In this paper http://www.dtc.umn.edu/~odlyzko/doc/arch/high.dim.spheres.pdf paper, the following section(Page 2, eq 1.1, eq 1.2 and eq 1.3) is given as an obvious fact. It is not so obvious to me, I donnot understand that how the average value of lattice points is the same as volume of the sphere. can anyone explain?
For $x = (x_1, ..., x_n) ∈ R^n$ , and $α > 0$, let $S(x, n, α)$ denote the n-dimensional sphere of radius $(αn)^{1/2}$ centered at $x$ $$S(x, n, α) = \left\{ z ∈ R_n: \sum_{i=1}^{n}(z_i − x_i)^2 ≤ αn \right\}$$ Let $N(x, n, α)$ denote the number of lattice points in $S(x, n, α)$, $$N(x, n, α) = \left\{ z ∈ R_n: \sum_{i=1}^{n}(z_i − x_i)^2 ≤ αn \right\}$$ We will study the dependence of $N(x, n, α)$ on $x$ for $α$ fixed and $n → ∞$. The average value of $N(x, n, α)$ as $x$ runs over the unit cube $0 ≤ x_1, ..., x_n ≤ 1$ is just the volume of the sphere $S(x, n, α)$, which equals
$$\frac{\pi^{n/2}(\alpha n)^{n/2}}{\Gamma(n/2 + 1)} $$ as $n \rightarrow \infty$
Imagine that instead of the hypersphere increasing in size as the radius goes up, the size of the hypersphere is fixed and instead the unit length goes down, kind of like you are zooming out. You can see how the density of the lattice points will increase until it "fills" the hypersphere. Since each lattice point corresponds with a unit hypercube, the lattice point count will approach the volume of the hypersphere as the precision of the points increases.