Let $M$ be an oriented $n$-dimensional manifold and $\omega\in \Omega^n(M)$ with compact support, $U\subseteq M$ open (if necessary you can assume that $U$ is the domain of a chart) and $$\operatorname{supp}(\omega)\subseteq U.$$ Since $U$ can be viewed as another oriented $n$-dimensional manifold and $\iota^*\omega\in \Omega^n(U)$, we may wonder whether $$\int_M\omega=\int_U\iota^*\omega$$ To do so, we first need to check that the RHS is well-defined and since I am working with Lee's Introduction to Smooth Manifolds this means that I need to check that $\iota ^*\omega$ has compact support. But is this always the case?
My failed attempt to show that $\iota^*\omega$ has compact support, in case you are interested: Since $\operatorname{supp}(\omega)\subseteq U$ it is tempting to conjecture that \begin{equation}\tag{1} \operatorname{supp}(\iota^*\omega)=\operatorname{supp}(\omega)=:A \end{equation} and since the subspace topology of $A\subseteq U$ equals the subspace topology of $A\subseteq M$ we would be done. If we introduce the set $$\{p\in M:\omega(p)\neq 0\}=\{q\in U:(\iota^*\omega)(q)\neq 0\}=:S,$$ then this boils down to showing that the closure of $S$ w.r.t. $U$ equals the closure of $S$ w.r.t. $M$. In particular, to obtain that \begin{equation}\tag{1} \operatorname{supp}(\iota^*\omega)\subseteq\operatorname{supp}(\omega) \end{equation} it suffices to show that $\operatorname{supp}(\omega)\in U$ is a closed subset that contains $S$ and this follows from $U\setminus S=U\cap (M\setminus U)$, but I wasn't able to show the other direction.