Consider two measurable sets $A_1, A_2 \subset \mathbb{R}^n$ and let $B$ be an open ball in $\mathbb{R}^n$. Assume that $A_1 \subset B$, $ A_2 \subset B$ and that for a certain $0 < \epsilon < 1 $ it holds $$ \frac{ \mathcal{L}^n (A_1) }{\mathcal{L}^n (B)} > 1- \epsilon, \quad \frac{ \mathcal{L}^n (A_2) }{\mathcal{L}^n (B)} > 1- \epsilon.$$ I have to prove that $$ \mathcal{L}^n (A_1 \cap A_2) > (1- 2 \epsilon ) \mathcal{L}^n (B).$$ The problem doesn't seem to be so difficult but I don't know how to proceed!
My idea is that the two estimates on $\mathcal{L}^n (A_1)$ and $\mathcal{L}^n (A_2)$ ensure that the sets $A_1$ and $A_2$ must be "big" subsets of $B$, I mean that the measure of $B \setminus A_1$ and of $B \setminus A_2$ are necessarily "small". Then the intersection $A_1 \cap A_2$ can't be empty because there's no "space enough" inside of $B$ such that the two sets $A_1$ and $A_2$ can be disjoint. On the contrary, the measure of the intersection can't be lower than a certain value (that depends on $\epsilon$).
Then I have also observed that if I show $$ \frac{ \mathcal{L}^n (A_1 \cap A_2) }{\mathcal{L}^n (B)} > \frac{ \mathcal{L}^n (A_1) }{\mathcal{L}^n (B)} \frac{ \mathcal{L}^n (A_2) }{\mathcal{L}^n (B)}, $$ then the proof is accomplished because the term on the right is greated than $ (1- \epsilon ) (1- \epsilon)> 1 - 2 \epsilon.$ Thank you for your help!!
Write $$\mathcal{L}^n(A_1\cup A_2)+\mathcal{L}^n(A_1\cap A_2)=\mathcal{L}^n(A_1)+\mathcal{L}^n(A_2)>(2-2\epsilon)\mathcal{L}^n(B).$$ This implies that $$\mathcal{L}^n(A_1\cap A_2)>(2-2\epsilon)\mathcal{L}^n(B)-\mathcal{L}^n(A_1\cup A_2), $$ and since $\mathcal{L}^n(A_1\cup A_2)\leq\mathcal{L}^n(B)$, this completes the estimate.