First given $M \subset \mathbb{R}^n$, and $x \in \mathbb{R}^n$, $r>0$, we define the blow up by: $$ \Phi_{x,r}(y) = \frac{y-x}{r}.$$
We say $x \in \mathbb{R}^n$ has a $k$ dimensional approximate tangent space $\pi_x$ if
$$ H^k \llcorner \Big(\Phi_{x,r}(M) \Big) = H^k \llcorner \Big(\frac{M-x}{r} \Big) \text{ converges weakly to } H^k \llcorner \pi_x,$$
in other words,
$$\lim_{r \to 0} \frac{1}{r^k} \int_M \varphi \Big( \frac{y-x}{r} \Big) dH^k(y) = \int_{\pi_x} \varphi dH^k $$
for all $\varphi \in C^0_c(\mathbb{R}^n).$
Now, there is a Theorem that says rectifiable sets have approximate tangent spaces $H^k$ a.e., and where they exist they are in fact classical tangent spaces.
My question is can we find a set $E$ and a point $x \in \mathbb{R}^n$ not necessarily in $E$, such that an approximate tangent space $\pi_x$ exists but a classical tangent space does not?
We will need to consider "unrectifiable sets" due to the theorem.
A set that comes to mind is the purely unrectifiable cantor-like set $C \subset \mathbb{R}^2$ which we define inductively (paraphrased from Leon Simon), for $n = 0, 1, ...$ by:
$C_n$ is the union of $4^n$ squares each with length $4^{-n}$, and $C_{n+1}$ is obtained from $C_n$ by replacing each square s of $C_n$ with $4$ squares of edges length $4^{-n-1}$ each sharing a corner with s, then $C_{n+1} \subset C_n$. Then let $C = \cap_{n=1}^\infty C_n$.
I dont see if or how I can make this example work, but my question stands for maybe some other set.