$A\cap(A+x) \neq \emptyset$ for a set $A$ with positive Lebesgue measure and $0<|x| < \delta$ can someone help me show that if $A$ is a measurable set with positive Lebesgue measure then there exists some $\delta >0$ such that $A\cap (A+x) \neq \emptyset$ whenever $|x|< \delta$?

I know that since the Lebesgue measure is invariant under translation $\lambda(A) = \lambda(A+x) \geq 0$. Also for a cover of open intervals of $A$ I think if $A \subset \bigcup_{i\in \mathbb{N}} (a_i, b_i) \rightarrow (A+x) \subset \bigcup_{i\in \mathbb{N}} (a_i+x, b_i+x)$
Then if I could show that $\bigcup_{i\in \mathbb{N}} (a_i, b_i) \cap \bigcup_{i\in \mathbb{N}} (a_i+x, b_i+x) \neq \emptyset$ everything would be ok, but I don't see how this intersection would look like.
Thanks for your help!
 A: I'm not shure whether  this proof is true, however I'm almost convinced. 
First of all you know that $\lambda(A)\gt0$. This implies that the measure of $A$ can be either finite or infinite.
We first assume that she is finite, because it is the complicated case.
The definition of Lebesgue measurable says that
$$\lambda(A)= inf \space(\lambda(U), A\subset U \space open)$$
This tells you that the set can be covered by the union of opens sets (the smallest you can find), i.e.
$$A= \bigcup_{n=1}^\infty{U_n}\space \space s.t \space \space A \subset U$$
Now if you take an $\alpha$ $\in$ $(0,1)$ you have that
$$\alpha \sum_{n=1}^\infty{|U_n|} \le\lambda(A)\le\lambda(A \cap \bigcup_{n=1}^\infty{U_n})\le \sum_{n=1}^\infty {\lambda(A\cap U_n)}$$
Now, this allows you to see that there is a $n_0$ such that $\alpha|U_{n_0}|\le\lambda(A\cap U_{n_0})$.
We will use this fact to demonstrate that $A \cap(A+x)\ne \emptyset$. In fact we can now take (for example) $\alpha= \frac {3}{4} $ and find an Interval $U$ such that
$$\frac{3}{4}|U|\le\lambda(A\cap U)$$
We now put a $\delta = \frac{1}{2}|U|$ (which, if you draw it on the real line, would be the center of the interval) and note that the set:
$$(A\cap U)\cup ((A\cap U) +x)\subset U\cup (U+x)$$ (remember that this is true because U is bigger than A by definition). Now we note also that since |x|$\lt \delta$ the set $U\cup (U+x)$ has measure $\lt \frac{3}{2}|U|$. We would like to use this fact to show that the set $(A\cap U)\cap ((A\cap U) +x) \neq \emptyset$. We will prove this by contradiction. In fact if we assume that $(A\cap U)\cap ((A\cap U) +x) = \emptyset$ we note that the two set are disjoint and therefore
$$\frac{3}{2}|U|>\lambda (U\cup (U+x))\ge \lambda ((A\cap U)\cap ((A\cap U) +x))=^{(because\space disjoint)} \lambda (A \cap U) + \lambda ((A \cap U) + x)= 2\lambda (A \cap U)= 2\times \frac{3}{4}|U|=\frac {3}{2}|U|  $$ Which is a contraddiction, hence the set can't be disjoint.
If $\lambda (A)$ is infinite than it's easier because you note that the assertion is true for every cover of open intervals $]-n,n[$. 
