Let $g:\mathbb{R}^n \to \mathbb{R}$ be a smooth real-valued function that decays like some positive power of $|x|^{-1}$ at infinity. Define the first order distribution $\mu $ by 
$$
\langle \mu , \phi \rangle = \int \limits \_{\mathbb{R}^n} \big ( \phi (g(x)) - \phi (0) \big ) \,dx , \qquad \phi \in C\_0^\infty (\mathbb{R}). 
$$ 
Let us for simplicity assume that $g$ is both positive as well as negative at some points so that 
$$
\operatorname{supp}\mu \subset [\inf g, \sup g]. 
$$ 
I'm wondering if we actually have equality rather than just $\subset $?



Support of a distribution
I think so.
Let $t\_0\ne 0$ be in that interval and note that the interval also equals the image of $g$. Assume $t\_0$ does not lie in the support of $\mu$.
Then there exists a neighborhood $U$ of $t\_0$ such that $\langle \mu,\phi\rangle=0$ for every $\phi$ supported in $U$.
By shrinking $U$ we can assume that there exists $\alpha>0$ such that $U\cap [-\alpha,\alpha]=\emptyset$.
Choose some $\epsilon>0$ such that the closed $\epsilon$-neighborhood $V$ of $t\_0$ is contained in $U$.
Now take a test function $\phi\ge 0$ supported in $U$ with $\phi\ge 1$ on $V$.
Then 
$$
\langle\mu,\phi\rangle \ge \int\_A 1\,dx,
$$
where $A=g^{-1}(V)$. As $A$ has positive measure, we have a contradiction.
As the support is closed, it also contains zero.


