Let $x \in \mathbb{R}^n$, $r_0 > 0$ and $\lbrace C_r \rbrace_{0 < r < r_0}$ be a family of Borel sets such that, for some $\beta > \alpha > 0$, $B(x,\alpha r) \subset C_r \subset B(x, \beta r)$ for $0 < r < r_0$ (for example, $C_r = x + rA,A$ open and bounded, $0 \in A$). Show that if $E \subset \mathbb{R}^n$ is a Lebesgue measurable set and $t \in \lbrace 0 , 1 \rbrace$, then $x \in E^{(t)}$ if and only if \begin{equation*} \lim_{r \to 0^+} \frac{|E \cap C_r|}{|C_r|} = t. \end{equation*} So far I got:
$\Rightarrow$: Let $x \in E^{(t)} = \lbrace x \in \mathbb{R}^n : \theta_n(E)(x) = t \rbrace$ be a point of density $t$, where \begin{equation*} \theta_n(E)(x) = \lim_{r \to 0^+} \frac{|E \cap B(x,r)|}{\omega_n r^n} \end{equation*} and \begin{equation*} \theta_n(E) = 1 \mbox{ a.e. on } E, \; \theta_n(E) = 0 \mbox{ a.e. on } \mathbb{R}^n \setminus E. \end{equation*} so \begin{equation}\label{here} \theta_n(E)(x) = \lim_{r \to 0^+} \frac{|E \cap B(x,r)|}{|B(x,r)|} \in \lbrace 0, 1 \rbrace. \end{equation} Let $\beta > \alpha > 0$ such that \begin{equation*} B(x, \alpha r) \subset C_r \subset B(x, \beta r) \end{equation*} for $0 < r < r_0$. From assume w.l.o.g. $t = 1$ and for $0 < r < r_0$ we get \begin{equation*} |E \cap C_r| \leq |E \cap B(x,\beta r)| \end{equation*} and \begin{equation*} |C_r| \geq |B(x, \alpha r)| = \omega_n \alpha^n r^n. \end{equation*} Hence for $0 < r < \beta r_0$ and $\tilde{r} := \frac{1}{\beta} r < r_0$ we get \begin{align*} \frac{|E \cap C_{\tilde{r}}|}{|C_{\tilde{r}}|} &\leq \frac{|E \cap B(x, \beta \tilde{r})|}{|B(x,\alpha \tilde{r})| } \\ &= \frac{|E \cap B(x,r)|}{|B\left(x,\frac{\alpha}{\beta}r \right)|} \\ &= \left( \frac{\beta}{\alpha} \right)^n \frac{|E \cap B(x,r)|}{\omega_n r^n}. \end{align*} Now for $r \to 0^+$ and therefore $\tilde{r} \rightarrow 0^+$ the right side converges to $\left( \frac{\beta}{\alpha}\right)^n$. Using $0 < r < \alpha r_0$ and $\tilde{r} = \frac{1}{\alpha} r$ we get \begin{align*} \frac{|E \cap C_{\tilde{r}}|}{|C_{\tilde{r}}|} &\geq \frac{|E \cap B(x, \alpha \tilde{r})|}{|B(x,\beta \tilde{r})| } \\ &= \frac{|E \cap B(x,r)|}{|B\left(x,\frac{\beta}{\alpha}r \right)|} \\ &= \left( \frac{\alpha}{\beta} \right)^n \frac{|E \cap B(x,r)|}{\omega_n r^n}. \end{align*} Hence \begin{equation*} \left( \frac{\alpha}{\beta} \right)^n \leq \lim_{r \to 0^+} \frac{|E \cap C_{\tilde{r}}|}{|C_{\tilde{r}}|} \leq \left( \frac{\beta}{\alpha} \right)^n. \end{equation*} This doesn't seem very helpful and I stuck on how to proof \begin{equation*} \lim_{r \to 0^+} \frac{|E \cap C_r|}{|C_r|} = t. \end{equation*} I would be thankful for every hint.
For the first part of your proof, what's the meaning in saying that $x\in E^{(t)}$ for $t\in\{0, 1\}$ then $\theta_n(E) = 1$ a.e. on $E$ and $\theta_n(E) = 0$ a.e. on $E^c$? The last equation you provided is just by definition since $x\in E^{(t)}$.
As for the proof, you assume wlog that $t = 1$. But what if you assume $t = 0$ instead?
Then from your proof it follows that $\lim_{r\to 0^+}\frac{|E\cap C_\tilde r|}{|C_\tilde r|} = \lim_{r\to 0^+} \frac{|E\cap C_r|}{|C_r|} = 0$, and this is what you wanted to show.