Disclaimer: I understand this theorem has been discussed before. I am not looking for a proof however, just a clarification of a mysterious step.
I was reading a proof of the following version of Steinhaus' Theorem, and got stuck on one step:
Theorem (Steinhaus): Let $E \subset \mathbb{R}$ and $m(E)>0$, where $m$ denotes Lebesgue measure on $\mathbb{R}$. Then the set $E-E$ contains an interval around zero.
Proof: By the Lebesgue differentiation theorem, or, equivalently, by the definition of a Lebesgue point, there exists an interval $I$ such that
$$m(I\cap E) \geq (1-\epsilon)\,m(I).$$
If the result were not true, there would exist a sequence $x_n \rightarrow 0$ with $x_n \not \in E-E$. Take $n$ large enough that $|x_n|<\epsilon\, m(I)$. Then $x_n +E \subset \mathbb {R} \setminus E$, but
$$\mathbf{m(I\cap (x_n+E))\geq (1-3\epsilon)\,m(I)}$$
(since Lebesgue measure is invariant under translations.) Since $E$ and $\mathbb R \setminus E$ are disjoint, this is a contradiction.
The part I am having difficulty with is in bold characters. I believe it follows from the following:
\begin{align*} m(I\cap (x_n+E)) & = m(x_n+(I-x_n)\cap E) \\ & = m((I-x_n)\cap E) \\ & \geq m(I) - 2|x_n| - m(I \setminus E) \\ & \geq (1-3\epsilon)\,m(I) \end{align*}
Is this correct? If not, could you point out the correct steps? Thanks!
Note: For the sake of completeness,
$$E-E:=\{x-y:\ x,y\in E \}.$$
HINT:
$$\mu (I \cap (x + E) ) = \mu ( (I- x) \cap E) \ge \mu ( I \cap E) - \mu ( (I \backslash (I - x) ) \cap E)\ge \mu ( I \cap E) - \mu(I \backslash (I - x) ) $$ and $\mu(I \backslash (I - x) )= |x|$ if $I$ fixed and $x$ small enough.
${\bf Added:}$ In fact, this seems to be true: if $\mu(I \cap E) \ge (\frac{1}{2} + \delta) |I| $ then $E - E$ will contain the interval $(-2\delta|I|, 2 \delta|I|)$.