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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 $X$.

HK Lee
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  • I don't think any examples exist, but any proof that no examples exist must use the finite area of $[0,1]^2$ in an essential way. If $S\subseteq[0,1]$ has "area" $\int_S{\frac{dx}{x}}$, then the area of $[0,1]$ is infinite. Moreover: there are ${a_j}{j=0}^\infty$ such that $a_0=1$ and $[a{j+1},a_j]$ has area $1$. Likewise, there exists $b_{n,j}\in(a_{n+j+1},a_{n+j})$ such that $[b_{n,j},a_{n+j}]$ has area $2^{-j}$. If $A_n=\bigcup_{j=0}^{\infty}{[b_{n,j},a_{n+j}]}$, then $A_n$ always has area $1$, but $A_n\to{0}$, which has area $0$. – Jacob Manaker Sep 17 '22 at 04:14
  • Here we use Euclidean Lebegue measure. – HK Lee Sep 17 '22 at 04:16
  • Yes, I know. I'm trying to isolate the properties of Lebesgue measure that rule out any examples; in order to do this, I need to look at other measures that satisfy some (but not all) of those properties. (Also, note that $[0,1]$ is still compact, so it's not compactness that's key here.) – Jacob Manaker Sep 17 '22 at 05:02

3 Answers3

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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 $\lim\ \sup_n\ \lambda (A_n)\leq \lambda (A)$

KCd
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HK Lee
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  • I don't follow the last paragraph. $d(A_j,A)$ is small if $A$ is contained in a tubular neighborhood of $A_j$ or if $A_j$ is contained in a tubular neighborhood of $A$. How do you rule out the second case? Moreover, tubular neighborhoods can increase measure by much more than $\epsilon$; consider a tubular neighborhood of $\left{(x,y):x\in\left{\frac{1}{4},\frac{1}{2},\frac{3}{4}\right},y\in\left[\frac{1}{4},\frac{3}{4}\right]\right}$. – Jacob Manaker Sep 17 '22 at 18:12
  • No ! Since $\lambda$ is Lebesgue measure so there is open set $V$ whose area can not exceed $\lambda(A)+\epsilon$. In your example, area of the tubular-neighborhood of your set can not exceed $\epsilon$ if tubular width is small. – HK Lee Sep 17 '22 at 18:32
  • OK, I see. I still don't follow how you rule out the case of $A_j\subseteq U_\epsilon(A)$, though. – Jacob Manaker Sep 17 '22 at 18:39
  • If we assume $A_j$ goes to $A$, then we have that $A_j$ is in $r$-tubular neighborhood of $A$. This is a definition of $A_j\rightarrow A$. Hence $A_j\subset U_r(A)\subset V$ for some open set $V$ if $r$ is sufficiently small. – HK Lee Sep 17 '22 at 18:42
  • Oh! Somehow I had my conjunctives mixed up. TY! – Jacob Manaker Sep 17 '22 at 18:45
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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

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

orangeskid
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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.

Jacob Manaker
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