Continuity of length and area 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
$$\{\text{$(t,0)$ $|$ $0 \leq t \leq 1$}\}$$
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 $2+\sqrt{2}$.
 A: 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.
A: 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.
