Rectifiability of space curve given rectifiability of projections Let $\gamma:[0,1] \to \mathbb{R}^3$ be rectifiable. Then the projections of $\gamma$ onto the planes perpendicular to the coordinate axes are rectifiable:
$$\int_{0}^{1} \left \| \dot{\gamma}(s) \right \| ds=\int_{0}^{1} \sqrt{x^2+y^2+z^2}ds<\infty$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{x^2+y^2}= \left \| \dot{\gamma}_{xy} \right \|$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{y^2+z^2}= \left \| \dot{\gamma}_{yz} \right \|$$
$$\sqrt{x^2+y^2+z^2} > \sqrt{z^2+x^2}= \left \| \dot{\gamma}_{zx} \right \|$$
with $\gamma_{ij}$ the projection on the $ij$-plane.
Does the converse hold?

$\gamma:[a,b]\to\mathbb{R}$ is rectifiable if $$\sup_{P} \sum_{i=1}^n |\gamma(t_i)-\gamma(t_{i-1})|<\infty,$$ where $P=\{t_i\}$ is a partition of $[a,b].$
If the projection over each coordinate plane is rectifiable then the curve is rectifiable. It is enough to have in mind the inequality
$$\sqrt{x^2+y^2+z^2}\le \sqrt{2}(\sqrt{x^2+y^2}+\sqrt{x^2+z^2}+\sqrt{y^2+z^2}).$$ Assume $x^2\le y^2\le z^2.$ Then
$$\sqrt{x^2+y^2+z^2}\le \sqrt{2y^2+2z^2}=\sqrt{2} \sqrt{y^2+z^2},$$ from where the inequality follows. 
