Let $\mathcal{D}(\mathbb{R})$ be the space of smooth functions with compact support and $\mathcal{D}'(\mathbb{R})$ the space of distributions (ie. continuous linear functionals on $\mathcal{D}$). There is a natural embedding $T: L^1_{\mathrm{loc.}}(\mathbb{R}) \to \mathcal{D}'(\mathbb{R})$: $$ T(f) = \varphi \mapsto \int_{\mathbb{R}} f(x) \, \varphi(x) \, \mathrm{d}x \qquad \forall f \in L^1_{\mathrm{loc.}} \quad \forall \varphi \in \mathcal{D} $$ The space $T(L^1_{\mathrm{loc.}}) \subset \mathcal{D}'$ of all distributions representable by a locally integrable function is called the space of regular distributions, all other distributions are called singular distributions. My question is: does every distribution $f \in \mathcal{D}'$ have a unique decomposition into the sum $f = f_{\mathrm{reg.}} + f_{\mathrm{sing.}}$of its regular and singular parts?
I intuitively feel that this should be true, but struggle to define such a decomposition more precisely, let alone prove it. The first problem one encounters is that given a smooth function $g(x)$, $f = g + \delta \,$ has many possible decompositions: $$ f = \underbrace{0}_{\mathrm{reg.}} + \underbrace{g + \delta}_{\mathrm{sing.}} \; = \underbrace{\tfrac{1}{2} g}_{\mathrm{reg.}} + \underbrace{\tfrac{1}{2} g + \delta}_{\mathrm{sing.}} = \underbrace{g}_{\mathrm{reg.}} + \underbrace{\delta}_{\mathrm{sing.}} = \underbrace{2g}_{\mathrm{reg.}} \; \underbrace{ - g + \delta}_{\mathrm{sing.}} $$ Obviously, the decomposition I want is $f_{\mathrm{reg.}} = g, \; f_{\mathrm{sing.}} = \delta$. Maybe a decomposition, where the support of $f_{\mathrm{sing.}}$ is equal to its singular support would be a good criterion? But I don't know, maybe there are distributions with non-discrete singular spectrum or something like that, when this approach would fail.
I know that there is a class of distributions called distributions of order zero that have a one-to-one correspondence with Radon measures (Carlsson, 2.6): $$ \forall T \in \mathcal{D}' \text{ of order 0 } \exists \text{ Radon measure } \mu \qquad T(\varphi) = \int_{\mathbb{R}} \varphi \; \mathrm{d}\mu \qquad \forall \varphi \in \mathcal{D} $$ Such a measure can be decomposed into $\mu_{\mathrm{reg.}}$ and $\mu_{\mathrm{sing.}}$ with respect to the Lebesgue measure by the Lebesgue's decomposition theorem. But this doesn't tell me anything about the decomposition of distributions of general order.