Alternative proof : If E and F have positive measures then E+F contains an interval The statement is following : 


  $E, F$ are measurable subsets in $\mathbb{R}$ which have positive
  measure. Then prove that $E+F=\left\{ x+y\ :\ x\in E,\ y\in F\right\}$
   contains an interval.


This is a problem 30 in Stein's Real analysis. I already read the proof of this by using convolusion which is here.(This problem is so-called Steinhaus theorem) Moreover, I proved that the previous problems number 28 and 29 in the book. 
My goal is applying the same technique to above one, using the technique in problem number 29 : 

First, by problem number 28, $ \exists I^F$ and $I^E$ such that $$m(E\cap I^{E})\geq{\displaystyle \frac{9}{10}m(I^{E})} \text{ and }  m(F\cap I^{F})\ge{\displaystyle \frac{9}{10}m(I^{F})}.$$ Let $E_{0}= E\cap I^{E}$ and $F_{0}= F\cap I^{F}$ and let $e_{0}, f_{0}$ : center of $I^{E}, I^{F}$, respectively.

Suppose $E_{0}+F_{0}$ does not contain an interval centered at $e_{0}+f_{0}$. Then $\exists a$ : small such that $E_{0}\cap(e_{0}+f_{0}+a-F_{0})=\phi$. But, $E_{0}\cup(e_{0}+f_{0}+a-F_{0})\subset I^{E}\cup(e_{0}+f_{0}+a-I^{F})$. Measure of left hand side is more than $\frac{9}{10}(m(I^{E})+m(I^{F}))$ but the measure of Right had side is slightly less than $\max(m(I^{E}),m(I^{F}))$.

And that's it. I stopped here. I can't make any contradiction.
I thought that if I can make measure of $I^E and I^F$ are equal, then I prove the problem but this one is also a big problem.

Anyone can comment on this?
This proof uses the  Lebesgue density theorem.  If $e$ and $f$ are points of density of $E$ and $F$, and $\epsilon &gt; 0$, there is $\delta &gt; 0$ such that $m([e-\delta,e+\delta] \backslash E)/(2\delta) &lt; \epsilon $ and
$m([f-\delta,f+\delta] \backslash F)/(2 \delta) &lt; \epsilon$.
I claim that $e + f + [-\delta/2,\delta/2] \subseteq E + F$.

Consider any $z \in e + f + [-\delta/2, \delta/2]$.  Let $s$ be chosen from the interval $e + [-\delta/2, \delta/2]$.  Thus $z-s \in f + [-\delta, \delta]$.  The set of $s \in e + [-\delta/2,\delta/2]$ for which $s \notin E$ has measure $&lt; 2 \delta \epsilon$.  The set of $s \in e + [-\delta/2, \delta/2]$ for which $z-s \notin F$ has measure $&lt; 2 \delta \epsilon$.  If $4 \delta \epsilon &lt; \delta$, removing these leaves a set of positive measure (and in particular at least one) in which $s \in E$ and $z-s \in F$, so that $z \in E + F$.
actually I think your idea could be extended by substituting $e+f$ to any real number, then you could move Ie until Ie is contained in another