Distribution Theory/Support and singular support

Definition (being zero on open sets):

Let $M$ be a smooth manifold, let $U\subseteq M$ be open, and let $T\in {\mathcal {D}}'(U)$. Let $V\subseteq U$ be an open subset. We say that $T$ is zero on $V$ iff for all $\varphi \in {\mathcal {D}}(V)$ we have $T(\varphi )=0$.

Proposition (distribution is zero on union of opens where it is zero):

Let $U\subseteq M$ ($M$ being a smooth manifold), and let $T\in {\mathcal {D}}'(U)$. Suppose that $T$ is zero on a family of open subsets $V_{\alpha }\subseteq U$ ($\alpha \in A$). Then $T$ is also zero on

$V:=\bigcup _{\alpha \in A}V_{\alpha }$.

Proof: Let $\varphi \in {\mathcal {D}}(V)$. Then $K:=\operatorname {supp} \varphi$ is a compact subset of $V$. Hence, extract a finite subcover $K\cap V_{\alpha _{1}},\ldots ,K\cap V_{\alpha _{n}}$. Then pick a finite partition of unity on $K$ of functions that are subordinate to $V_{\alpha _{1}},\ldots ,V_{\alpha _{n}}$ (using that $K\cap \bigcup _{i\neq j}(V\setminus V_{\alpha _{i}})$ is a compact subset of $V_{\alpha _{j}}$ and convolving an indicator function on that with a mollifier of sufficiently small support), and use linearity of $T$. $\Box$

Definition (support):

Let $T\in {\mathcal {D}}'(U)$, where $U$ is an open subset of a smooth manifold. The support of $T$ is the set

$U\setminus \bigcup V$,

where the union ranges over all open $V\subseteq U$ on which $T$ is zero.