Let $\mathscr{M}(\mathbb{R}^n)$ the vector space of all Radon measures on $\mathbb{R}^n$. Let $\mathscr{H}^d$ denote the Hasudorff $d$-dimensional measure for $d\leq n$. If $S\subset \Omega$ is such that $\mathscr{H}^d(S)<\infty$ then \begin{gather*} \nu:= \mathscr{H}^d \llcorner S\\ \nu(A)= \mathscr{H}^d(A\cap S) \end{gather*} defines a Radon measure, i.e. $\nu\in \mathscr{M}(\mathbb{R}^n)$. Similarly given a set $S$ and a $\mathscr{H}^d\llcorner S$ summable function $g$ the formula \begin{equation} \gamma(A)=\int_{A\cap S} g \;d\mathscr{H}^d\qquad \nu = g\;\mathscr{H}^d\llcorner S \end{equation} defines a Radon measure.
QUESTION
To me it seems reasonable to consider the set of all measures of the type $\gamma $ defined above as a "base" for all the Radon measures. Consequently is it true that given a Radon measure $\mu\in\mathscr{M}(\mathbb{R}^n)$ there exists a family of sets $\{S_d\}_{d\in [0,\infty)}$ and a family of functions $\{g_d\}_{d\in [0,\infty)}$ such that \begin{equation*} \mu=\int_{0}^\infty \;\;g_d\mathscr{H}^d\llcorner S_d \;\;d\mathscr{L}^1(d). \end{equation*}
The latter is then equivalent, in some sort of way, to: is there a measure $\mu \in \mathscr{M}(\mathbb{R}^n)$ which is singular to all Hausdorff measures?
UPDATE
I found a work from C.D.Culter "The Hausdorff measure distribution of finite measures in euclidean space", in which this kind of decomposition is treated. https://cms.math.ca/openaccess/cjm/v38/cjm1986v38.1459-1484.pdf