Define the set $A \subset \mathbb R^2$ by proceeding as follows. Let $A_0$ be a closed equilateral triangular region of side 1. $A_1$ are the three equilateral triangular regions of side $\frac 1 3$ in the corners of $A_0$. In general $A_{j+1}$ are the equilateral triangular regions in the corners of each triangle of $A_j$ and the side of each triangular region in $A_{j+1}$ is a third of the side of each triangular region of $A_j$. Now we define $$A := \bigcap_{j=0}^\infty A_j \; .$$
This construction process is shown here (the thin lines are not part of $A_j$):
Now I want to show, that $$\mathcal H^1(A) = 1 \; ,$$ where $\mathcal H^1(A)$ is the Hausdorff measure of $A$ with $s = 1$. I think that I've managed to show that $\mathcal H^1(A) \leq 1$, but I have problems to show that $\mathcal H^1(A) \geq 1$. My idea was to consider consider the projection $p: \mathbb R^2 \to \mathbb R$, defined by $(x,y) \mapsto x$. By construction we have $p(A_j) = [0,1]$ for each $j \in \mathbb N$. Furthermore $p$ is Lipschitz with $\text{Lip}(p) = 1$. If I can show that $p(A) = [0,1]$, I'm finished, because then we have $$1 = \lambda^1(p(A)) = \mathcal H^1(p(A)) \leq \text{Lip}(p) \mathcal H^1(A) = \mathcal H^1(A) \; .$$ So my question is, if this idea is correct and does it hold that $p(A) = [0,1]$? If so, why does it hold?
If $A_n$ is a nested sequence ($A_{n+1}\subseteq A_n$) of compact sets and $y \in p(A_n)$ for all $n$, where $p$ is a continuous map from $A_1$ to a Hausdorff space, then I claim $y \in p(\bigcap_n A_n)$. Namely, let $x_n \in A_n$ with $p(x_n) = y$. The sequence $x_n$ must have a limit point $x$, and it's easy to prove that $x$ is in all $A_n$ and $p(x) = y$..