Length of non-horizontal curve Let $M$ be a sub-Riemannian space. 
Consider a smooth curve $\gamma:[0,1]\to M$ such that 
$\dot\gamma(t)\not\in  H_{\gamma(t)}$, where $H_{\gamma(t)}$ is the horizontal subbundle ( i.e. $\gamma$ is totally non-horizontal curve).
Is it obvious that the curve is not rectifiable or has infinite length?
 A: Here is a rigorous proof that non-horizontal curves are not rectifiable.
First, recall that, given a metric space $(M,\delta)$ and a mapping $\gamma:[0,1]\to M$, the $\delta$-length of $\gamma$ is, by definition,
$$
\ell_\delta(\gamma) = \sup\left\{\delta\bigl(\gamma(0),\gamma(t_1)\bigr)+\delta\bigl(\gamma(t_1),\gamma(t_2)\bigr)+\cdots+\delta\bigl(\gamma(t_m),\gamma(1)\bigr)\ \bigl|\ 0< t_1 <t_2<\cdots<t_m<1 \right\}.
$$
If $\ell_\delta(\gamma)<\infty$, one says that $\gamma$ is $\delta$-rectifiable. 
Second, a basic result in Riemannian geometry is this: If $g:TM\to\mathbb{R}$ is a Riemannian metric on a connected manifold $M$ and $\delta_g:M\times M\to [0,\infty)$ is the associated distance metric, then, for any piecewise $C^1$-mapping $\gamma:[0,1]\to M$, one has
$$
\ell_{\delta_g}(\gamma) = \int_0^1 g\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t.
$$
Now, suppose that $M$ is a manifold with a smooth plane field $H\subset TM$ with the property that each pair of points in $M$ can be joined by some $H$-curve, i.e., a piecewise $C^1$-curve $\gamma:[0,1]\to M$ such that $\gamma'(t)$ lies in $H_{\gamma(t)}$ for all $t\in[0,1]$.  Suppose that $h:H\to\mathbb{R}$ is a smooth function that restricts to be a positive definite quadratic form on $H_x$ for each $x\in M$.  Then one can define a metric $\delta:M\times M\to[0,\infty)$ by the formula 
$$
\delta(x,y)=\inf\left\{ \int_0^1 h\bigl(\gamma'(t)\bigr)^{1/2}\ \mathrm{d}t\ \bigl|\ \gamma:[0,1]\to M\ \text{is an $H$-curve}, \gamma(0)=x,\gamma(1)=y\right\}.
$$
Proposition: $\ell_\delta(\gamma) = \infty$ for any $\gamma:[0,1]\to M$ that is piecewise $C^1$ but not an $H$-curve.
Proof:  Choose a smooth splitting $TM = K\oplus H$ where $K$ is a smooth plane field (necessarily of positive rank, or else all curves are $H$-curves and there is nothing to prove).  Let $k:K\to \mathbb{R}$ be a smooth function that restricts to be a positive definite quadratic form on each $K_x$ for $x\in M$.  Consider the family of Riemannian metrics $g_n = n\,k \oplus h$ on $M$, and let $\delta_n$ be the distance metric on $M$ associated to $g_n$.  Then it follows directly from the definitions that $\delta(x,y)\ge \delta_n(x,y)$ for all $x,y\in M$.  Consequently, it follows (again from the definitions) that
$$
\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)
$$
for all maps $\gamma:[0,1]\to M$.  Now suppose that $\gamma:[0,1]\to M$ is piecewise $C^1$ but not an $H$-curve. Writing $\gamma'(t) = a(t) + b(t)$, where $a(t)$ lies in $K_{\gamma(t)}$ and $b(t)$ lies in $H_{\gamma(t)}$, one has that $a(t)$ is non vanishing for $t$ in an open subset of $[0,1]$.  Consequently, $\int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t > 0$.  But then, for all $n$, one has
$$
\ell_{\delta_n}(\gamma) = \int_0^1 \left(\,n\,k\bigl(a(t)\bigr) + h\bigl(b(t)\bigr)\,\right)^{1/2}\,\mathrm{d}t 
\ge \sqrt{n}\ \int_0^1 k\bigl(a(t)\bigr)^{1/2}\,\mathrm{d}t.
$$
Thus, the inequality $\ell_\delta(\gamma) \ge \ell_{\delta_n}(\gamma)$ for all $n$ implies that $\ell_\delta(\gamma) = \infty$.
A: Length is only defined for horizontal curves on a subRiemannian manifolds.
This is due to the subRiemannian (or subFinsler) idea that you are only allowed move along a subbundle of the tangent bundle.
For other curves it is natural to take the length to be infinite if you need to assign them a length.
One can approximate a subRiemannian metric by a sequence of Riemannian metrics which blow up in nonhorizontal directions.
Now if you fix a curve and take the limit of its lengths with respect to these approximating Riemannian metrics, the limit length can only be finite if the curve is horizontal everywhere.
Since $M$ is a metric space, you can of course define the length of a smooth curve $\gamma:[0,1]\to M$ in the usual way as a supremum of sums of distance between points.
If the curve is nonhorizontal, this supremum is infinite for the following reason.
Let $\Delta\subset TM$ be the subbundle where the subRiemannian metric lives (I denote by $TM$ the whole tangent bundle).
Iterated brackets of $\Delta$ generate the whole tangent bundle.
If $\dot\gamma(t)\in[\Delta,\Delta]\setminus\Delta$, then $d(\gamma(t+\epsilon),\gamma(t))\approx\epsilon^{1/2}$ for small $\epsilon$.
For more brackets you get smaller powers, so you should get $d(\gamma(t+\epsilon),\gamma(t))\gtrsim\epsilon^{1/2}$ when $\dot\gamma(t)\in TM\setminus\Delta$.
This scaling leads to infinite length.
I believe this argument is fairly obvious for those working with subRiemannian structures.
