilon$-Ramsey with respect to $A$ if for every $E \subseteq B$, then there is a $\nu$ in $P(B)$ such that $P(A) * \nu \subseteq P(B)$ and $$ |\mu * \nu (E) - \nu (E)| < \epsilon $$ for all $\mu$ in $P(A)$. Notice that in some sense $E$ is defining a partition of $P(B)$ and we are postulating the existence of a copy of $P(A)$ in $P(B)$ which is homogeneous for $E$ up to an error of $\epsilon$.

It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$ |f(\mu * \nu) - f(\nu)| < \epsilon $$ where $f$ has been extended linearly to $P(B)$.

We say that $(G,*)$ is Ramsey if for every finite subset $A \subseteq G$ and every $\epsilon > 0$, there is a finite subset $B$ of $G$ with is $\epsilon$-Ramsey with respect to $A$. Notice that if $B$ satisfies that for every $E \subseteq B$ there is a $\nu$ in $P(B)$ such that $$ |g * \nu (E) - \nu (E)| < \epsilon $$ for all $g$ in $A$, then $B$ is contained in a finite set which is $\epsilon$-Ramsey (we need only to replace $B$ by $A * B \cup B$).

To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n < \infty)$ be an increasing sequence of finite sets covering $G$. Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. Otherwise there would be a $\nu$ in $P(B_n)$ such that $g * \nu$ is in $P(B_n)$ for each $g$ in $A$ and $$ |g * \nu (E \cap B_n) - \nu (E \cap B_n)| < \epsilon $$ Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. This order makes $T$ into an infinite finitely branching tree. By König's lemma, $T$ has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.

Now the claim is that the Ramsey property of a discrete group is equivalent to its amenability. That amenability implies the Ramsey property follows from Følner's characterization of amenability. Also observe that $G$ is amenable provided that for every $\epsilon > 0$, every finite list $E_i$ $(i < n)$ of subsets of $G$, and $g_i$ $(i < n)$ in $G$, there is a finitely supported $\mu$ such that $$ |\mu (g_i * E_i) - \mu (E_i) | < \epsilon. $$ Set $B_{-1} = \{1_G\} \cup \{g^{-1}_i :i < n\}$ and construct a sequence $B_i$ $(i < n)$ such that $B_{i+1}$ is $\epsilon/2$-Ramsey with respect to $B_i$.

Now inductively construct $\nu_i$ $(i < n)$ by downward recursion on $i$. If $\nu_j$ $(i < j)$ has been constructed, let $\nu_i \in P(B_i)$ be such that $$ |\mu * \nu_{i} * \ldots * \nu_{n-1} (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 $$ for all $\mu$ in $P(B_{i-1})$. Set $\mu = \nu_0 * \ldots * \nu_{n-1}$. If $i < n$, then since $\nu_0 * \ldots * \nu_{i-1}$ and $g_i^{-1} * \nu_0 * \ldots * \nu_{i-1}$ are in $P(B_{i-1})$, $$ |g_i^{-1} * \mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 $$ $$ |\mu (E_i) - \nu_i * \ldots * \nu_{n-1} (E_i)| < \epsilon/2 $$ and therefore $|\mu (g_i * E_i) - \mu (E_i)| < \epsilon$.

Observe that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$ |f(\mu \* \nu) - f(\nu)| < \epsilon $$ where $f$ has been extended linearly to $P(B)$. To see this, notice that $B$ is assumed to satisfy this statement for $f: B \to \{0,1\}$. Furthermore, $f \mapsto f(\mu \* \nu) - f(\nu)$ is linear and $\{0,1\}^B$ is the set of extreme points of $[0,1]^B$. Thus the maxima and minima of this map are realized at elements of $\{0,1\}^B$.

It can be shown with an argument similar to the one below that if $B$ is $\epsilon$-Ramsey with respect to $A$, then for every $f:B \to [0,1]$ there is a $\nu$ in $P(B)$ such that $$ |f(\mu \* \nu) - f(\nu)| < \epsilon $$ where $f$ has been extended linearly to $P(B)$.

To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n < \infty)$ be an increasing sequence of finite sets covering $G$. Let $T$ be the tree whose $n$th level consists of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Here we order the tree by putting $E \leq_T E'$ if $E = E' \cap B_n$, where $E$ is on the $n$th level of $T$. Since this tree is finitely branching and infinite, it has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.

To connect this to the question, suppose that $G$ is not Ramsey, as witnessed by a finite $A \subseteq G$ and $\epsilon > 0$. I claim there is a set $E \subseteq G$ such that for every $\mu \in P(G)$, there is a $g \in A$ such that $|\mu(E \cdot g) - \mu (E)| \geq \epsilon/2$. Let $B_n$ $(n < \infty)$ be an increasing sequence of finite sets covering $G$. Let $T_n$ be the set of all subsets $E$ of $B_n$ which witness that $B_n$ is not $\epsilon$-Ramsey with respect to $A$. Observe that if $E$ is in $T_{n+1}$, then $E \cap B_n$ is in $T_n$. Otherwise there would be a $\nu$ in $P(B_n)$ such that $g \* \nu$ is in $P(B_n)$ for each $g$ in $A$ and $$ |g \* \nu (E \cap B_n) - \nu (E \cap B_n)| < \epsilon $$ Such a $\nu$ would also witness that $E$ is not in $T_{n+1}$. Define $T = \bigcup_n T_n$ and order $E \leq_T E'$ if $E = E' \cap B_m$ where $E$ is in $T_n$. This order makes $T$ into an infinite finitely branching tree. By König's lemma, $T$ has an infinite path whose union is some $E \subseteq G$. If there were a measure $\mu$ which was $\epsilon/2$-invariant for $E$ with respect to translates by elements of $A$, there would be a finitely supported $\nu$ which was $\epsilon$-invariant for $E$ with respect to translates in $A$. But this would be a contradiction since then the support of $\nu$ would be contained in some $B_n$ and $\nu$ would witness that $E \cap B_n$ was not in $T$.