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


$$\lim\_{n \mapsto \infty} \text{area}(R\_n) = \text{area}(U)?$$




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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


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


and


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


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


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


and


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


and


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


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



Continuity of length and area
**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](http://www.numdam.org/item?id=ASNSP_2002_5_1_1_33_0) 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.


