Let $(G,\mathcal G)$ be a measurable group and $\nu^{\ast k}$ denote the $k$th convolution power of a probability measure $\nu$ on $(G,\mathcal G)$ for $k\in\mathbb N$.
Remember that a probability measure $\mu$ on $(G,\mathcal G)$ is called infinitely divisible if for all $k\in\mathbb N$, there is a probability measure $\nu_k$ on $(G,\mathcal G)$ with $\mu=\nu_k^k$.
A priori, it is not clear whether or not the $\nu_k$ are uniquely determined by $\mu$ and hence whether or not the $k$th convolution root $\mu^{\ast1/k}:=\nu_k$ is well-defined.
Are we able to show well-definedness and extend the convolution root to arbitrary (nonnegative) real numbers?
It's clear to me how we can show well-definedness and construct the desired $t$th convolution root for arbitrary $t\ge0$ whenever we assume that $G$ is a normed $\mathbb R$-vector space and $\mathcal G=\mathcal B(G)$, since then the whole problem can be reformulated in terms of characteristic functions.
Remark: I'm especially interested in convolution semigroups $(\mu_t)_{t\ge0}$, i.e. families of probability measures on $(G,\mathcal G)$ satisfying $\mu_{s+t}=\mu_s\ast\mu_t$ for all $s,t\ge0$. It obviously holds $\mu_t=\mu_{t/k}^{\ast k}$ for all $k\in\mathbb N$ and hence $\mu_t$ is infinitely divisible for all $t\ge0$. As before, I would like to show that $\mu_t^{\ast1/k}:=\mu_{\frac tk}$ is well-defined and we can extend this definition to arbitrary convolution roots.
Maybe we need at least to assume that $G$ is a metric space and $\mathcal G=\mathcal B(G)$. Then we could assume that $(\mu_t)_{t\ge0}$ is "continuous", i.e. $\mu_t\xrightarrow{t\to0+}\delta_0$ weakly (which is equivalent to $\mu_t(B_\varepsilon(0))\xrightarrow{t\to0+}1$ for all $\varepsilon>0$).
