Let $B_1(0)\subset \mathbb{R}^n$ be the open unit ball, let $A\subseteq \mathbb{R}^n$ be a Borel set with positive and finite Borel measure such that $\lambda(A)=\lambda(B_1(0))$, where $\lambda$ denotes the Lebesue measure.
Let $X\triangle Y$ denote the symmetric difference of two sets $X,Y\in\mathbb{R}^n$. Let $$\frac{\lambda(A\triangle B_1(0))}{\lambda(B_1(0))}<\frac{1}{4}.$$
How to conclude that $\lambda(A\cap B_1(0))>\frac{3}{4}\lambda(B_1(0))?\tag1$
I am not sure if we additionally have to assume that $\operatorname{dist}(A,B_1(0))=\inf\{|x-y|:x\in A, y\in B_1(0)\}>0$ to conclude (1); because if $\operatorname{dist}(X,Y)>0$ then $\lambda(X\cup Y)=\lambda(X)+\lambda(Y)$.
It is $$\frac{\lambda(A\triangle B_1(0))}{\lambda(B_1(0))}\le\frac{\lambda(A\setminus B_1(0))+\lambda(B_1(0)\setminus A)}{\lambda(B_1(0))}=\frac{\lambda(A)+\lambda(B_1(0))-2\lambda(B_1(0)\cap A)}{\lambda(B_1(0))}$$
$$=\frac{2\lambda(B_1(0))-2\lambda(B_1(0)\cap A)}{\lambda(B_1(0))}=2-\frac{2\lambda(B_1(0)\cap A)}{\lambda(B_1(0))},$$ from here I don't know how to go further.
However, if $\operatorname{dist}(A,B_1(0))>0,$ then $\frac{\lambda(A\triangle B_1(0))}{\lambda(B_1(0))}=\frac{\lambda(A\setminus B_1(0))+\lambda(B_1(0)\setminus A)}{\lambda(B_1(0))}$ and therefore $2-\frac{2\lambda(B_1(0)\cap A)}{\lambda(B_1(0))}<\frac{1}{4}$. From here, I get $\lambda(A\cap B_1(0))>\frac{7}{8}\lambda(B_1(0))$, but not (1)…..
Best Answer
Abbreviate $B_{1}\left(0\right)$ as $B$ and prescribe measure $\mu$ by $D\mapsto\frac{\lambda D}{\lambda B}$ on Borel subsets of $\mathbb{R}^{n}$.
Then $\mu B=1$.
If $\mu\left(A\Delta B\right)<\frac{1}{4}$ and $\mu\left(A\cap B\right)\leq\frac{3}{4}$ then $\mu\left(A\cup B\right)=\mu\left(A\Delta B\right)+\mu\left(A\cap B\right)<1$
But we also have $\mu(A\cup B)\geq\mu B=1$, so a contradiction is found.
So we conclude that: $$\mu\left(A\Delta B\right)<\frac{1}{4}\implies\mu\left(A\cap B\right)>\frac{3}{4}$$