# a Lebesgue measure calculation

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.

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