Minkowski's theorem generalization 
Let  $\Omega$ be a compact subset of $\mathbb R^n$ with $\mu(\Omega) > n$. Then there exists $p \in \mathbb R^n$ such that $|(\Omega + p) \cap \mathbb Z^n| \geq n+1$. In other words, some translate of $\Omega$ contains at least $n+1$ points with only integer coordinates.

The way to do this would be by contradiction, but I have a doubt I'd like to clear anyway.
Let's define $x\sim y$ in $\mathbb R^n$ if $x - y \in \mathbb Z^n$. This is an equivalence relation. Now, note that if there exists $y$ such that $[y] \cap \Omega$ has more than $n+1$ points, then we are done. 
Note that for all $y$ , there is a unique $x \in B := [0,1) \times ...\times [0,1)$ such that $x \sim y$.
Let's assume the contrary of the conclusion. This means, that for all $x \in B$, we have $|[x] \cap \Omega| \leq n$. 
This allows us to define an surjective map from $S := \cup_{i=0}^{n-1} (B + (i,0,...,0)) \to \Omega$ (the left hand set is just $[0,n) \times [0,1) \times ... \times [0,1)$) as follows: for each $x \in \Omega$ we have $[x] \cap \Omega = \{x = x_0,...,x_{k-1}\}$ , where $k \leq n$, now just map $[y] + (i,...,0) \to x_i$ where $y \in B$ is the unique $[y] = [x]$. 
Basically, I've sort of indexed the translates of every point in $\Omega$ using the set $S$.
If we can show that this map is measurable, than $\Omega$ has volume smaller than $S$, which means that its volume is less  than or equal to $n$, contradiction. 

But how do I show that the above map is measurable? Also , this proof does not use the fact that $\Omega$ is compact, so is there something that I am missing?

Note :As a generalization of Minkowski's theorem, I think the tag "algebraic number theory" suits this question also.
 A: Here's a proof of your statement.
I'll change $\mathbb{R}^n$ to $\mathbb{R}^k$ (I believe that you didn't intend the dimension to be equal to $n$ in question).
Denote $m = \sup_{p\in\mathbb R^k} |(\Omega + p)\cap \mathbb{Z}^k|$ and $d = \lceil\mathrm{diam}(\Omega)\rceil$. Let $N\in \mathbb{N}$ and consider $A=[0,N)^k-\Omega = \{x-y\mid x\in [0,N)^k,y\in\Omega\}$. Clearly, $\mu(A)\le (N+d)^k$. Then, denoting $T =[0,N)\cap \mathbb{Z}^k$,
$$
\int_{A}\sum_{x\in T}\mathbf{1}_{\Omega + p}(x) dp \le \int_{A}\big|(\Omega + p)\cap \mathbb{Z}^k\big| dp \le m \mu(A)\le m(N+d)^k.
$$
However, using that $-\Omega+x \subset A$ for any $x\in T$, we have
$$
\int_{A}\sum_{x\in T}\mathbf{1}_{\Omega + p}(x) dp = \sum_{x\in T}\int_{A}\mathbf{1}_{-\Omega + x}(p) dp = \sum_{x\in T} \mu\big((-\Omega+x)\cap A\big)\\
= \sum_{x\in T} \mu(-\Omega+x) = \mu(\Omega) N^k. 
$$
Therefore, 
$$
m \ge \frac{\mu(\Omega)N^k}{(N+d)^k}.
$$
Letting $N\to\infty$, we obtain $m\ge \mu(\Omega)$, whence $m\ge n+1$.

Concerning the measurability of the maps in question, it depends on how you enumerate $\Omega\cap[x]$: for some enumerations, the maps might be measurable, but they are not in general. It might be possible to apply measurable selection argument, but I should think.
