# Uniform bound on the measure of $\Omega_\delta = \Omega \cap \delta\mathbb Z^d$ if $\Omega$ is an open bounded set with Lipschitz boundary

Let $$\Omega \subset \mathbb R^d$$ be an open bounded set with Lipschitz boundary. Let us consider $$\Omega_\delta = \Omega \cap \delta\mathbb Z^d$$ for $$\delta >0$$. I want to say that the measure of $$\Omega_\delta$$ is bounded by a constant which is independent of $$\delta$$ and somehow depending only on the original $$\Omega$$.

How can this be proved? I guess it is a classical result in numerical analysis, but I can't find a reference.

My motivation is this: suppose that we prove a discrete Poincaré inequality in $$\Omega_\delta$$. Then I want to get out of it the "continuous" Poincaré inequality without the measure of the domain blowing up in the limit. Does it make sense?

• The set $\Omega$ is open where? And what measure do you mean? And you probably mean something different than “$\Omega\cap \delta\mathbb{Z}$”, since $\delta\mathbb{Z}$ is a countable subset of $\mathbb R$. Commented Oct 16, 2022 at 10:50
• @PietroMajer Yes, sorry. I meant $\Omega \subset \mathbb R^d$ and the typo is that the post should read $\mathbb Z^d$
– Hiro
Commented Oct 16, 2022 at 10:53
• @PietroMajer And I suppose that the appropriate mesure would be the one that counts the points in $\Omega_\delta$
– Hiro
Commented Oct 16, 2022 at 10:54
• It seems a bit odd to expect a bound independent of $\delta$, no? I would think that $\lvert \Omega \cap \delta \mathbf{Z}^d \rvert$ would grow like $\delta^{-d}$. Commented Oct 16, 2022 at 11:15
• @LeoMoos My motivation is this: suppose that we prove a discrete Poincaré inequality in $\Omega_\delta$. Then I want to get out of it the "continuous" Poincaré inequality without the measure of the domain blowing up in the limit. Does it make sense?
– Hiro
Commented Oct 16, 2022 at 11:34