# Sum converging to an integral - Riemann sum?

Let $$\epsilon > 0$$ and $$L >1$$ such that $$\frac{L}{2\epsilon} \in \mathbb{N}$$. Take $$\Lambda_{\epsilon, L} :=\epsilon\mathbb{Z^{d}}/L\mathbb{Z}^{d}$$. Suppose $$f \in C^{1}(\mathbb{R}^{d}/L\mathbb{Z}^{d},\mathbb{C})$$ and define a sum: $$\epsilon^{d}\sum_{x \in \Lambda_{\epsilon, L}}f(x)$$ Is is possible to prove that: $$\epsilon^{d}\sum_{x \in \Lambda_{\epsilon, L}}f(x) \stackrel{\epsilon \to 0}\to \int_{\mathbb{R}^{d}/L\mathbb{Z}^{d}}f(x)dx$$ using, idk, Riemann sums or something like this? I'm really stuck here. Any help would be appreciated.

• Shouldn't that be $\epsilon^d\sum_{x \in \Lambda_{\epsilon, L}}f(x)$? For positive $f$, as $\epsilon \to 0$ both $\frac 1{\epsilon^d}$ and $\sum_{x \in \Lambda_{\epsilon, L}}f(x)$ go to $\infty$. With that correction, the sum is indeed a Riemann sum for the integral. – Paul Sinclair May 24 at 22:19
• @PaulSinclair you are completely right! I will edit it. Thanks! – IamWill May 24 at 23:19

With the correction, it is a Riemann sum. For purposes of integration, we can equate the hyper-torus with $$[0,L)^d$$. Since $$f$$ is continuous and is being integrated on a compact set (integral is the same for $$[0,L]^d$$), the integral converges. As such all Riemann sums must converge to the same limit.
A Reimann sum consists of dividing up the domain in non-overlapping regions summing over all the regions the product of the function at a sample point in the region by the region's area. The lattice points of $$\Lambda_{\epsilon,L}$$ are equally spaced. By the identification with $$[0,L)^d$$, they are the points $$(n_1, n_2, ..., n_d)\epsilon$$ for integers $$0 \le n_i <\frac L\epsilon$$. We can consider that the sample point for a cell $$[n_1\epsilon, (n_1+1)\epsilon]\times \ldots \times[n_d\epsilon,(n_d+1)\epsilon]$$ whose $$d$$-volume is $$\epsilon^d$$. The Riemann sum is therefore $$\sum_{x \in \Lambda} f(x)\epsilon^d$$
Since the norm of this partition $$\sim \epsilon$$, it goes to $$0$$ as $$\epsilon \to 0$$. Therefore the sum must converge to the integral as $$\epsilon \to 0$$ as well.
FYI - I have no idea why the $$2$$ is there in the requirement that $$\frac L{2\epsilon}$$ is integer. What is useful is that $$\frac L\epsilon$$ is integer. But even that could be dropped and the result would still hold. It would just take some additional argument.
• Thank you so much! I think the $2$ is here to demand $\frac{L}{\epsilon}$ to be an even number, so the lattice $\Lambda_{\epsilon, L}$ is symmetric about the origin. – IamWill May 25 at 16:25