Prove that Koch's Snowflake Curve is Jordan Measurable Obviously we can not prove this fractal is Jordan Measurable with standard proofs, making some rectangles and talk them smaller than $\varepsilon$ or something. The problem is of course in the infinite long boundary.
I thought maybe to use induction, but I'm not sure on what (the amount of straight lines? The amount of extra lines per step?) and how. Anybody an idea?
 A: 
A bounded subset of $\mathbb R^n$ is Jordan measurable iff its boundary has Lebesgue measure $0$. 

(See wikipedia.) The Koch snowflake $K$ is bounded, and its boundary equals $K$ because $K$ has empty interior and is closed. Hence $K$ is Jordan measurable iff $\partial K = K$ has Lebesgue measure $0$.

The image of a smooth map $f \colon \mathbb R^n \longrightarrow \mathbb R^{n+1}$ has Lebesgue measure $0$ in $\mathbb R^{n+1}$.

Let $D_n$ be the subset of $\mathbb R^2$ formed by drawing the $n$th iteration of the Koch snowflake. It is the union of the images of $3 \cdot 4^n$ smooth maps (which are just straight lines), and by finite additivity of the Lebesgue measure, $D_n$ has Lebesgue measure $0$ for all $n \in \mathbb N$. Since $K \subset \bigcup_{n=1}^\infty D_n$, it follows from monotonicity and countable additivity of the Lebesgue measure that $K$ has Lebesgue measure $0$. 

UPDATE: here's a way around the Lebesgue measure.
Let $D_n$ be as above. Each of the $3 \cdot 4^n$ segments of $D_n$ has length $\tfrac{s}{3^n}$ if our original equilateral triangle had side length $s$. We form a (finite) open cover of $D_n$ by enclosing each of its segments in an open rectangle $R_n$ which has length $\tfrac{s}{3^n} + l_n$ and width $w_n$. Any numbers $l_n, w_n > 0$ will do (we'll choose them later). Since $K$ is closed and bounded, it's compact, and $\bigcup_{n=1}^\infty R_n$ is an open cover of $K$, so it reduces to a finite subcover $R_{n_1}, \dots, R_{n_k}$ for some $k \in \mathbb N$. Then the Jordan measure of $K$ is bounded above via
$$|K| \leq \sum_{j=1}^k |R_{n_j}| \leq \sum_{n=1}^\infty |R_n| = \sum_{n=1}^\infty 3 \cdot 4^n w_n \left( \frac{s}{3^n} + l_n \right).$$
Given any $\varepsilon > 0$, if we take $w_n = \tfrac{\varepsilon}{2s \cdot 3 \cdot 4^n}$ and $l_n = \tfrac{s}{3^n}$, then the above becomes $|K| \leq \tfrac{\varepsilon}{2} < \varepsilon$ using geometric series, and it follows that $K$ has Jordan measure $0$ and is Jordan measurable.
