Let $\mu$ be a Borel probability measure on $\mathbb{R}^n$ and define the associated distribution on $C_c^\infty(\mathbb{R}^n)$ by $$T_\mu(f) = \int_{\mathbb{R}^n}f(x)\mu(dx).$$ Suppose that there exists $s > 0$ such that for any multi-index $\alpha$ the distribution $D^\alpha T_\mu(f) := (-1)^{|\alpha|}T_\mu(D^\alpha f)$ lies in the Sobolev space $H^{-s}$ (the dual space of $H^s$) in the sense that $$|D^\alpha T_\mu (f)| \le C(\alpha)\|f\|_{H^s}$$ so that $D^\alpha T_\mu$ extends to a bounded linear function on $H^s$ by density.
Question: How do we show that $\mu$ has a smooth density with respect to the Lebesgue measure?
Roughly speaking, since each derivative of $\mu$ lies in the fixed negative sobolev space $H^{-s}$ we should have something like $\mu$ living in any Sobolev space $H^m$. Then we apply Sobolev embedding to get the smoothness. How does one make this precise and get that $\mu$ actually has a density with respect to Lebesgue measure?