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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.

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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 $n \ge 2$.

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  • $\begingroup$ I am also interested in this calculation and I almost got it with your hints. An embarrassing question: Why is $(1-(\frac{1}{2})^n)\alpha_n=\frac{3}{4}\lambda(B_1(0))$? $\endgroup$ – Sabrina G. Oct 5 '18 at 7:24
  • $\begingroup$ $(1 - (1/2)^n) \ge 3/4$ if $n \ge 2$. $\endgroup$ – Umberto P. Oct 5 '18 at 11:00
  • $\begingroup$ yes, this is clear. $\endgroup$ – Sabrina G. Oct 5 '18 at 12:21

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