$Proposition:$ If $A,B \subseteq \mathbb{R}$ lebesgue measurable sets such that $0<μ(A),μ(B)< \infty$,then $A+B$ contains an interval.
I found a proof of this proposition using Steinhauss Theorem and i would like if someone can explain me some steps in this proof that i don't understand.
$PROOF:$
Let $G\supset A$ ,$G= \bigcup_{n=1}^{\infty}I_n$ a union of intervals with rational endpoints such that $μ(G) \leqslant \frac{4}{3}μ(Α)$.
We see that $A=\bigcup_{n=1}^{\infty}(A \cap I_n)$ and we can easily deduce that
$$\sum_{n=1}^{\infty}μ(I_n) \leqslant \frac{4}{3}μ(Α) \leqslant \frac{4}{3}μ\sum_{n=1}^{\infty}μ(A \cap I_n)$$.
Thus exists $n_1 \in \mathbb{N}$ such that $μ(A \cap I_{n_1}) \geqslant \frac{3}{4} μ(I_{n_1})$.$(1)$
Applying the same argument for the set $B$ we can find $n_2 \in \mathbb{N}$ such that $μ(B \cap I_{n_2}) \geqslant \frac{3}{4} μ(I_{n_2})$.$(2)$
Because of the fact that $μ(I_{n_1}),μ(I_{n_2}) \in \mathbb{Q}$, we can partition $I_{n_1},I_{n_2}$ into $m$ and $n$ respectively consecutive subintervals
$Step 1:$ From $(1),(2)$ we can find intervals $I_0,J_0$ such that $μ(Ι_0)=μ(J_0)$ such that $$μ(B \cap J_0) \geqslant \frac{3}{4} μ(J_0)$$
and also $$μ(A \cap I_0) \geqslant \frac{3}{4} μ(I_0)$$
$Step2:$ In other words there exists an interval $I$ centered at $0$ and $x,y \in \mathbb{R}$ such that $$μ(B-y \cap I) \geqslant \frac{3}{4} μ(I)$$
$$ μ(A-x \cap I) \geqslant \frac{3}{4} μ(I)$$
It follows that $μ(B-y \cap A-x \cap I) \geqslant \frac{1}{2} μ(I)>0$
$Step3:$ Let $C=(A-x) \cap (B-y)$ If we apply Steinhauss theorem to $C-C$ we deduce that $A-B$ contains an interval.
Finally we deduce that $A-(-B)=A+B$ contains an interval.
Can someone explain me step1 and step2??
Thank you in advance.