 Solution

A smooth rank-$r$ smooth distribution $D\subset TM$ is involutive distribution if the Lie bracket of vector fields of two local sections remains a section. It is an integrable distribution if every point lies on an immersed $r$-dimensional integral manifold with tangent spaces equal to $D$. The Frobenius theorem says these conditions are equivalent, and gives local coordinates with $D=\mathrm{span}({\partial }_1,\dots ,{\partial }_r)$.

For necessity, fields tangent to an integral manifold have brackets tangent to it: they annihilate functions vanishing on the manifold, and so does their commutator. For sufficiency, induct on $r$. The rank-zero case is immediate. Straighten a nonvanishing local section using the flow-box theorem to obtain ${\partial }_1\in D$. Choose a frame ${\partial }_1,Y_2,\dots ,Y_r$ with the $Y_i$ having no ${\partial }_1$ component. Involutivity gives ${\partial }_{1}Y=AY$ for the column of these fields and a smooth matrix $A$. Solve the matrix ordinary differential equation ${\partial }_{1}B=-BA$, with $B=I$ on $x_1=0$. It stays invertible, and $Z=BY$ satisfies ${\partial }_{1}Z=0$. On the transverse slice, the $Z_i$ span an involutive rank-$(r-1)$ distribution. The induction hypothesis supplies adapted slice coordinates; extend them independently of $x_1$. These give the required rank-$r$ coordinate distribution and its integral manifolds.

For the real Heisenberg group, multiplication isDifferentiating left translation at the identity gives its left-invariant frame of the real Heisenberg group:Then $[E_1,E_2]=E_3$ and the other basis brackets vanish. Since $E_3\notin \mathrm{span}(E_1,E_2)$, the Heisenberg horizontal distribution is not involutive, hence not integrable by the Frobenius theorem.