3.3

Part a.) Under the given conditions, we know that $\forall \epsilon > 0$ there exists $r > 0$ such that: $$\frac{m(E \cap [-r,r])}{m([-r,r])} \geq (1-\epsilon)$$ Now, simply observe (by the translation / rotation invariance of the Lebesgue Measure): $$m(-E \cap [-r,r]) = m(E \cap [-r,r]) \geq (1-\epsilon)2r$$ Thus, for $\epsilon < \frac{1}{2}$: $$0 < (1-2\epsilon)2r \leq m(-E \cap E \cap [-r,r])$$ ...and thus, since these sets have positive measure, we can always find a sequence of $x_n's$ that satisfy the condition.

Part b.) It suffices to simply consider the $cE$ case, where $c > 1$. Since: $$\frac{m(cE \cap [-r,r])}{m([-r,r])} = \frac{m(E \cap [-\frac{r}{c}, \frac{r}{c}])}{m([-\frac{r}{c},\frac{r}{c}])}$$ We have: $$m(cE \cap [-r,r]) = \frac{m([-r,r])}{m([-\frac{r}{c},\frac{r}{c}])}m\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) = \ldots$$ $$\ldots = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big)$$ Thus, like in part a, we have that since $\forall \epsilon > 0$ there exists an $r_0 > 0$ such that $\forall r \in (0,r_0)$: $$m(E \cap [-r,r]) > (1 - \epsilon)2r$$ Therefore, since $\frac{r}{c} < r$, we have: $$m(cE \cap [-r,r]) = cm\Big(E \cap \Big[-\frac{r}{c},\frac{r}{c}\Big]\Big) > c(1 - \epsilon)2\frac{r}{c} = \ldots$$ $$\ldots = (1 - \epsilon)2r > 0$$ ...when $\epsilon < \frac{1}{2}$. Thus, under these settings, the estimate: $$m(cE \cap E \cap [-r,r]) > (1 - 2\epsilon)2r$$ ...holds, $\implies$ we can choose a sequence of $x_n's$ satisfying the desired condition.

To see that this condition holds for $0 < c < 1$, observe that if we simply define $cE = F$, that: $\frac{1}{c}F = E$, and the statement holds per the previous argument.

Now, since we know it works for $c > 0$, and $c = -1$ (from Part a.), simply combine the arguments to see that this clearly works for $c \in \mathbb{R} \backslash \lbrace 0 \rbrace$.