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Can't you just use the Lyapunov convexity theorem directly?

As usual, identify $\ell^\infty(G)$ with $C(\beta G)$, and work with $\beta G$ the Stone-Cech compactification. As this is a compact Hausdorff space, if $\mu$ is a regular measure on $\beta G$ then an atom of $\mu$ must be a point. So we can decompose $\mu$ as something in $\ell^1(\beta G)$ together with an atom-less measure, say a member of $M_c(\beta G)$ (continuous measures).

(Left) translation by members of $G$ give automorphisms of $\beta G$, and hence leave $\ell^1(\beta G)$ and $M_c(\beta G)$ invariant. I claim that nothing in $\ell^1(\beta G)$ can be left invariant. Let $\mu\in\ell^1(\beta G)$ be left invariant. Write $\beta G$ as the disjoint union of $G$-orbits, say $\bigcup_i G u_i$. Then $\mu$ must be supported on finite orbits (else we couldn't sum the coefficients, so we wouldn't be in $\ell^1$). If $u\in\beta G$ with $Gu$ finite, then there is $s\not=e$ in $G$ with $su=u$. Realise $u$ as an ultrafilter. Let $A\subseteq G$ be maximal with $A\cap s^{-1}A=\emptyset$. This means that if $r\not\in A$ then there is $t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$, which implies that $t=r\in s^{-1}A\cup\{s^{-1}r\}$, that is, $sr\in A$. So $r\not\in A\implies sr\in A \implies r\in s^{-1}A$, so $G=A\cup s^{-1}A$. So Zorn implies there is $A\subseteq G$ with $A \cap s^{-1}A=\emptyset$ and $A\cup s^{-1}A=G$. Then either $A\in u$ so $A\in su$ so $s^{-1}A\in u$, contradiction; or $s^{-1}A\in u$ so $A\in su=u$ contradiction.

So I (hope!) I've shown that actually for any $u\in\beta G$, the orbit map $G\rightarrow\beta G; s\mapsto su$ is injective.

In particular, invariant means live in $M_c(\beta G)$, and so are atom-less, and so now we can just apply Lyapunov.

Edit: As Valerio points out, this shows that $X={ X=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq\beta G \text{ is Borel}}$ Borel}\}$ is a convex set in$[0,1]^n$. Now, each$A\subseteq G$induces the clopen set $O_A={ O_A=\{ u\in\beta G: A\in u }$, \}$, and these sets $O_A$ form a base for the topology. Now each $\mu_i$ is regular, so given $\epsilon>0$ and $A\subseteq\beta G$ Borel, we can find $B,C\subseteq G$ with $O_B \subseteq A\subseteq O_C$ and with $\mu_i(C)-\mu_i(B)<\epsilon$, for all $i$ (under the obvious abuse of notation). (This follows as any open set is a union of sets of the form $O_C$, and then approximate with a finite union.) So $Y={ Y=\{ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq G}$ G\}$ is a subset of$X$, and is dense in$X$. I don't see right now why$Y$need be convex. 2 added 672 characters in body Can't you just use the Lyapunov convexity theorem directly? As usual, identify$\ell^\infty(G)$with$C(\beta G)$, and work with$\beta G$the Stone-Cech compactification. As this is a compact Hausdorff space, if$\mu$is a regular measure on$\beta G$then an atom of$\mu$must be a point. So we can decompose$\mu$as something in$\ell^1(\beta G)$together with an atom-less measure, say a member of$M_c(\beta G)$(continuous measures). (Left) translation by members of$G$give automorphisms of$\beta G$, and hence leave$\ell^1(\beta G)$and$M_c(\beta G)$invariant. I claim that nothing in$\ell^1(\beta G)$can be left invariant. Let$\mu\in\ell^1(\beta G)$be left invariant. Write$\beta G$as the disjoint union of$G$-orbits, say$\bigcup_i G u_i$. Then$\mu$must be supported on finite orbits (else we couldn't sum the coefficients, so we wouldn't be in$\ell^1$). If$u\in\beta G$with$Gu$finite, then there is$s\not=e$in$G$with$su=u$. Realise$u$as an ultrafilter. Let$A\subseteq G$be maximal with$A\cap s^{-1}A=\emptyset$. This means that if$r\not\in A$then there is $t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$, which implies that $t=r\in s^{-1}A\cup\{s^{-1}r\}$, that is,$sr\in A$. So$r\not\in A\implies sr\in A \implies r\in s^{-1}A$, so$G=A\cup s^{-1}A$. So Zorn implies there is$A\subseteq G$with$A \cap s^{-1}A=\emptyset$and$A\cup s^{-1}A=G$. Then either$A\in u$so$A\in su$so$s^{-1}A\in u$, contradiction; or$s^{-1}A\in u$so$A\in su=u$contradiction. So I (hope!) I've shown that actually for any$u\in\beta G$, the orbit map$G\rightarrow\beta G; s\mapsto su$is injective. In particular, invariant means live in$M_c(\beta G)$, and so are atom-less, and so now we can just apply Lyapunov. Edit: As Valerio points out, this shows that$X={ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq\beta G \text{ is Borel}}$is a convex set in$[0,1]^n$. Now, each$A\subseteq G$induces the clopen set$O_A={ u\in\beta G: A\in u }$, and these sets$O_A$form a base for the topology. Now each$\mu_i$is regular, so given$\epsilon>0$and$A\subseteq\beta G$Borel, we can find$B,C\subseteq G$with$O_B \subseteq A\subseteq O_C$and with$\mu_i(C)-\mu_i(B)<\epsilon$, for all$i$(under the obvious abuse of notation). So$Y={ (\mu_1(A),\cdots,\mu_n(A)) : A\subseteq G}$is a subset of$X$, and is dense in$X$. I don't see right now why$Y$need be convex 1 Can't you just use the Lyapunov convexity theorem directly? As usual, identify$\ell^\infty(G)$with$C(\beta G)$, and work with$\beta G$the Stone-Cech compactification. As this is a compact Hausdorff space, if$\mu$is a regular measure on$\beta G$then an atom of$\mu$must be a point. So we can decompose$\mu$as something in$\ell^1(\beta G)$together with an atom-less measure, say a member of$M_c(\beta G)$(continuous measures). (Left) translation by members of$G$give automorphisms of$\beta G$, and hence leave$\ell^1(\beta G)$and$M_c(\beta G)$invariant. I claim that nothing in$\ell^1(\beta G)$can be left invariant. Let$\mu\in\ell^1(\beta G)$be left invariant. Write$\beta G$as the disjoint union of$G$-orbits, say$\bigcup_i G u_i$. Then$\mu$must be supported on finite orbits (else we couldn't sum the coefficients, so we wouldn't be in$\ell^1$). If$u\in\beta G$with$Gu$finite, then there is$s\not=e$in$G$with$su=u$. Realise$u$as an ultrafilter. Let$A\subseteq G$be maximal with$A\cap s^{-1}A=\emptyset$. This means that if$r\not\in A$then there is $t\in (A\cup\{r\}) \cap (s^{-1}A\cup\{s^{-1}r\})$, which implies that $t=r\in s^{-1}A\cup\{s^{-1}r\}$, that is,$sr\in A$. So$r\not\in A\implies sr\in A \implies r\in s^{-1}A$, so$G=A\cup s^{-1}A$. So Zorn implies there is$A\subseteq G$with$A \cap s^{-1}A=\emptyset$and$A\cup s^{-1}A=G$. Then either$A\in u$so$A\in su$so$s^{-1}A\in u$, contradiction; or$s^{-1}A\in u$so$A\in su=u$contradiction. So I (hope!) I've shown that actually for any$u\in\beta G$, the orbit map$G\rightarrow\beta G; s\mapsto su$is injective. In particular, invariant means live in$M_c(\beta G)\$, and so are atom-less, and so now we can just apply Lyapunov.