After some thought, I think the core of the question is: Can it be proved without using measure theory (lebesgue) that if f is continuous and g is riemann integrable then the composition $f\circ g$ is riemann integrable?
considering a function $\gamma:[a,b]\to \mathbb{R}^n$ integrable in the following sense:
exists $x_0 \in \mathbb{R}^n$ that forall $\varepsilon >0 $ exists $\delta >0$ such that forall $\Pi = \lbrace t_i \rbrace_{0\leq i \leq k}$ partition of $[a,b]$ that $$\max_{0\leq i \leq k-1}\left|| t_{i+1}-t_i \right\| = \lVert \Pi \rVert<\delta$$ it's worth that forall $P = \lbrace p_i \rbrace_{0\leq i \leq k-1}$ choice of intermediate nodes for $\Pi$ is true that $$ \lVert x_0 - \sum_{i=0}^{k-1} \gamma(p_i)\cdot(t_{i+1} -t_i ) \rVert < \varepsilon$$
Is true that then $|\gamma|:[a,b] \to \mathbb{R}$ ($ |\gamma|(t)=|\gamma(t)|$) is Riemann integrable?
I have understood that the definition that I put for integrability for curves is equivalent to that the coordinate functions are Riemann integrable (in particular, this definition implies that the coordinate functions of the curve are bounded functions)
Also, I understand that a function is Riemann integrable if is bounded and the set of discontinuities has measure zero. With this I see possible to put together a demonstration of this implication: Prove that the integrability of the curve implies continuity at almost all points, and this implies continuity at almost all points of the function by taking its module.
Assuming this reasoning is correct, is there a way to prove this implication without using Lebesgue's theory?