Given a locally compact abelian group $G$ with Haar measure $m$, it is well-known that there exist measures $\mu\in M(G)$ which are singular (with respect to $m$), but the convolution product $\mu\ast\mu$ is absolutely continuous.

Is there any condition characterizing the measures $\mu$ which satisfy the following (opposite) $tauberian$ property: given $\nu \in M(G)$, $\mu\ast\nu$ absolutely continuous implies $\nu$ absolutely continuous?

Given a locally compact abelian group $G$ with Haar measure $m$, it is well-known that there exist measures $\mu\in M(G)$ which are singular (with respect to $m$), but the convolution product $\mu\ast\mu$ is absolutely continuous.

(Q.1) Is there any condition characterizing the measures $\mu$ which satisfy the following (opposite) $tauberian$ property: given $\nu \in M(G)$, $\mu\ast\nu$ absolutely continuous implies $\nu$ absolutely continuous?

If $\mu_1$ is invertible in $M(G)$ and $\mu_2$ is absolutely continuous, then $\mu=\mu_1+\mu_2$ satisfies this tauberian property.

(Q.2) Are there further examples?