Lemma 2.14 (Ruzsa's covering lemma) For any additive sets $A,B$ with common ambient group $Z$, there exists an addtitive set $X_+\subseteq B$ with $$B\subseteq A-A+X_+;\quad |X_+|\leq \dfrac{|A+B|}{|A|}; \quad |A+X_+|=|A||X_+|$$ and similiarly there exists an additive set $X_-\subseteq B$ with $$B\subseteq A-A+X_-;\quad |X_-|\leq \dfrac{|A-B|}{|A|}; \quad |A-X_-|=|A||X_-|$$ Covering lemmas such as the one above are convenient for a number of reasons. Firstly, they allow for easy computation of iterated sum sets. For instance, if one knows that $$A+B\subseteq A+X$$ the one can immediately deduce that $$A+nB\subseteq A+nX \ \ \text{for} \ \ n\geq 0.$$
This is an excerpt from Tao-Vu book and I was trying to apply Ruzsa's covering lemma in order to prove that if $A+B\subseteq A+X$, then $A+nB\subseteq A+nX$. However, I failed to do that.
Can anyone show the proof please? Thank you so much!
You don't need Ruzsa's covering lemma for this, just repeatedly apply the hypothesis.
More explicitly, by induction on $n$ - $A+(n+1)B = (A+nB)+B\subseteq A+nX+B=A+B+nX\subseteq A+X+nX=A+(n+1)X$.