Lebesgue Measure in ${R}^m$ I am trying to solve the following problem.
Problem Statement
If $E \subset \mathrm{R}^m$ and $\lambda_m\left(E\right) >1 $, where $\lambda_m$ is the Lebesgue measure on ${R}^m$, then there are two distinct points $x, x^1 \in E$ such that the differences between the corresponding coordinates are integers.
Some Thoughts towards proof
Since Lebesgue measure is translation invariant, we can relocate the set $E$ while preserving it's measure, such that one of it's vertex relocates to the origin/zero vector in the $Z^m$ point lattice/Grid. The relocation causes an equal shift for all points in $E$ where the shift is a vector $s \in [0, 1)^m$. 
I suppose now, if I can show that there is a point in this relocated set with the Euclidean norm as an integer, I would be done, but I am not quite sure how to proceed further. 
Any help would be much appreciated.

[A rewording without talking about quotients]
If $x\in\mathbb{R}$ then $x-\lfloor x\rfloor\in[0,1)$ is an integer translation.
Apply this translation to all coordinates of all points of $E$. The transformation consists in translating each size-one cube $[a_1,a_1+1)\times...\times[a_n,a_n+1)$ with $a_1,...,a_n\in\mathbb{Z}$ to the cube $[0,1)^n$. If $E$ were mapped injectively (no part of $E$ falls on top of another), then the image of $E$ would have the same volume as $E$. But $E$ has volume $>1$, while $[0,1)^n$ has volume $1$. It doesn't fit. 
Therefore $E$ is not mapped injectively by this transformation. Therefore at least two of its points differ by an integer translation.
Actually this shows that there must be set of this type of points of positive measure. So, there are a lot of them.
