Measure of the set of all $x$ such that $A-x$ contains infinitely many integers Let $A$ have finite Lebesgue measure and let $B$ be the set of all $x$ in $\mathbb R$ such that $A-x$ contains infinitely many integers. Prove that $m(B) = 0$.
As soon as I saw infinitely many, I thought I should write $B$ in a $\bigcap_{n=1}^{\infty}\bigcup_{k\geq n}B_k$ form. Then I hoped to prove that $\sum_iB_i <\infty$ which by using the first Borel-Cantelli lemma would give the result I wanted. So I have: $$x \in B \Leftrightarrow (\forall k\in \mathbb Z)(\exists t\in \mathbb Z): t< k \text{ or } t> k \text{ and } x = a-t  \text{ for } a \in A \Leftrightarrow$$ $$x \in \bigcap_{k \in \mathbb Z} [(\bigcup_{t > k}A-t) \bigcup (\bigcup_{t<k}A-t)]$$ 
But this does not look like what I wanted and moreover I'm not even sure that it is correct. Also, now I think that writing set $B$ in such a form, would give me a set of the form $A-t$ with $t\in Z$ as a $B_i$. But then, that set has measure equal to the measure of $A$, and so the only way for $\sum_i B_i$ to converge, would be if $A$ had measure $0$, which is not an assumption (although it could be true, I don't know). I would appreciate any help.
 A: It's enough to show that $\lambda(B\cap [0,1])=0$ because $B+j=B$ for each integer $j$. A Borel-Cantelli like argument does the job.  
A: I think I have solved it with the help of the advice of Davide Giraudo and Niels Diepeveen. I hope I am correct!
First of all I will prove that $\mu_L(B\cap [0,1]) = 0$. To do this, I want to write $B$ in another way: 
$$ x \in B\cap [0,1] \Leftrightarrow $$
$$(\forall n \in \mathbb{N})(\exists k \in \mathbb{N}) \text{ such that }  k > n \text{ and } x \in [0,1] \text{ and } (x\in A - k \text{ or } x \in A+k)$$
$$ \Leftrightarrow  x \in \bigcap_{n \geq 1} \bigcup_{k > n} (((A-k)\cup (A+k)) \cap [0,1] ) $$
$$\Leftrightarrow x \in \{((A-i)\cup (A+i)) \cap [0,1]  \text{; i.o.}\} $$
Now I need to prove that $$\sum_{i=1}^{\infty} \mu_L(((A-i)\cup (A+i)) \cap [0,1]) < + \infty $$ But this is true because
\begin{align*}
\sum_{i=1}^{\infty} \mu_L(((A-i)\cup (A+i)) \cap [0,1]) &\leq \sum_{i=1}^{\infty}[ \mu_L((A-i)\cap[0,1]) +\mu_L((A+i) \cap [0,1])] \\
&=\sum_{i=1}^{\infty}[ \mu_L(A\cap[i,i+1]) + \mu_L(A\cap[-i,-i+1])] \\
&=\mu_L(A\cap[1,+\infty) + \mu_L(A\cap(-\infty , 0] \leq 2\mu_L(A)
\end{align*}
Where I use that $\mu_L((A-i)\cap[0,1]) = \mu_L((A\cap[i,i+1))$.
This means that the sum converges and by the First Borel-Cantelli lemma $\mu_L(B\cap [0,1]) = 0$. But then 
\begin{align*}
\mu_L(B) &= \mu_L(\bigcup_{k \in \mathbb{Z} } ( B \cap (k, k+1))) \\
&= \sum_{k \in \mathbb{Z} } [B\cap (k, k+1) ]\\
&= \sum_{k \in \mathbb{Z} } [(B+k) \cap (0,1) \\
&= \sum_{k \in \mathbb{Z} } [B \cap [0,1]] \\
&= \sum_{k \in \mathbb{Z} } 0 = 0
\end{align*}
and therefore $\mu_L(B) = 0$.
