Does there exist a rectifiable curve with infinite indefinite integral? Since $\gamma\in C^1$ implies that $\gamma$ is rectifiable and that $\int_{a}^{b}\left|\gamma'(t)\right|dt=\Lambda(\gamma)<\infty$, I was wondering if there would be any counterexample in a more general setting, e.g. $\gamma'\in\mathcal{R}$.
In addition, I saw a lot of solutions claiming that the divergence of the indefinite integral $\int_{a}^{b}\left|\gamma'(t)\right|dt$ simply implies that $\gamma$ is not rectifiable. This didnt't make much sense to me as the equation $\int_{a}^{b}\left|\gamma'(t)\right|dt=\Lambda(\gamma)$ holds only if $\gamma'\in C^1 \left(\Rightarrow \gamma'\in\mathcal{R}\left[a,b\right]\right)$, which is not the case.
More explicitly, would there exist a curve $\gamma:\left[a,b\right]\rightarrow\mathbb{R}^k$ which satisfies the following conditions?

*

*$\gamma$ is differentiable and $\gamma'$ is Riemann integrable on all $\left[a,c\right]$ with $c<b$.

*$\gamma$ is rectifiable.

*$\int_{a}^{b}\left|\gamma'(t)\right|dt=\lim_{c\to b}\int_{a}^{c}\left|\gamma'(t)\right|dt=\infty$.

 A: Short answer.
No there is no such $\gamma$ satisfying those 3 conditions.

Long Answer.
Usually, theorems are true in much more general circumstances than you may have been exposed to, and that is indeed the case here. For example, the following statement is true:

Theorem 1.
If $\gamma:[a,b]\to\Bbb{R}^n$ is differentiable at every point of $[a,b]$, and $\gamma'$ is Riemann-integrable on $[a,b]$, then $\gamma$ is rectifiable and
\begin{align}
L(\gamma)=\int_a^b|\gamma'(t)|\,dt
\end{align}

Here, we're working with a proper Riemann-integral on the RHS. Even though the statement involves only Riemann integrals, I do not know how to prove this without more machinery involving absolute continuity/Lebesgue integrals, and especially the Lebesgue version of the FTC. In the above theorem, proving rectifiability of $\gamma$ and the inequality $\leq$ are elementary. The tough part is proving the inequality $\geq$.
Let me just mention how the proof goes. First, define the total variation function $\Gamma:[a,b]\to[0,\infty]$ as
\begin{align}
\Gamma(x):=L(\gamma|_{[a,x]}).
\end{align}
Now, let $\{t_0,\dots, t_N\}$ be an arbitrary partition of $[a,x]$; then
\begin{align}
\sum_{i=1}^N|\gamma(t_{i})-\gamma(t_{i-1})|&=\sum_{i=1}^N\left|\int_{t_{i-1}}^{t_i}\gamma'(t)\,dt\right|\tag{FTC}\\
&\leq \sum_{i=1}^N\int_{t_{i-1}}^{t_i}|\gamma'(t)|\,dt\\
&=\int_a^x|\gamma'(t)|\,dt
\end{align}
Thus, by taking the supremum over all the partitions, we have that $\Gamma(x):=L(\gamma|_{[a,x]})\leq \int_a^x|\gamma'(t)|\,dt$. In particular, this is true for $x=b$, thus it proves the rectifiability of $\gamma$ and the inequality $\leq$ mentioned in the theorem (in particular, it also shows $\Gamma$ is real-valued; i.e doesn't take the value $\infty$).
To prove the reverse inequality, let us temporarily make the extra hypothesis that $\Gamma$ is differentiable on $[a,b]$ and $\Gamma'$ is Riemann-integrable on $[a,b]$. Then, for any $t\in (a,b)$ and sufficiently small $h>0$, we have by the additivity of lengths of paths that
\begin{align}
|\gamma(t+h)-\gamma(t)|&\leq L(\gamma|_{[t,t+h]})\\
&=L(\gamma|_{[a,t+h]})-L(\gamma|_{[a,t]})\\
&=\Gamma(t+h)-\Gamma(t)
\end{align}
Dividing by $h$ and taking the limit as $h\to 0^+$ shows that $|\gamma'(t)|\leq \Gamma'(t)$. Integrating from $a$ to $x$ and using the FTC (the baby Riemann version) shows that
\begin{align}
\int_a^x|\gamma'(t)|\,dt&\leq \int_a^x\Gamma'(t)\,dt=\Gamma(x)-\Gamma(a)=\Gamma(x),
\end{align}
since $\Gamma(a)=0$.
This proves the theorem under the extra hypothesis of $\Gamma$ being differentiable with Riemann-integrable derivative. It turns out that this is unnecessary, but actually proving the theorem without these assumptions is much harder and I do not know how to do it without Lebesgue integrals and the notion of absolute continuity, and the corresponding Lebesgue-integral version of the Fundamental theorem of calculus.
For completeness, here's how the proof goes. We only have to assume $\gamma:[a,b]\to\Bbb{R}^n$ is absolutely continuous (for short, $\gamma$ is AC). Note also that Wikipedia only defines absolute continuity for functions with $\Bbb{R}$ as the target space, but the same definition can be made for $\Bbb{R}^n$-valued functions. From here, it is not too hard to show that $\Gamma$ is also AC. Hence (see theorem 7.20 in Rudin's RCA, which again holds for $\Bbb{R}^n$-valued functions by slight modifications), both $\gamma$ and $\Gamma$ are differentiable a.e on $[a,b]$ and the FTC holds. At any point $t\in (a,b)$ where $\gamma$ and $\Gamma$ are differentiable, we prove as above that $|\gamma'(t)|\leq \Gamma'(t)$. Since this is true for a.e $t\in [a,b]$, the Lebesgue-version of the FTC now implies
\begin{align}
\int_a^x|\gamma'(t)|\,dt&\leq \int_a^x\Gamma'(t)\,dt=\Gamma(x)-\Gamma(a)=\Gamma(x),
\end{align}
since $\Gamma(a)=0$.
So, the proof idea is actually very simple, but getting the technical details right under the weak assumptions is the difficult part.

To summarize, one has the following strong theorem:

Theorem 2.
If $\gamma:[a,b]\to\Bbb{R}^n$ is AC, then $\gamma$ is rectifiable and
\begin{align}
L(\gamma)&=\int_a^b|\gamma'(t)|\,dt
\end{align}
(using a Lebesgue integral on the RHS).

Theorem 1 is a special case of theorem 2 (since the hypotheses of theorem 1 imply those of theorem 2; see Rudin's RCA theorem 7.20 for this, and also because if the Riemann integral exists, so does the Lebesgue integral, and the two are equal).
In view of theorem 1 (the proof of which we had to take a long detour), it follows that there is no curve $\gamma$ satisfying the three conditions you pose.
