Area function is continuous on a set of compact sets in $[0,1]^2$

Question

Consider $X=[0,1]^2\subset \mathbb{R}^2$. If $H_X$ is a set of all compact sets in $X$, then we can define a metric $d$ on $H_X$ i.e. Hausdorff metric $d$ :

For $A,\ B\in H_X$, then $d(A,B)=R$ iff there exists a smallest $R$ s.t. $R<r$, $r$ is arbitrarily close to $R$ and $U_r(A)$ contains $B$ and $U_r(B)$ contains $A$, where $U_r(C)=\{ a\in X$|$ |a-c|\leq r$ for some $c\in C\}$ and $|\ -\ |$ is Euclidean distance.

If $A_n =\{ (x,y)\in X| y=\frac{i}{n},\ 0\leq i\leq n\}$, then $A_n$ goes to $X$. Hence if $Area$ is Euclidean Lebesgue measure, $$ \lim_n\ {\rm Area}\ (A_n)=0 < {\rm Area}\ X=1 \ (1)$$

Question : I want to know whether or not there is an example opposite to $\ast$. Is there an example $A_n$ with $A_n\rightarrow A$ s.t. ${\rm Area}\ A <\lim_n\ {\rm Area} \ A_n\ (2)$ ?

Remark : a. If we consider a length function on a set of continuous maps from unit interval to $X$, then we have $(2)$ but not $(1)$.

b. Note that area function is continuous on a set of all convex subsets in $X$.

Answer 1

For $A$ a subset of the metric space $X$ we have

$$A_{\epsilon} \colon = \{ x \in X \ | \ d(x, A) \le \epsilon\}$$

is a closed subset of $X$ and $$\bigcap_{n \ge 1} A_{\frac{1}{n}} = \bar A$$

In your case you also have

$$\lim_{n \to \infty} \mu(A_{\frac{1}{n}}) = \mu(\bar A)$$

Answer 2

If $\lambda$ is Lebesgue measure, then it is Caratheodory so that all compact sets in $X$ are measurable. Hence if $A$ is compact, then for $\varepsilon>0$, there is open set $V$ containing $A$ s.t. $$ \lambda (V-A)<\varepsilon $$

If $U_r(B)$ is a $r$-tubular neighborhood of a subset $B$, then there is $r>0$ s.t. $V$ contains the tubular neighborhood $U_r(A)$ : If not, there is a sequence $a_i$ not in $V$ s.t. $d_E(a_i,A)\rightarrow 0$ and $a_i\rightarrow a\in A$ where $d_E$ is a Euclidean distance.

Since $V$ is open so $V$ contains an open ball of center $a$ which is a contradiction.

So if $d(A_i,A)\rightarrow 0$ where $d$ is a Hausdorff metric, then $$\lambda (A_i)\leq \lambda (V) \leq \lambda (A)+\varepsilon $$ for all sufficiently large $i$. Hence we have $\lim\ \sup_n\ \lambda (A_n)\leq \lambda (A)$

Answer 3

No, no such sequences exist.

Let $K=\bigcup_n{\bigcap_{m\geq n}{A_m}}$. If $1_S(x)=\begin{cases}1&x\in S\\0&x\notin S\end{cases}$, then $$1_K(x)=\liminf_{n\to\infty}{1_{A_n}(x)}$$

Since $[0,1]^2$ has finite area, $1_{[0,1]^2}$ is integrable; by the dominated convergence theorem $$\lambda(K)=\int{1_K\,d\lambda}=\int{\liminf_{n\to\infty}{1_{A_n}}\,d\lambda}=\liminf_{n\to\infty}{\int{1_{A_n}\,d\lambda}}=\lim_{n\to\infty}{\lambda(A_n)}$$

But $\lim_n{A_n}$ (in the sense of Hausdorff distance) must contain $K$, as follows from the hint for Step (2) of this answer.