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@Bill Johnson: Clearly $X\cap \hat X^*$ is continuously embedded in $\ell_2$. Hence $\ell_2$ is continuosly embedded in $X^*+\hat X^{**}$, and (I think) the result is clear in the reflexive case.
@Ramiro de la Vega: You are right: $K^\mathcal{U}$ is not the set-ultraproduct, but the ultracoproduct (following $\mathcal{U}$) in the sense of Bankston [J. Symbolic Logic 52 (1987) 404-424]. The set ultrproduct is dense in $K^\mathcal{U}$.
@Asaf, the decomposition I considered is $\mu=\mu_d +\mu_c$, discrete part plus continuous part. Then the continuous (no atoms) part can be decomposed in absolutely continuous plus singular continuous.
CONTINUATION: It was proved by R. Doss [Studia Math. 45 (973), 111-117] that for every continuous measure $\mu$ there exists a non absolutely continuous measure $\mu$ such that $\mu *\nu$ is absolutely continuous. Therefore a measure satisfying the tauberian property has non-zero atomic part.
Every $\mu\in M(G)$ can be decomposed as $\mu= \mu_d + \mu_c$ where $\mu_d$ is the discrete (atomic) part and $\mu_c$ is the continuous part (no atoms).\newline
$(M(G),*)$ is a commutative Banach algebra. $\mu$ invertible means that there is $\nu$ such that $\mu * \nu =\delta_e$, where $e$ is the unit of $G$ ($\delta_e$ is the unit of $M(G)$), and $\mu_2$ a.c. wrt m means $A$ Borel subset of $G$ and $m(A) =0$ implies $\mu_1(A)=0$; equiv., there is $f_1 \in L_1(m)$ such that $d\mu_1 = f_1\cdot d m$.
