I was reading the proof of Proposition 15.5 of Lee's Introduction to smooth manifold which states that
Let $M$ be a smooth $n$-form with or without boundary. Any nonvanishing $n$-form $\omega$ on $M$ determines a unique orientation of $M$ for which $\omega$ is positively oriented at each point. Conversely, if $M$ is given an orientation, then there is a smooth nonvanishing $n$-form on $M$ that is positively oriented at each point.
Lee proved only the first statement. Actually, he proved the second statement very roughly saying 'by the usual partition-of-unity argument (see problem 13-2),...'. I didn't solve problem 13-2 so I wonder if my proof is correct.
Suppose $M$ is oriented. Let $\Lambda_+^nT^*M\subset\Lambda^nT^*M$ be an open subset of positively oriented $n$-covectors. Note that $\Lambda^n_+T^*_pM$ is an open half line of $\Lambda^nT^*_pM$ which is convex. For each $p\in M$, there is an open subset $U_p\subset M$ such that there is a positively oriented local frame $(E_i):U_p\to TM$. Let $(\epsilon^i):U_p\to T^*M$ be a dual coframe on $U_p$ to $(E_i)$. Then $\epsilon_p=\epsilon^1\wedge\cdots\wedge\epsilon^n$ is a positively oriented nonvanishing smooth $n$-form on $U_p$. Let $\{\psi_p\}$ be a smooth partition of unity subordinate to $\{U_p\}$. Then define $\epsilon =\sum_p \psi_p\epsilon_p$. Since $\psi_p\geq 0$ for each $p$ and $\sum_p\psi_p =1$, by convexity assumption, $\epsilon\in \Lambda^n_+T^*M$ as desired.
Definition of orientable manifold:
