Let $x \in \mathbb R^n$, $s>0$ and $\{C_r\}_{0<r<s}$ be a family of borel sets such that $B(x, \alpha r) \subset C_r \subset B(x, \beta r)$ for some fixed $0< \alpha < \beta$. Show that if $E$ is a Lebesgue measurable set and $t \in \{0,1\}$ then $$\lim_{r \to 0^{+}} \frac{\lambda(E \cap C_r)}{\lambda(C_r)}=t \Leftrightarrow \lim_{r \to 0^{+}} \frac{\lambda(E \cap B(x,r))}{\omega_n r^n}=t$$ But can be fails for some $t \in (0,1)$.
For see the limits i try to use the monotonicity of the measure and a change of variable e.g $t=\alpha r$, but i cant conclude. I think to i can study these problem via cases and using the theorem of lebesgue density points of $E$.
EDIT: I can conclude the case when $t=0$ but the case when $t=1$ not work with the change of variable, topologically i understand what happen, but i cant justify.
Any hint or help will be very grateful.
FShrike has already pointed out in the comments that the case $t=1$ follows from $t=0$ by replacing $E$ with $\mathbb R^n\backslash E$.
To obtain counterexamples for $t\in (0,1)$, we can argue as follows. The trick will be realizing that the condition $$B(x, \alpha r) \subset C_r \subset B(x, \beta r)$$ means we only really have to consider a discrete set of scales. We'll let $x=0$ and $s=1$, as all the ideas easily generalize.
For any set $A\subseteq \mathbb R^n$ and $r>0$, denote by $rA$ the set $rA=\{ra\mid a\in A\}$. Let $B=B(0,1)$ be the unit ball.
Let $S\subseteq B\backslash \frac{1}{2}B$ be any measurable subset of measure $\lambda (S) \in (0,\lambda(B\backslash \frac{1}{2}B))$.
Let $$E=\bigcup_{k=0}^\infty \frac{1}{2^k}S,$$
and define $C=S\cup \frac{1}{2}B$.
Since $\lambda(E)=\lambda(C\cap E)=\lambda(B\cap E)\in (0,\lambda(B))$, and $\lambda(C)<\lambda(B)$, we have $$0<\frac{\lambda(B\cap E)}{\lambda(B)}<\frac{\lambda(C\cap E)}{\lambda(C)}<1.$$
Moreover, for each $k\in\mathbb N$, it is easy to see from the scale invariant nature of our construction that
$$\frac{\lambda(\frac{1}{2^k}B\cap E)}{\lambda(\frac{1}{2^k}B)}= \frac{\lambda(B\cap E)}{\lambda(B)}$$ and $$\frac{\lambda(\frac{1}{2^k}C\cap E)}{\lambda(\frac{1}{2^k}C)}= \frac{\lambda(C\cap E)}{\lambda(C)}.$$
Now, for each $r>0$ we define $k_r:=\lfloor\log_2\frac{1}{r}\rfloor$, so that $r\leq \frac{1}{2^{k_r}}<2r$, and define $C_r:=\frac{1}{2^{k_r}}C$. Then $C_r$ contains $\frac{1}{2^{k_r+1}}B\supseteq \frac{r}{2}B$ and is contained in $\frac{1}{2^{k_r}}B\subseteq 2rB$, and so satisfies the requirements.
But then $$\lim_{r\to 0}\frac{\lambda(C_r\cap E)}{\lambda(C_r)}=\frac{\lambda (C\cap E)}{\lambda(C)}=:t\in (0,1)$$ (in fact the expression is constant in $r$), whereas for the subsequence $r_k=\frac{1}{2^k}$ we have $$\lim_{k\to \infty}\frac{\lambda(r_kB\cap E)}{\lambda(r_kB)}=\frac{\lambda (B\cap E)}{\lambda(B)}\neq t.$$
Finally, note that we can compute $t$ explicitly as $$t=\frac{\lambda(C\cap E)}{\lambda(C)}=\frac{\lambda(E)}{\lambda(C)} = \frac{\sum_{k=0}^\infty \frac{1}{2^{nk}}\lambda(S)}{\lambda(S)+\frac{1}{2^n}\lambda(B)} = \frac{\left(\frac{2^n}{2^n-1}\right)\lambda(S)}{\lambda(S)+\frac{1}{2^n}\lambda(B)} = \frac{\frac{2^n}{2^n-1}}{1+\frac{\lambda(B)}{2^n\lambda(S)}},$$ so $t$ varies between $0$ and $1$ as $\lambda(S)$ varies between $0$ and $\lambda(B\backslash \frac{1}{2}B)$, hence we obtain counterexamples for all $t\in (0,1)$.