Let $E\subset \mathbb{R}^n$ be a Borel set with $\lambda(E)=\lambda(B_1(0))$, where $\lambda$ denotes the Lebesgue measure and $B_1(0)$ the open unit ball. Fix $x\in\mathbb{R}^n$. Let $F\subset \mathbb{R}^n$ be a Borel set with finite and positive Lebesgue measure such that $E\triangle F\subset\subset B_{\frac{1}{2}}(x)$, i.e. the symmetric difference $E\triangle F$ is compactly contained in $B_{\frac{1}{2}}(x)$.
Why is $\lambda(F)\ge \frac{3}{4}\lambda(B_1(0))$?
I come this far: It is $\lambda(E\triangle F)\le \lambda(B_{\frac{1}{2}}(x))= \lambda(B_{\frac{1}{2}}(0))$ and $\lambda(E\triangle F)=\lambda(E\setminus F)+\lambda (F\setminus E)$, however, I don't know how to proceed.
I appreciate any help.
Best Answer
Let $\alpha_n = \lambda(B_1(0))$.
Since $\lambda(E) = \alpha_n$ you have $\lambda(E \setminus F) + \lambda(E \cap F) = \alpha_n$.
However $\lambda(E \setminus F) \le \lambda(E \triangle F) \le \lambda(B_{\frac 12}(x)) = \left( \frac 12 \right)^n \alpha_n$.
You can arrange these to find $\lambda(E \cap F) \ge \left( 1 - \left( \frac 12 \right)^n \right) \alpha_n$ giving you what you need if $n \ge 2$.