Is there a rectifiable curve whose arc-length isn't continuous?

Question

Let $\mathrm{V}$ be a normed space. We say that $f:[a,b] \to \mathrm{V}$ is rectifiable if $\sup\limits_P \Gamma_f(P) < \infty$ where $P = (s_i)_{i=0,\ldots,m}$ are increasing families of points with $s_0=a$ and $s_m=b$, and $\Gamma_f(P) = \sum\limits_{i=1}^m \|f(s_i) - f(s_{i-1})\|$ is the arc-length of the poligonal curve with vertices the points of $P$.

1.-2. below provide a rough sketch of why this makes sense intuitively:

1. If $P$ and $Q$ are two increasing families of points starting at $a$ and ending at $b$, and if $P$ is a subfamily of $Q$ then $0 \leq \Gamma_f(P) \leq \Gamma_f(Q)$. Intuitively, the finer the polygonal curve we use to approximate $f$, the better the approximation.
2. Therefore, the quantity $\mathscr{L}_f(a,b) = \sup\limits_P \Gamma_f(P)$ always exists, it can be either $+\infty$ or $<\infty$, in the second case we get rectifiability and the real number $\mathscr{L}_f(a,b)$ is the arc-lenght of $f$.
3. For all $[x,y] \subset [a,b]$, $\mathscr{L}_f(x,y) \leq \mathscr{L}_f(a,b)$. Therefore, if a curve is rectifiable, so is every restriction to a subinterval; if a restriction has infinite arc-length, so does the whole curve.
4. If $\mathscr{L}_f(a,b) = 0$ then $f$ is constant on $[a,b]$.
5. If $f$ is a rectifiable curve, define $s_f:[a,b] \to \mathbf{R}$ via $s_f(x) = \mathscr{L}_f(a,x)$. Then $s_f$ is well-defined and non-decreasing; if $f$ is non-constant on every open subinterval of $[a,b]$, then $s_f$ is increasing.

The main theorem regarding rectifiability follows (and, in fact, it is often that this theorem is used tacitly to define arc-length solely for $\mathscr{C}^1$-curves).

Theorem. Let $\mathrm{V}$ be complete (i.e., $\mathrm{V}$ is a Banach space) and $f$ be a curve with continuous first derivative. Then $f$ is rectifiable and $\mathscr{L}_f(a,b) = \int\limits_a^b \|f'\|$.

If $f$ is $\mathscr{C}^1$ on $[a,b]$, we apply the previous theorem on $[a,x]$ to obtain $s_f(x) = \int\limits_a^x \|f'\|$.

Corollary. $s_f$ is continuously differentiable if $f$ is $\mathscr{C}^1$, in fact, $s_f'(x) = \|f'(x)\|$.

I want a proof OR a counter-example of the following:

$s_f$ is continuous whenever $f$ is (only) assumed rectifiable.

I only have ancillary observations to this:

1. $s_f$ is non-decreasing, so it only can have jump-discontinuities.
2. Suffice to show $s_f$ is right continuous, for then, we can consider a curve $g$ traversed in the opposite direction, the right continuity of $s_g$ will imply the left continuity of $s_f$.
3. The example of Example of non-rectifiable curve with finite arc length integral is related but of no use since we will be assuming that $f$ is rectifiable.

Answer 1

I will post an answer now based on Hagen vol Eitzen's comment above. As is commonly done, I will call $P$ a "partition" of a compact interval, say $[a,b]$, if $P$ is an increasing finite family of points in $[a,b]$ starting at $a$ and ending at $b$ (i.e., as above).

Theorem. Let $f$ be a rectifiable left-continuous curve. Then $s_f$ is left-continuous.

Proof. Suppose that $s_f$ were left discontinuous at some $x \in (a, b]$. Then there is a partition $P$ of $[a,x]$ such that $\Gamma_f(P) \in \left( \sup\limits_{a \leq t < x} s_f(t), s_f(x) \right)$. Suppose $P = (p_0, \ldots, p_m)$, then $p_{m-1} < x = p_m$. Let $Q(q) = (p_0, \ldots, p_{m-1}, q)$ for $q \in (p_{m-1},x)$. Then $\varphi(q) = \Gamma_f(Q(q)) \leq s_f(q) \leq \sup\limits_{a \leq t < x} s_f(t)$ for all such $q$. By definition, $\varphi(q) = \sum\limits_{k=1}^{m-1} \| f(p_k) - f(p_{k-1}) \| + \|f(q) - f(p_{m-1})\|$ and the left-continuity of $f$ implies that of $\varphi$, so $\lim\limits_{q \uparrow x} \varphi(q) = \Gamma_f(P)$ but we also must have $\lim\limits_{q \uparrow x} \varphi(q) \leq \sup\limits_{a \leq t < x} s_f(t) < \Gamma_f(P)$, a contradiction. QED

Remark. Right-continuity of $s_f$ cannot follow if $f$ is not right-continuous as Nate Eldredge's comment shows. Consider a left continuous curve $f(t) = 0$ for $t \in [-1,0]$ and $f(t) = v$ for $t \in (0,1]$. Now let $P$ denote a partition of $[-1,x]$; if $0 < x$ then there is one $k$ and only one such that $p_{k-1} \leq 0 < p_k$. Then $\Gamma_f(P) = \sum \|f(p_i) - f(p_{i-1})\|$ will sum only zeroes except when the exceptional index $k$ exists, so $s_f(t) = 0$ for $t \in [-1,0]$ and $s_f(t) = \|v\|$ for $t \in (0,1]$. Actually, a direct proof as above of right-continuity is awkward as the inequalities that appear aren't in the right direction, rather is better to invert direction of traversal in the curve.

Theorem. Let $f$ be a rectifiable right-continuous curve. Then $s_f$ is right-continuous.

Proof. Let $u(t) = a+b-t$; note $u$ is it own inverse and it is decreasing and continuous. Define $g(t) = f(u(t))$ (for $t \in [a,b]$) the canonical opposite of $f$. We prove now that $s_f(x) + s_g(u(x))$ is a constant, the constant being the arc-length of $f$. Let $G$ and $D$ be partitions of $[a,x]$ and $[a,u(x)]$, then $u(D)$ is a partition of $[u(u(x)),u(a)] = [x,b]$ and therefore $u(D) = (x, D_0)$ and $(G,D_0)$ is a partition of $[a,b]$ with $x$ a member of it. Reciprocally, if $(g_0, \ldots, g_m, d_1, \ldots, d_n)$ is a partition of $[a,b]$ with $g_m = x$ then $G = (g_0, \ldots, g_m)$ defines a partition of $[a,x]$ and $u(D) = (x, d_1, \ldots, d_n)$ defines a partition of $[x,b]$ so $D = u(u(D))$ defines a partition of $[a, u(x)]$. Having all these, $\Gamma_f(G) + \Gamma_g(D) = \Gamma_f((G,D_0))$ and taking the supremum over $(G,D)$ is the same as taking the supremum over $(G,D_0)$ (since $x$ is fixed), so $s_f(x) + s_g(u(x))$ is the arc-length of $f$, which is constant. The right-continuity of $f$ implies (and in fact is equivalent to) the left-continuity of $g$ so $s_g(\cdot)$ is left-continuous by the theorem above, this implies the right-continuity of $s_f$. QED

To sum up intuitively: the continuity (including side-continuity or the lack thereof) of a curve is inherited by its arc-length function.