We know that if $f:[a,b] \rightarrow \mathbb{R}^{n}$ is a $C^1$ path then $$\ell(f)=\int_{a}^{b}\|f'(t)\|dt.$$
Moreover, in the proof of this result we use explicitly the continuity of $f'.$
I'm trying to find an example of a differentiable function (at every point of $[a,b]$) such that $f':[a,b] \rightarrow \mathbb{R}^n$ is not continuous and $$\ell(f) > \int_{a}^{b} \|f'(t)\|dt.$$
I found this answer, but if I'm not mistaken, the Cantor function is not differentiable at every point.
Can anyone help me with this example?
Such a curve does not exist. When $f$ is differentiable, we always have the bound $$ \ell(f) \leq \int_a^b \|f'(t)\|\,dt. $$ To see this, consider a partition $a = t_0 < \dots < t_n = b$ of $[a, b]$. Using the fundamental theorem of calculus, we can write $$ f(t_{i + 1}) - f(t_i) = \int_{t_i}^{t_{i + 1}} f'(t)\,dt, $$ where each side of the equation is vector-valued. By the triangle inequality for vector-valued integrals, we get $$ \|f(t_{i + 1}) - f(t_i)\| = \left\|\int_{t_i}^{t_{i + 1}} f'(t)\,dt\right\| \leq \int_{t_i}^{t_{i + 1}}\|f'(t)\|\,dt. $$ It follows that $$ \sum_i \|f(t_{i + 1}) - f(t_i)\| \leq \int_a^b\|f'(t)\|\,dt. $$ Taking a supremum of the LHS, we obtain the inequality above.
Edit: As pointed out in the comments, we have to be careful when applying the fundamental theorem of calculus in the above proof when $f'$ is not Riemann integrable. It turns out that the (second) fundamental theorem of calculus holds iff $f$ is absolutely continuous (see Theorem 7.18 of Rudin's Real and Complex Analysis). Moreover, if $f$ is everywhere differentiable and $f'$ is $L^1$, then $f$ is absolutely continuous (see Theorem 7.21 of Rudin). Therefore, the only case where the fundamental theorem of calculus fails is when $f'$ is not $L^1$, in which case the RHS of the inequality above is $\infty$.