Property singular support of the convolution of distributions Let $u \in \mathcal{D}'(\Omega)$ and $U \subset \Omega$ open.
By definition we say that $u \in \mathcal{E}(U)$ if $\exists u(x) \in \mathcal{E}(U)$ such that 
\begin{align*}
\displaystyle \langle \varphi , u \rangle = \int_U u(x) \varphi(x) dx , \forall \varphi \in \mathcal{D}(U)
\end{align*}
and singular support of $u \in \mathcal{D}'(\Omega)$  is the set of all points $x \in \Omega$ such that not exists an open neighborhood $U\subset \Omega$ wich $u_{|U}$ is regular function, i.e.
\begin{align*}
\displaystyle \mathrm{singsupp}(u) = \lbrace x \in \Omega : \nexists U \subset \Omega : u_{|U} \in \mathcal{E}(U) \rbrace
\end{align*}
and then if $U_{max}$ is maximum open of $\Omega$ with the distribution $u \in \mathcal{E}(U_{\max})$ then
\begin{align*}
\displaystyle \mathrm{singsupp}(u) = \Omega \setminus U_{max}
\end{align*}
I'm trying to make a little better the proof that if $u$ or $v$ is a distrubution with compact support then
\begin{align*}
\mathrm{singsupp}(u \ast v) \subset \mathrm{singsupp}(u) + \mathrm{singsupp}(v)
\end{align*}
this proof it can be found in "Distribution Theory" by Duistermaat, page 130. It uses a regular version of the lemma Urysohn (Lemma 2.19 page 29), or as here Proof of regular version of the Urysohn lemma 
As mentioned, I'm trying to better express this demonstration (and this proof is slightly different).
Proof.
We place $L=\mathrm{singsupp}(u)$ and $M=\mathrm{singsupp}(v)$. Consider  $\mathrm{Int}(L)\subset L$ and $\mathrm{Int}(M) \subset M$, by Urysohn lemma $\forall \delta > 0$ small
\begin{align*}
\displaystyle  \exists \psi \in \mathcal{D}(\mathbb{R}^n)  :  \mathrm{supp}(\psi) \subset L(\delta) \subset \mathrm{Int}(L)
\end{align*}
with $\psi(x)=1$ $\forall x \in L(\delta)$ and $0 \leq \psi(x) \leq 1$ $\forall x \in \mathrm{Int}(L)$, where $L(\delta)$ is a compact. Similary
\begin{align*}
\displaystyle  \exists \eta \in \mathcal{D}(\mathbb{R}^n)  :  \mathrm{supp}(\eta) \subset M(\delta) \subset \mathrm{Int}(M)
\end{align*}
with $\eta(x)=1$ $\forall x \in M(\delta)$ and $0 \leq \eta(x) \leq 1$  $\forall x \in \mathrm{Int}(M)$, where $M(\delta)$ is also a compact set. We define $u_1=\psi u$ and $u_2=(1-\psi)u$, and also $v_1=\eta v$ and $v_2=(1-\eta)v$, then $u=u_1+u_2$ and $v=v_1+v_2$, with $\mathrm{supp}(u_1) \subset \mathrm{supp}(\psi) \subset L(\delta)$, and $\mathrm{supp}(v_1) \subset \mathrm{supp}(\eta) \subset M(\delta)$, therefore $u_1, v_1 \in \mathcal{E}'(\mathbb{R}^n)$ are distribution with compact support.
Moreover $u_2 , v_2 \in \mathcal{E}(\mathbb{R}^n)$ (WHY?).
Consequently of the four terms in following convolution, only the last three are regular functions
\begin{align*}
\displaystyle u \ast v = u_1 \ast v_1 + u_1 \ast v_2 + u_2 \ast v_1 + u_2 \ast v_2
\end{align*}
this implies that
\begin{align*}
\displaystyle \mathrm{singsupp}(u \ast v) &= \mathrm{singsupp}(u_1 \ast v_1) \\
&\subset \mathrm{supp}(u_1 \ast v_1) \\
&\subset \mathrm{supp}(u_1) + \mathrm{supp}(v_1) \\
&\subset L(\delta)+M(\delta) \\
&\subset L+M
\end{align*}
Thanks in advance for any comments or corrections.
 A: I do not have the book of Duistermaat available at the moment, so I cannot compare your proof with the proof in the book. At the moment I see two problems with the proof: 
1) Since you have $L(\delta)\subseteq L$ and not $L\subseteq L(\delta)$, I do not see why $(1-\psi)u$ should be a smooth function.
2) One of your distributions $u$ or $v$ might have non-compact singular support. Assume $u$ is the one where $\operatorname{sing supp} u$ is non-compact. Since $\operatorname{supp}\psi$ is compact there is no chance that $1-\psi$ can vanish on the whole singular support.
I think that these two problems are the reason why in the "standard proof" (e.g. given in the book of Hörmander) only the compactly supported distribution is split in a smooth and a singular part. In addition usually it is assumed that
$$
\psi(x)=1 \text{ for } x\in K \text{ and } \operatorname{supp}\psi\subset B(K,\delta)
$$
since then obviously $(1-\psi)u\in\mathcal{E}(\mathbb{R}^{n})$. On the other hand, in this case you have to show that also the convolution of the singular part with $v$ is smooth on a certain open set.
