General links between Hausdorff convergence and Hausdorff dimension 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.
 A: 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 $mes(A)=0$.
