Almost complex manifolds are orientable I want to verify the fact that 

every almost complex manifold is orientable. 

By definition, an almost complex manifold is an even-dimensional smooth manifold $M^{2n}$ with a complex structure, i.e., a bundle isomorphism $J\colon TM\to TM$ such that $J^2=-I$, where $I$ is the identity.
For every tangent space $T_pM$, there exists tangent vectors $v_1,\cdots, v_n$ such that $(v_1,\cdots, v_n,Jv_1,\cdots,Jv_n)$ is a real ordered basis for $T_pM$. I've checked that any two such bases are related by a matrix with positive determinant.
So for each tangent space $T_pM$ I fix such a basis. Hence, I've obtained a possibly discontinuous global frame $X_1,\cdots,X_n,JX_1,\cdots,JX_n$. To show this frame determine an orientation on $M$, I've been trying to construct for each $p\in M$ a coordinate neighborhood $(U,x^1,\cdots,x^{2n})$ such that
$$(dx^1\wedge\cdots dx^{2n})(X_1,\cdots,X_n,JX_1,\cdots JX_n)>0$$ for every point in $U$.
How do I show that? Thanks!
 A: I finally figured this out, so I want to post my own answer, since I couldn't find a satisfactory answer myself online.
It suffices to prove that for every $p\in M$, there exists a local frame $(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ in some open neighborhood $U$ of $p$.
Fix $p\in M$ and choose an ordered basis $(v_1,\cdots,v_n,Jv_1,\cdots,Jv_n)$. Choose sections $\sigma_1,\cdots,\sigma_n$ such that $\sigma_i(p)=v_i$ for each $1\leq i\leq n$. Since $(\sigma_1|_p,\cdots,\sigma_n|_p,J\sigma_1|_p,\cdots,J\sigma_n|_p)$ is linearly independent,
$$\omega=\sigma_1\wedge\cdots\sigma_n\wedge(J\sigma_1)\wedge\cdots\wedge(J\sigma_n)\neq0$$
at $p$. Since $\omega$ is continuous, it is nonzero for some neighborhood $U$ of $p$. In other words,$(\sigma_1,\cdots,\sigma_n,J\sigma_1,\cdots,J\sigma_n)$ is linearly independent in $U$, so that it is a local frame.
A: Here's a slightly different way of looking at it. First, note that we can choose a Riemannian metric $g$ on $M$ that satisfies $g(X,Y) = g(JX, JY)$ (start with any Riemannian metric $h$ and then define $g(X,Y) = h(X,Y) + h(JX, JY)$). You can then show that $\omega(X,Y) = g(X,JY)$ is skew-symmetric: 
\begin{align} \omega(X,Y) & = g(X,JY) = g(JY, X) 
\\& = g(J^2Y, JX) = -g(Y, JX) = -\omega(Y,X)
\end{align}
and is therefore a 2-form. $\omega$ is also non-degenerate, since $\omega(X, -) = - g(JX, -)$ and a metric is non-degenerate by definition. Then since $\omega$ is non-degenerate, its top exterior power $\omega^n = \omega\wedge \dots \wedge \omega$ is nowhere zero. A nowhere vanishing top form defines an orientation on $M$, so we are done. 
A: I see you have added an answer - and Charlie's is fine too. 
Here is another way... 
First of all - you have checked this, but for completeness - if we endow a real vector space $V$ of dimension $2n$ with a complex structure (i.e.,view it as a $C=\mathbb R[J]$ module), then we have chosen an orientation. For, suppose  $v_1,\cdots, v_n$ and $w_1,\cdots, w_n$ are $C$-bases, and  $g $ is the transition matrix ($g$ has  entries in $C$, or  $g = a + Jb$, with $a$, $b$ real) between the two, with determinant $d\in C$. Then  $g,$ viewed as a transformation over the reals, has determinant  $\det_{\mathbb R}g =d\bar d$, and hence is positive. 
Proof: We wish to calculate $\det_{\mathbb R} g$. Tensor $V$ with  $\mathbb C$ - that is, extend scalars. The determinant does not change:$${\det}_{\mathbb R}g = {\det}_{\mathbb C} (g\otimes 1).$$
So we wish to calculate $\det_{\mathbb C} (g\otimes 1)$:  $g\otimes 1$ preserves $V_{\pm i}$, the $\pm i$-eigen-spaces of $J$. Thus, since $g\otimes 1$ acts as $a\pm ib$ on $V_{\pm i}$, and $V\otimes {\mathbb C}=V_i\oplus V_{-i}$, $${\det}_ {\mathbb R} g=\det g\otimes 1 = \det g\otimes 1|_{V_i}\cdot \det g\otimes 1|_{V_{-i}} = \det (a +ib) \cdot \det (a-ib) >0.$$
Therefore, since
$$ w_1\wedge\cdots \wedge w_n \wedge J w_1\wedge\cdots \wedge J w_n = {\det}_ {\mathbb R} g\ v_1\wedge\cdots \wedge v_n \wedge J v_1\wedge\cdots \wedge J v_n,$$
the orientation is well-defined.
To answer the question proper...
By assumption, $J$ is a global tensor, and thus each $T_p(M)$ is endowed with a $J$ structure, and hence (by the previous argument), an  orientation,  independent of charts. To bring this in line with the original formulation in your question: suppose 
$$\phi: U\subset M\to \mathbb R^{2n},$$ is a chart, and $p\in U$.
Then $\partial/\partial x_1|_p,\cdots ,\partial/ \partial x_{2n}|_p$ - obtained from $\phi$ (or $\phi^{-1}$), is an oriented basis. If the orientation fails to match the $J$ orientation, modify $\phi$ by swapping the first two coordinates. By  continuity of $J$, the (modified, if needed) chart will give you an oriented frame on all of $U$ that matches that of $J$. The manifold $M$ is  therefore orientable - that is the transition function jacobian determinants  will be positive.
A: This answer summarizes Charlie Cifarelli's nice answer using only words:


*

*A paracompact manifold can be endowed with a Riemannian metric.

*A paracompact, almost complex manifold can be endowed with an almost Hermitian metric.

*An almost Hermitian manifold is an almost symplectic manifold.

*An almost symplectic manifold has a canonical volume form, and is hence orientable.
