I am trying to do the following exercise. Consider $E= \bigcup_{j=1} ^{\infty} I_j$ with $I_j$ one dimensional segments in $\mathbb{R}^n$ with lengths $l_j$ such that:
- $\sum_{j=1} ^{\infty} l_j < \infty$
- $E \subset B(0,1)$ and dense inside of it
- Also the directions of the $I_j$ are dense (meaning in every open set inside $B(0,1)$ you can find a countable set of directions of $I_j$ which are dense in $S^{n-1}$.
Now, $E$ is clearly 1-rectifiable. So the theory tells me that in $\mathcal{H}^1 $-a.e. point of $E$ I should have a 1-approximate tangent space which is a line. In particular, given $x \in I_j$ with $v_j$ its direction, I should have that
$$ Tan ^1 (E, x) = Span \{v_j\}.$$
My definition of approximate tangent space is as follows: given $x \in E \subset \mathbb{R}^n$ I say that $E$ has d-approximate tangent space $\pi \in G(d,n )$ (grassmanians) if $\rho ^{-d} \mu_{x, \rho} \to \mathcal{H}^d |_\pi$ as $\rho \to 0$ in duality with $C_c (\mathbb{R}^{n})$ functions (i.e. weak-* convergence of measure), where
$$ \mu_{x, \rho}= (\Phi_{x, \rho})_* \mu \text{ with } \Phi_{x,r} (y)= (y-x)/r ; \; \mu = \mathcal{H}^d |_{E} \text{ and } (\Phi_{x, \rho})_* \text{ denoting the push forward.}$$
For d-rectifiable sets the above property is true $\mathcal{H}^d$ -a.e., so in particular this is true for the set $E$ of my example. The question is: how do I show it by hand? Let me write here what I have done. Let's say $n=2$ to make the notation simpler and consider without loss of generality $0 \in I_1$ with $I_1= (-1,1) \times \{0\}$ and such that no other segment intersect the point. This is clearly not restrictive modulo translations and rotations. I want to show that given $\mu = \mathcal{H}^1 |_E$:
$$ r^{-1} (\Phi_{0,r})_* \mu \to \mathcal{H}^1 |_ \pi \text{ with } \pi= Span \{(1,0)\}.$$
I take a function $g \in C_c (\mathbb{R}^2)$ and I get:
$$ r^{-1} \int_{\mathbb{R}^2} g(x) d (\Phi_{0,r})_* \mu = r^{-1} \int_{\mathbb{R}^2} g(x/r) d \mu = r^{-1} \sum_{j=1} ^{+\infty} \int_{I_j} g(x/r) d \mathcal{H}^1 .$$
Now, for $I_1$ you change variables and obtain exactly the limit, for the other ones you can do the same, but the problem is that I don't see why $r^{-1} \sum_{j=2} ^\infty \int_{I_j} g(x/r) d \mathcal{H}^1$ goes to zero as $r \to 0$. Assuming for example that $Supp(g) \subset B(0,1)$, when you change variables you only considering the intersections $I_j \cap B(0,r)$, and the key fact surely is that $g$ has compact support (since if you take $g \equiv 1$ the conclusion is false). I may be missing some easy fact, but I can't think how to deal with the density of the segments.