Lebesgue measure on the Real numbers, set of positive Lebesgue measure Question: Prove that if $E$ is a set of real numbers with Lebesgue measure $m(E)>1$, then there are two distinct points $\alpha$ and $\beta$ in $E$ such that $\alpha - \beta$ is an integer. Hint: Let $E_{n}=(E-n) \cap [0,1)$.
What I've done: for each $n\in \mathbb{Z}$, put $E_{n}=(E-n) \cap [0,1)$.
Then by translation,  $E_{n}=(E\cap [n,n+1))-1$ and $E=\bigcup_{n=-\infty}^{\infty} E\cap (n,n+1)=\bigcup_{n=-\infty}^{\infty} E_{n}+n$. Each $n$ is measurable and whenever $n\neq m$, $E_{n}\cap E_{m}=\emptyset$. Also, we have $E_{n}\subset [0,1)$ for each n and so $\bigcup_{n=-\infty}^{\infty}E_{n}\subset[0,1)$. By the translation invariance of Lebesgue measure, $m(E_{n}+n)=m(E_{n})$ for each n. Now, suppose that $E_{n}+n \cap E_{m}+m =\emptyset$ if $n\neq m$. By the sigma-additivity of measure, $m(E)=m(\bigcup_{n=-\infty}^{\infty}E_{n}+n)=\sum_{n=-\infty}^{\infty}m(E_{n}+n)=\sum_{n=-\infty}^{\infty}m(E_{n})=m(\bigcup_{n=-\infty}^{\infty}E_{n})\leq m([0,1))=1$ (*)
which contradicts $m(E)>1.$
Therefore, there must be some m,n in $\mathbb{Z}$ such that $E_{n}+n \cap E_{m}+m \neq\emptyset$. 
Take z in $E_{n}+n \cap E_{m}+m $. We can find two points $\alpha\in E_{n}\subset E$ and $\beta\in E_{m}\subset E$ with $\alpha\neq \beta$ such that $z=\alpha+n=\beta+m$, i.e., $\alpha-\beta=m-n$, which is an integer.
My question is the following: Is it justifiable to use sigma-additivity of the measure in equality (*) when n ranges from $-\infty$ to $\infty$? If not, Is there any other way around to avoid this step, but reach the same conclusion as above? Any suggestion or correction is very much appreciated. Thank you! 
 A: First , I will modify your proof slightly.
Let $E_{n}=(E-n) \cap [0,1)$ for each $n\in \mathbb{Z}$. Then by translation,  $E_{n}=(E\cap [n,n+1))-n$. Then $E=\bigcup_{n=-\infty}^{\infty} (E\cap [n,n+1))=\bigcup_{n=-\infty}^{\infty} (E_{n}+n)$. If $E_{n} \cap E_{m} \neq \emptyset$ for some $n,m$ satisfying $n\neq m$, we can easily see that $ \exists$ two distinct $\alpha,\beta \in E$ such that $\alpha-\beta \in \mathbb{Z}.$ Indeed, $\exists x \in E_n \cap E_m$ for some $n,m$ with $n \neq m,$ then $\alpha=x+n \in E_n +n \subseteq E$ and $\beta=x+m \in E_m +m \subseteq E.$ Thus $\alpha-\beta \in \mathbb{Z}$ and thus the proof is complete. 
Now, we assume, on the contrary, that $E_{n} \cap E_{m} = \emptyset$ for any $n,m$ satisfying $n\neq m$.
For each $n$, $E_n+n$ is measurable. By the subadditivity, the translation invariance and the sigma-additivity, $m(E)=m(\bigcup_{n=-\infty}^{\infty}(E_{n}+n))\le\sum_{n=-\infty}^{\infty}m(E_{n}+n)=\sum_{n=-\infty}^{\infty}m(E_{n})=m(\bigcup_{n=-\infty}^{\infty}E_{n})\leq m([0,1))=1$, which contradicts $m(E)>1.$
For your question, the answer is yes. You should check the definition of doubly infinite series.
Doubly infinite series 
$$
\sum_{k = -\infty}^\infty a_k,
$$
is a pair of two (ordinary infinite) series, 
its negative part 
$\sum_{k = -1}^\infty a_{-k}$
and its positive  part $\sum_{k = 0}^\infty a_k$.
A doubly infinite series converges if each of its constituent parts (positive and negative), converges; the sum of the series is then, by definition, the sum of the limits of the two series. 
In the proof, $\sum_{n=-\infty}^{\infty}m(E_{n})=\sum_{n=-\infty}^{-1}m(E_{n})+\sum_{n=0}^{\infty}m(E_{n})=m(\bigcup_{n=-\infty}^{-1}E_{n})+m(\bigcup_{n=0}^{\infty}E_{n})=m(\bigcup_{n=-\infty}^{\infty}E_{n})$.
