Distributions and measures 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
 A: To answer the first question: yes, at least locally. That is, given a distribution $u$ of order $0$, compactly supported for simplicity, the "order 0" condition asserts that $u$ factors through the space $C^o_c(U)$ of continuous compactly supported functions _with_the_corresponding_topology_. Then invoke the Riesz representation theorem to obtain the corresponding measure. Reduce to this case by a locally finite smooth partition of unity.
I cannot instantly answer the second question, but would try to reduce it to the compactly-supported case by a partition of unity argument.
I think the comment deals with the question about the "singular continuous" part. 
A: 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."
