The staircase paradox, or why $\pi\ne4$ What is wrong with this proof?

Is $\pi=4?$

$\newcommand{\Reals}{\mathbf{R}}$There are a number of good answers to this primordial question, but none mentions the usual definition of arc length: If $\gamma:[a, b] \to \Reals^{n}$ is a continuous path, the arc length of $\gamma$ is the supremum, taken over all partitions $(t_{i})_{i=0}^{n}$ of $[a, b]$, of
$$
\sum_{i=1}^{n} \|\gamma(t_{i}) - \gamma(t_{i-1})\|.
\tag{1}
$$
This supremum always exists as a positive extended real number. If $\gamma$ is piecewise continuously-differentiable, the arc length is finite, and given by
$$
\int_{a}^{b} \|\gamma'(t)\|\, dt.
$$
(The integral is computed by partitioning $[a, b]$ into finitely many intervals on which $\gamma'$ is continuous and summing the contributions.)
Here, the circle can be (smoothly) parametrized by
$$
\gamma(t) = (\tfrac{1}{2}\cos t, \tfrac{1}{2}\sin t),\qquad 0 \leq t \leq 2\pi.
$$
The point is, the troll's approximation by "Manhattan" polygons does not give the supremum of (1), nor is there any reason to expect it should since none of the vertices lie on the circle.

On the subject, there is a genuinely vexing issue for surfaces in $\Reals^{3}$: The analog of (1) (form a triangular approximation whose vertices lie on the surface, sum the areas of the triangles, and take the supremum) is infinite even for a bounded portion of a right circular cylinder, a surface as smooth as one could hope. Spivak's Comprehensive Introduction to Differential Geometry has a nice diagram illustrating what goes wrong. Intuitively, crush a paper cylinder so it looks like an accordion bellows, and imagine this is done in such a way that the vertices of the crushed cylinder lie on another cylinder whose area we wish to approximate. The area of the bellows can be made as large as we like.
