Integrability of almost product structure An almost produsct structure on a smooth manifold $M$ is a pair $(D,D')$ of complementary distributions. One says that the structure is integrable if there is a local map $(U,\phi^i,\psi^\alpha)$ for every point on $M$ with property: $$D=span \left\{\frac{\partial}{\partial \phi^i} \right\}$$
$$D'=span \left\{\frac{\partial}{\partial \psi^i} \right\}.$$ Generally the integrability of an almost product structure is more than an integrability of both distributions in any sence you like (involutivity or existance of a foliated atlas), yet I have faced the fallowing statement in $\textit{Foliations and Geometric Structures - A. Bejancu, H. R. Farran}$:

if $D$ is a distribution of codimension one, then both $D$ and $D'$ are integrable and we may consider foliated charts whose coordinates $(x^1, ..., x^{m−1}, t)$. [...] In this case we can choose a natural frame field
  $\left\{
\frac{∂}{
∂x^i}
, \frac{∂}{
∂t} \right\}$
  on M such that
  $\frac{∂}{
∂x^i}
∈ Γ(D)$ and $\frac{∂}{
∂t}
∈ Γ(D').$

I have been thinking for a while yet can't see why can we assume also that the last canonical vector field spans the last distribution?  
 A: First, pick a coordinate chart $$y = (y^1, ... , y^m) \,\, \in \, \,\mathbb{D}^{(m-1)} \times (-\delta, \delta)  \,\, \subset \,\, \mathbb{R}^m$$ with $\mathbb{D}^{(m-1)} = \{\hat{y} = (y^1, ..., y^{m-1}) \, : \,  -1 < y^j < 1\}$ and $y^m \in (-\delta,\delta)$, where the codimesnion one foliation $F_D$ of the distribution $D$  looks like  $$F_D = \{ \,\mathbb{D}^{(m-1)} \times\, \{y_0^m\} \,\, : \,\, y^m_0 \in (-\delta,\delta)\, \}$$ i.e. $D = \text{span}\left\{\frac{\partial}{\partial y^1}, ... , \frac{\partial}{\partial y^{m-1}}\right\}$. In such a chart, the other foliation $D'$, which is in fact a line field, is panned by a vector field $$Y(y) = \sum_{j = 1}^{m}  \, Y^j(y) \, \frac{\partial}{\partial y^j} = \left( \, \sum_{j = 1}^{m-1}  \, Y^j(y) \, \frac{\partial}{\partial y^j} \, \right) \, + \, Y^m(y) \, \frac{\partial}{\partial y^m}$$ transverse to $D$, meaning that $Y^m(y) = Y^m(y^1,..., y^m)$ is never zero. Therefore, one can rescale the vector field $Y(y)$ by dividing it by $Y^m(y)$ and get the vector field
$$X(y) = \frac{1}{Y^m(y)} \, Y(y) = \left( \, \sum_{j = 1}^{m-1} \, \,\frac{ Y^j(y)}{Y^m(y)} \,\, \frac{\partial}{\partial y^j} \, \right) \, + \,  \frac{\partial}{\partial y^m}$$
which after setting $X^j(y) = \frac{ Y^j(y)}{Y^m(y)} $ for all $j=1, ...,m-1$ looks like
$$X(y) = \left( \, \sum_{j = 1}^{m-1} \, \,X^j(y) \,\, \frac{\partial}{\partial y^j} \, \right) \, + \,  \frac{\partial}{\partial y^m}$$ or if we split the variables into $y = (\hat{y}, y^m)$, where $\hat{y} = (y^1, ..., y^{m-1})$
$$X(\hat{y}, y^m) = \left( \, \sum_{j = 1}^{m-1} \, \,X^j(\hat{y}, y^m) \,\, \frac{\partial}{\partial y^j} \, \right) \, + \,  \frac{\partial}{\partial y^m}$$ The new vector field $X(y) = X(\hat{y}, y^m)$ still spans the line field $D'$. A phase curve of $X(\hat{y}, y^m)$, starting form a point $(\hat{y}, y^m)$, looks like $$t \, \mapsto \,\big(\,\hat{y}(t \, | \,y^m, \, \hat{y}), \, y^m + t\,\big)$$ where $\hat{y}(y^m  \, | \,y^m, \, \hat{y} ) = \hat{y}$. Whit this special construction in mind, define the diffeomorphism $$\Phi \, : \, \mathbb{D}^{(m-1)} \times (-\delta, \delta) \, \to \, \mathbb{D}^{(m-1)} \times (-\delta, \delta) $$ by 
$$\Phi \, : \, (x,t) \, \mapsto \, \big(\hat{y}(t  \, | \,0, \, x ) , \,  t\,\big)$$ where $x = (x^1,... , x^{m-1}) \in \mathbb{D}^{(m-1)}$. You may have to shrink the domain of the map $\Phi$, I am being a bit sloppy and I am abusing notation. Then 
$$\Phi_*\left(\frac{\partial}{\partial t}\right) = \frac{d}{dt} \Phi(x, t) = \frac{d}{dt} \hat{y}(t  \, | \,0, \, x )  + \frac{dt}{dt} \, \frac{\partial}{\partial y^m} = X(\hat{y}, y^m) = X(y).$$ Notice that if we take an $m-1$ dimensional slice $\mathbb{D}^{(m-1)} \times \{t\}$ the diffeomorphism $\Phi$ maps it to $\Phi\big(\mathbb{D}^{(m-1)} \times \{t\}\big) = \mathbb{D}^{(m-1)} \times \{t\}$, which is a leaf of the foliation $F_D$, because $\Phi(x,t) = (\hat{y},t)$. Therefore, in the new coordinate chart $\Phi$ with local coordinates $(x^1,..., x^{m-1}, t)$, the distribution $D$ of the foliation $F_D$ is still spanned by $D = \text{span}\left\{\frac{\partial}{\partial x^1}, ... , \frac{\partial}{\partial x^{m-1}}\right\}$ while the transverse one-dimesnional distribution $D'$ is now spanned by $D' = \text{span}\left\{\frac{\partial}{\partial t}\right\}$.
