Hello,


After reading the [previous post](https://mathoverflow.net/questions/17732/difference-between-measures-and-distributions), I still have some doubts. Let's consider everything on $R$ to avoid complications. 


1. Can we say that any distribution $\mu\in\mathcal{D}'(R)$ of zero order is a signed radon measure?
2. 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?
3. For a radon measure $\mu$ on $R$, can we apply [the Lebesgue's decomposition](http://en.wikipedia.org/wiki/Lebesgue's_decomposition_theorem) 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



Distributions and measures
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. 


