Indefinite Riemann integral of $f\in C(\Bbb R^n)\cap L^1(\Bbb R^n)$ using bounded lattices as test sets

Question

Let $f\in C(\mathbb{R}^n)\cap L^1(\mathbb{R}^n)$, and for $k,l\in\mathbb{N}$, define the bounded lattice $$\Lambda_{k,l}=\{x\in\frac{1}{k}\mathbb{Z}^n\ :\ \lvert x\rvert_\infty\le l\}$$ and let $\Lambda_k=\Lambda_{k,k}$, so it is not so difficult to see that $$\int_{[-l,l]^n}f(x)\,\mathrm{d}x=\lim_{k\to\infty}\frac{1}{k^n}\sum_{x\in\Lambda_{k,l}}f(x)$$ and I wish to show that $$\int_{\mathbb{R}^n}f(x)\,\mathrm{d}x=\lim_{k\to\infty}\frac{1}{k^n}\sum_{x\in\Lambda_k}f(x)$$ but I'm struggling to get any uniform convergence of $$\lim_{k\to\infty}\frac{1}{k^n}\sum_{x\in\Lambda_{k,l}}f(x)$$ with respect to $l$. Is this statement true? If so, is there some easy way to prove it, and if not, is there some simple counterexamples (e.g. some function whose value is concentrated on $\mathbb{Q}^n$)?

Answer 1

Let's look at the simple case of $\mathbb R.$ Think of $m=\pm 1,\pm 2,\dots$ and $f$ as a sequence of triangles of height $m^2$ centered over $\pm m$ with bases equal to $1/m^4.$ Then $\int_{-\infty}^\infty f $ converges to $1/1^2 + 1/2^2 + \cdots .$ But

$$\frac{1}{k}\sum_{x\in \Lambda_k}f(x) \ge \frac{1}{k}f(k) = \frac{1}{k}k^2 = k\to \infty.$$

If you assume $f$ is nonnegative and decreasing on $[0,\infty),$ nonnegative and increasing on $(-\infty,0],$ then your desired result holds.