Let $C_n$ be a sequence of rectifiable simple closed curves in $\mathbb{R}^2$ that converge to a rectifiable simple closed curve $D$ in the Hausdorff topology. It is easy to construct examples where

$$\limsup_{n \mapsto \infty} \text{length}(C_n) \neq \text{length}(D).$$

Question 0: I actually do not know any examples where the limit on the LHS does not exist. Can anyone provide one?

Question 1: In all the examples I know of, we have

$$\limsup_{n \mapsto \infty} \text{length}(C_n) \geq \text{length}(D).$$

Must this always hold?

Question 2: If the $C_n$ bound convex regions, must we have

$$\lim_{n \mapsto \infty} \text{length}(C_n) = \text{length}(D)?$$

Question 3: Let $R_n$ be the region bounded by $C_n$ and let $U$ be the region bounded by $D$. Must we have

$$\lim_{n \mapsto \infty} \text{area}(R_n) = \text{area}(U)?$$

Edit: In the original version, I had the inequality in Question 1 backwards (thanks to Emil Jeřábek and kaleidoscop for pointing that out!). This has now been corrected. The examples I had in mind were a little easier than the ones in kaleidoscop's answer. Namely, let $C_n$ consist of the union of the sets

$$\{\text{$(t,0)$ $|$ $0 \leq t \leq 1$}\}$$


$$\{\text{$(1,t)$ $|$ $0 \leq t \leq 1$}\}$$

with a "staircase" that starts at $(0,0)$, goes $1/n$ up, then goes $1/n$ to the right, then $1/n$ up, then $1/n$ to the right, etc, ending at $(1,1)$. Each $C_n$ has length $4$. However, the $C_n$ converge to the union of the sets

$$\{\text{$(t,0)$ $|$ $0 \leq t \leq 1$}\}$$


$$\{\text{$(1,t)$ $|$ $0 \leq t \leq 1$}\}$$


$$\{\text{$(t,t)$ $|$ $0 \leq t \leq 1$}\},$$

which has length $2+\sqrt{2}$.

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    $\begingroup$ For Q0, just define $C'_{2n}=C_n$, $C'_{2n+1}=D$. Concerning Q1, all examples I can immediately think of have the opposite inequality. Are you sure you wrote it the intended way? $\endgroup$ – Emil Jeřábek Jun 11 '15 at 17:11
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    $\begingroup$ Concerning question 1, like said Emil, it is probably the other inequality. See the answer to this post,mathoverflow.net/questions/72002/… where the limit is strictly larger, but the curves are not rectifiable, I'm not sure one can "rectify" this example. $\endgroup$ – kaleidoscop Jun 11 '15 at 17:22
  • $\begingroup$ You're both right, I got the inequality in the first question backwards. Thanks for pointing that out! $\endgroup$ – Blair Jun 11 '15 at 19:49

The answer of question 1 is positive. In fact, you can even replace the $\limsup$ by a $\liminf$: this is Golab's theorem that the $1$-dimensional Hausdorff measure is lower semi-continuous on the set of compact sets of the plane. I did not find a good reference to this theorem, but it is cited in a paper by Raphaël Cerf which seems to contain other relevant information.

I think the answer to question 2 is also positive. First, $D$ must bound a convex domain; for each $\varepsilon>0$ we can then find two convex approximations $D_i,D_o$ of $D$, one on the inside and one on the outside, which are disjoint from $D$ and at Hausdorff distance less than $\varepsilon$ from $D$, and of length $\varepsilon$-close to the length of $D$. Now, when $C_n$ is close enough to $D$, it must lie in the annulus bounded by $D_i$ and $D_o$. Using the projections $P_n$ to the domain bounded by $C_n$ and $P_i$ to the domain bounded by $D_i$, and recalling they are $1$-Lipschitz, we get (denoting lengths by $|\cdot|$): $$ |D|-\varepsilon \le |D_i| = |P_i(C_n)| \le |C_n| = |P_n(D_o)| \le |D_o| \le |D|+\varepsilon $$

I also think the answer to question 3 is positive. I will not sketch a proof, but it should be doable using the regularity of Lebesgue measure and and the fact that the Minkowski content of a domain with rectifiable boundary is the length of its boundary. Maybe you will need to adapt the proof of that fact, but the key-word "Minkowski content" should get you going.

  • $\begingroup$ Thanks! I would be very interested in any additional comments you have about Question 3 if you have time. $\endgroup$ – Blair Jun 14 '15 at 3:01

I'm not sure I understand everything because some questions seem weird, so let me try this answer:

Let $D$ be the segment $[0,1]$, and $C_n$ the graph of the function $f_n(x)=n^{-1}sin(n^3x)$. Then the Hausdorff distance between $D$ and $C_n$ is $n^{-1}$, but the length of $C_n$ goes to infinity. This can be seen heuristically by the fact that the length of $C_n$ is $n^3$ times the length of a small sinusoid that scales like $n^{-2}$.

I'm not sure I understand question $0$, because if you take $C_n$ for even $n$ and $D$ for odd $n$ you have no limit...

If you want a closed curve you can just glue such curves together to form a loop.

Questions 2 and 3 are more tricky, but let me first know if that answers your first questions, to make sure I got it right.

  • $\begingroup$ Thanks! You are correct that I had the inequality in the first question wrong (and I'm a little ashamed that I didn't see how easy Question 0 was; I inserted it as an afterthought while writing the rest). $\endgroup$ – Blair Jun 11 '15 at 19:58

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