This is a question from Hoffstein cryptography book.
I'm trying to show that
$\lim_{R \to \infty}$$\frac{{\#(\mathbb{B}_R(\mathbf{0})}\cap L)}{{Vol(\mathbb{B}_R(\mathbf{0}))}}$$=\frac{1}{Vol(\mathcal{F})}$ for a lattice L and its fundamental domain F
Following the questions, I've already proved that
$\bigcup_{\mathbf{v}\in L ,{\mathcal{F}+\mathbf{v}}\subset{\mathbb{B}_R(\mathbf{0})}}$$(\mathcal{F}+\mathbf{v})\subset{\mathbb{B}_R(\mathbf{0})}\subset$ $\bigcup_{\mathbf{v}\in L ,{\mathcal{F}+\mathbf{v}}\cap{\mathbb{B}_R(\mathbf{0})}\neq\emptyset}{({\mathcal{F}+\mathbf{v}})}$
Taking volumes, I've also proved
$\#\{\mathbf{v}\in L:\mathcal{F}+\mathbf{v}\subset{\mathbb{B}_R(\mathbf{0})}\}\cdot Vol(\mathcal{F}) \leq Vol ({\mathbb{B}_R(\mathbf{0})})\leq \#\{\mathbf{v}\in L:(\mathcal{F}+\mathbf{v})\cap{\mathbb{B}_R(\mathbf{0})}\neq\emptyset\}\cdot Vol(\mathcal{F}).$
Now I'm struggling with this part: Prove that the number of translates $\mathcal{F}+\mathbf{v}$ that intersect ${\mathbb{B}_R(\mathbf{0})}$ without being entirely contained within ${\mathbb{B}_R(\mathbf{0})}$ is comparatively small compared to the number of translates that are entirely contained within ${\mathbb{B}_R(\mathbf{0})}$.
I understand this is true by instinct in geometric way, but I don't know how to prove this mathematically. Can someone help me with this?
Thank you a lot in advance.
Let $$L_R = \bigcup_{v \in B_R(0) \cap L} (v+\mathcal{F})$$ then $$B_{R-c}(0) \subset L_R \subset B_{R+c}(0)$$ where $c$ is large enough such that $\mathcal{F} \subset B_c(0)$.
Then $Vol(L_R) = Vol(\mathcal{F})|B_R(0) \cap L|$ and $\lim_{R \to \infty} \frac{Vol(B_{R-c}(0))}{Vol(B_{R}(0))} = 1=\lim_{R \to \infty} \frac{Vol(B_{R+c}(0))}{Vol(B_{R}(0))}$ so $$1=\lim_{R \to \infty} \frac{Vol(L_R)}{Vol(B_{R+c}(0))}=\lim_{R \to \infty} \frac{Vol(\mathcal{F})|B_R(0) \cap L|}{Vol(B_{R}(0))}$$