Let $m,n$ be integers with $m\le n$ and $\mu$ be a Radon measure on $\mathbb{R}^n$ which satisfies the following conditions:
- $\mu (B_r (x))=\omega_m r^m$ for every $r>0$ and $x\in\mathrm{supp}\,\mu$, where $\omega_m$ is the volume of the unit ball in $\mathbb{R}^m$,
- the family of measures $\{ r^{-m}\mu_r \}_{r>0}$ converges weakly to $\left. \mathcal{H}^m \right|_V$ as $r\to\infty$, that is, $\lim_{r\to\infty}\int_{\mathbb{R}^n}\varphi\, d(r^{-m}\mu_r)=\int_V \varphi\, d\mathcal{H}^m$ for all continuous function $\varphi$ with compact support, where $\mu_r (A)=\mu (rA)$, $V$ is a $m$-dimensional subspace and $\mathcal{H}^m$ be the $m$-dimensional Hausdorff measure.
I want to show that for every $y\in V$ with $|y|=1$, there exists a sequence $\{ x_j \}_j$ in $\mathrm{supp}\,\mu$ such that $x_j/|x_j| \to y$ and $|x_j| \to\infty$, but I have no idea.
Any help?
Choose a sequence $r_k\rightarrow\infty$ as $k\rightarrow\infty$. By weak convergence, there exists a sequence $\{z_k\}$ such that $z_k\in$ supp $\mu_{0,r_k}$ and $z_k\rightarrow y$.
Since the supports of $\mu$ and $\mu_{0,r_k}$ are related by supp $\mu_{0,r_k} = \frac{1}{r_k}$ supp $\mu$, for each $z_k$ we have the existence of $x_k\in$ supp $\mu$ such that $x_k = r_k z_k$.