If $B\subset\mathbb{R}^n$ is measurable and $(x_l)_{l\in\mathbb{N}}$ is a bounded family, so $(B + x_l)$ is pairwise disjoint, then $\mu(B)=0$. Assignment:

Show that: If $B\subset\mathbb{R}^n$ is Lebesgue-measurable and if there is a bounded family $(x_l)_{l\in\mathbb{N}} \subset \mathbb{R}^n$ so that the family $(B + x_l)_{l\in\mathbb{N}}$ is pairwise disjoint, it follows that $B$ is a null set. ($\mu(B)=0$)

What I've got is:
$$\mu(B) ≤ \mu(B+x_l) ≤ \mu\left(\bigcup_{l\in\mathbb{N}}B+x_l\right)\leq\sum_{l\in\mathbb{N}}\mu(B+x_l)$$
(I think the last inequality might be a true equality, since the $B+x_l$ are pairwise disjoint.) I thought it would make sense to use that every countable set is a null set, but I'm not sure if that's it. Especially because I did not really use the pairwise disjoint - property at all, which seems essential in the assignment.
I'd appreciate any help.
 A: 
Hint: Assume that $B$ is bounded. Why does the general case follow?

Firstly, the first inequality is (of course) an equality, since $B$ is already measurable and the lebesgue-measure is variant on arbitrary translation. As your hint suggests, we can assume without loss of generality that $B$ is bounded. This is because, if $B$ was not bounded we can consider the cover $\bigcup_{m\in\mathbb{N}}B\cap [-m,m]^n$ of B. Since $[-m,m]^n$ is bounded, we have that for every $m\in\mathbb{N}$ the set $B\cap[-m,m]^n$ is bounded. The cover is countable. As a result we can use the theorem (that is still to be proved for bounded sets), after using basic inequalities, on each of the (null) sets.
Now, if $B$ is bounded, there exist a $a,b\in\mathbb{R}^n$ such that $B\subset[a,b]^n$ and since $(x_l)_{l\in\mathbb{N}}$ is bounded as well, there are also $c,d$ with $x_l \in [c,d]^n$ for all $l\in\mathbb{N}$. It follows for arbitrary $l$ that $(B+x_l) \subset [a+c,b+d]^n$. Using that the family is pairwise disjoint, yields:
$$0≤\mu(B)≤\sum_{l\in\mathbb{N}}\mu(B) =\sum_{l\in\mathbb{N}}\mu(B+x_l)≤\mu([a+c,b+d]^n) < \infty$$
We observe that the inequality is only true iff $\mu(B)=0$.
