Check that the following curves $\gamma : [0, 1] \rightarrow \mathbb{R^2}$ are not rectifiable
$(a)\:\: \gamma(0) = (0,0)$ and $\gamma(x) = (x, xsin(1))$ for $x \neq 0$.
$(b)\:\: \gamma$ is a space-filling curve: by this we mean that the image of the continuous map $\gamma:[0, 1] \rightarrow \mathbb{R^2}$ is the unit square
$S=\{ (x,y) \in \mathbb{R^2}\: | \:0 \leq x \leq 1, 0 \leq y \leq 1\}$.
So we've found the first part of the problem (how to prove that that function is non-rectifiable), but we're having trouble conceptualizing the space-filling curve and how to apply the partitions that we would sum up to get the length, and so prove that it's non-rectifiable.
Best Answer
If $\gamma$ is a space filling curve, then it cannot have a finite length. For a rigorous proof, consider for a given $N$ the circles at the $N^2$ points of the form $(\frac{k}{N},\frac{l}{N})$ with $k,l = 0,1,\dots,N-1$, with radius being very small, say $\frac{1}{N^2}$. Each such circle contains a point on the curve. On the other hand, a curve connecting two such circles has length $\simeq \frac{1}{N}$, so the total length of the curve would have to be at least $\simeq N^2 \cdot \frac{1}{N} = N$.
A rectifiable curve has, by definition, finite length. Hence, a space-filling curve cannot be rectifiable.