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Let $A_n \subset \mathbb{R}^2$ be a closed subset for all $n\in \mathbb{N}.$ We assume that all $A_n$ have same Hausdorff dimension (we can even assume that $\dim(A_n) = 1$ for all $n$.) We also have $A_n \subset A_{n+1}$ for all $n$.

Assume moreover that there is a subset $A\subset \mathbb{R}^2$ such that $A_n \to A$ where the convergence is the convergence obtained thanks to Hausdorff distance. What can we say about the Hausdorff dimension of $A$ ?

Are there general theorems which make links between Hausdorff dimension and Hausdorff distance ? Can we switch the limit and the dimension ?

I would also like to ask the question with the Lebesgue measure. For example, if all the $A_n$ are such that $\text{mes}(A_n) =0$, do we have $\text{mes}(A) =0$ ?

Thank you for any help/reference.

C. Dubussy
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    I don't believe there's any connection whatsoever. The collection of finite sets is dense in the Hausdorff metric space. Thus, you can approximate any set you want as closely as you want with a finite set, which has measure zero. – Mark McClure Feb 02 '17 at 11:34
  • Thank you, this answer well enough my question about measure theory. Do you have any information with Hausdorff dimension ? – C. Dubussy Feb 02 '17 at 11:48
  • I think this answers the question specifically for dimension. Any set in $\mathbb R^n$ can be approximated by sets of any given dimension $d$ where $0\leq d \leq n$. – Mark McClure Feb 02 '17 at 12:43

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Unfortunately Hausdorff dimension does not behave well under Hausdorff convergence, essentially because Hausdorff distance is not good at detecting "holes".

The only exception I know is Golab's theorem, look for example at the question Lower semi-continuity of one dimensional Hausdorff measure under Hausdorff convergence: if $A_n$'s are connected then $$H^1(A)\leq\liminf_n H^1(A_n).$$ Therefore if the $1$-dimensional measures are uniformly bounded and $A_n\subset A_{n+1}$ then $A$ has dimension $1$, but you need all these assumptions.

As an example with $A_n$ $1$-dimensional, connected but not with uniformly bounded measure: take a continuous curve $\gamma:[0,1]\to\mathbb R^2$, whose image can be made of any dimension between $1$ and $2$. Take a sequence $\gamma_i$ of smooth curves converging uniformly to $\gamma$ (for instance by convolution), and consider $$A_n=\bigcup_{i=1}^n \gamma_i([0,1]).$$ Then $A_n$ has dimension $1$ but $A_n\to \gamma([0,1])\cup \bigcup_{i=1}^\infty \gamma_i([0,1])$ which has the same dimension as $\gamma([0,1])$.

For a more spectacular failure (without connectedness), take any compact set $K\subset \mathbb R^2$, which can be taken of any dimension between $0$ and $2$. $K$ is separable, being a subset of a separable space. Take a dense subset $\{d_i\}_{i\in\mathbb N}\subset K$, and define $$A_n=\bigcup_{i=1}^n \{d_i\}.$$ Then $A_n$ is finite for every $n$, thus of zero dimension, but $A_n\to K$ in the Hausdorff distance.

This last example also shows that even if $mes(A_n)=0$ it does not follow that $mes(A)=0$.

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Del
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  • When you talk about sigma-additivity, you mean that the limit of the $A_n$ for the Hausdorff distance is necessarily $\cup_n A_n$ ? – C. Dubussy Feb 02 '17 at 12:20
  • @MarkMcClure Sorry, I totally misread that request :) In any case the same example with finite points shows that $mes(A)$ is not necessarily zero – Del Feb 02 '17 at 13:04
  • @C.Dubussy Not necessarily! I will edit my post – Del Feb 02 '17 at 13:05
  • Oh ok, now it's perfectly clear. Thanks for your counter-example. – C. Dubussy Feb 02 '17 at 13:15
  • Do you know a kind of equivalent of Golab's theorem but this time for Lebesgue measure and Hausdorff convergence ? – C. Dubussy Feb 09 '17 at 12:39
  • @C.dubussy what would be a possible statement? – Del Feb 09 '17 at 20:00
  • Let $A_n \subset A_{n+1}$ a sequence of subsets with nice topological properties (for example they are union of segments) with $\text{mes}(A_n) = 0$ for all $n$ and $A_n \to A$ for Hausdorff distance. Then $\text{mes}(A) = 0$. – C. Dubussy Feb 09 '17 at 20:27
  • @C.dubussy if they are connected and 1 dimensional you can apply Golab. If they are not connected, in general the measure of the limit can be positive: the last construction with points in the answer is an example (you can also take segments instead of points) – Del Feb 10 '17 at 15:07
  • But Golab theorem is only for $H^1$ right ? Not for the Lebesgue measure or other measure. – C. Dubussy Feb 10 '17 at 18:18
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    @C.dubussy yes, but finite $H^1$ implies zero Lebesgue measure, so if you now that they are connected and with bounded $H^1$ measure, then the limit has finite $H^1$ measure, and in particular has zero Lebesgue measure. – Del Feb 10 '17 at 18:45
  • Thank you very much I didn't knew this property. (I'm a beginner with these kind of measures) – C. Dubussy Feb 10 '17 at 18:51