Compact embedding of space of signed Radon measures into Sobolev space $W^{-1,q}$ from Evans paper; Does it work in one space dimension? Background: I work on a PDE problem where I have some approximating sequence of measure-valued functions and I need to compactly embed it into some negative Sobolev space $W^{-m,q}$ on the bounded interval in $\mathbb{R}$. I am mostly interested in the spaces where $q=2$. I found only one such embedding in the one theorem from the paper:
Evans - Weak convergence methods for nonlinear partial differential equations, 1990. 
Theorem 6 (Compactness for measures, page 7):  Assume the sequence $\{\mu_k\}_{k=1}^{\infty}$ is bounded in $\mathcal{M}(U)$, $U \subset \mathbb{R}^n$. Then $\{\mu_k\}_{k=1}^{\infty}$ is precompact in $W^{-1,q}(U)$ for each $1 \leq q <1^*$.
Here $\mathcal{M}(U)$ represents space of signed Radon measures on $U$ with finite mass, $U \subset \mathbb{R}^n$ is an open, bounded, smooth subset of $\mathbb{R}^n, n \geq 2$ and $1^*=\frac{n}{n-1}$ represents a Sobolev conjugate.
The identical theorem (Lemma 2.55, page 38) is given in the book: Malek, Necas, Rokyta, Ruzicka - Weak and Measure-valued Solutions to Evolutionary PDEs, 1996, with a difference that instead of $1 \leq q <1^*$, in there is written $1 \leq q <\frac{n}{n-1}$ (here it isn't written explicitly that $n\geq 2$).
My question: does the Theorem 6 works in one dimension ($n=1$)? That is do we have a compact embedding of space $\mathcal{M}(U)$ into the space $W^{-1,q}(U)$, where $U \subset \mathbb{R}$? 
And additionaly: 


*

*I assume that if we have compact embedding into $W^{-1,q}(U)$, then we have it also in the $W^{-m,q}(U), m\geq 1$?

*Are there any other measure spaces (e.g. space of finite positive measures $\mathcal{M}_+$, space of probability measures with finite first moment $Pr_1$, etc.) that are compactly embedded into some negative Sobolev spaces $W^{-m,q}(U)$?


I think that if we use definition of the Sobolev conjugate: $\frac{1}{p^*}=\frac{1}{p}-\frac{1}{n}$, we get for $p=1,n=1$ the $\frac{1}{1^*}=\frac{1}{1}-\frac{1}{1}\Rightarrow 1^*=\infty$. So we would have that theorem 6 (maybe) works for every $1 \leq q < \infty$ (and then for $q=2$ also)? If we use $p^*=\frac{np}{n-p}$ we would have for $n=1,$ $p^*=\frac{p}{1-p}$ and here we could not take $p=1$ and get $p^*$.  
I usually do not deal with the measure-valued and negative Sobolev spaces, so I don't know much about them. Help with this would be great and I definitely need it. And any additional reference besides the two mentioned above would be nice. Thanks in advance.
 A: Here is a partial answer, which has to do with dual compact embeddings: If the embedding between (resonable) Banach spaces $X\subset\subset Y$ is compact then the dual embedding is compact too, $Y^*\subset\subset X^*$.
This is useful here since the space of Radon measures is the dual of continuous bounded functions, $\mathcal M(U)=(C_b(U))^*$.
Now for $p>n$ we have that $W^{1,p}$ is continuously embedded into some Hölder space $C^\alpha$ (for some $\alpha\equiv \alpha(n,p)$). By the Arzelà-Ascoli theorem this shows that the embedding
$$
W^{1,p}(U)\subset\subset C_b(\bar U)
$$
is compact too.
As a consequence we have that the embedding
$$
\mathcal M(U)\subset\subset W^{-1,q}(U)
$$
provided that $q=p'$ is such that $p>n$, ie for all $q<1^*=\frac{n}{n-1}$ (this is exactly why the cirical $1^*$ exponent appears in your Theorem 6).
As for the second part of the question: since the embedding $W^{m,p}\subset W^{1,p}$ is trivially continuous for $m>1$, the reversed embedding $W^{-1,q}\subset W^{-m,q}$ is continuous. Then the composition of "compact$\circ$continuous = compact" $\mathcal M\subset\subset W^{-1,q}\subset W^{-m,q}$ also gives compactness for
$$
\mathcal M(U)\subset\subset W^{-m,q}(U).
$$
