Let $f:\mathbb{R}^n\to \mathbb{R}$ be a smooth and bounded function which tends to 0 at infinity. Define, for $t>0$, the distribution 
$$
\nu (t) = \int \limits \_{f(x)\ge t} dx,
$$
and (in the distributional sense) the positive measure
$$
\mu = -\frac{d\nu }{dt}.
$$
By a change of variables we see that 
$$
\langle \mu , \phi \rangle \_{\mathcal{D}', \mathcal{D}}=\int \limits \_{\mathbb{R}^n} \phi (f(x))\,dx \quad \text{for }\phi \in C\_0^\infty (\mathbb{R}\_+).
$$
In the text I'm reading it is stated (without motivation or proof) that 
$$
\operatorname{supp}\operatorname{sing}\_A \mu = \operatorname{supp}\operatorname{sing}\_A \nu , 
$$
where $\operatorname{supp}\operatorname{sing}\_A $ denotes the analytical singular support of a distribution (i.e. the smallest closed set outside of which the distribution is real analytic). In general we have $\operatorname{supp}\operatorname{sing}\_A u' \subset \operatorname{supp}\operatorname{sing}\_A u$ but why do we have equality in a case like this? 



Why equality of singular supports?
$\nu$ is a distribution on the real line and the operator $P=d/dt$ is elliptic with constant coefficients. In that case we have
$$
\text{singsupp $\nu$}=\text{singsupp $P\nu$}
$$
for the $C^\infty$ singular support as well as for the analytic singular support. The same equality holds for wave-front-sets (smooth and analytic) and this implies the previous equality.


