TITLE: Distributions and measures
QUESTION [6 upvotes]: Hello,
After reading the previous post, I still have some doubts. Let's consider everything on $R$ to avoid complications.  

Can we say that any distribution $\mu\in\mathcal{D}'(R)$ of zero order is a signed radon measure?
Since $\mu\in\mathcal{D}'(R)$ which is non-negative on non-negative test functions $C_c^\infty(R)$ is a positive radon measure, it is natural to ask what is corresponding part for the Schwartz distribution $\mu\in\mathcal{S}'(R)$ which is non-negative on non-negative test functions $\mathcal{S}(R)$? Intuitively, it is the radon measure whose mass grows slowly at infinity. Is there a name for this measure?
For a radon measure $\mu$ on $R$, can we apply the Lebesgue's decomposition locally with  a compact set fixed (say $K=[-a,a]$)? Then $\mu_K=\mu_{ac}+\mu_{sc}+\mu_{pp}$. The absolutely continuous part $\mu_{ac}$ corresponds to an absolutely continuous function; the pure point part $\mu_{pp}$ corresponds to sum of delta functions. How about the singular continuous part $\mu_{sc}$? 
EDIT: This question can also be put in the following way: Let $f$ be a singular function, what can we say $\int f\psi d x$ with $\psi\in\mathcal{D}(R)$?

Thank you for your help! :-)
Best
Anand

REPLY [6 votes]: About the second question:
The Schwartz space $\mathcal{S}(R)$ is a Frechet space, i.e. it's topology is given by a countable sequence of seminorms.  Any Schwartz distribution (continuous linear functional on $\mathcal{S}(R)$) has to bounded by one of these seminorms.  If the Schwartz distribution is of "order zero" then the only relevant seminorms are $\|f\|_n = \sup (1+|x|)^n |f(x)|$.  If $T\in \mathcal{S}'(R)$ is bounded by $\|\ldots\|_n$, then this means $T$ is of the form $\langle T, f\rangle = \int (1+|x|)^{n} f(x) \mu(dx)$ for some finite measure on $R$.
So I would think measures which are Schwartz distributions are measures "of polynomial growth."