Equivalence of different criteria for Lebesgue measurability Let $E \subset \mathbb{R}^d$ be a set. Two different criteria for Lebesgue measurability of $E$ is given by:
$(i)$ (Outer approximation by open) For every $\epsilon>0$, one can contain $E$ in an open set $U$ with $m^*(U \setminus E) \leq \epsilon$.
$(ii)$ (Almost open) For every $\epsilon>0$, one can find an open set $U$ such that $m^*(U \Delta E) \leq \epsilon$.
The problem is to show the equivalence of these two statements. Now, $(i) \Rightarrow (ii)$ is of course trivial. But I'm having some difficulty in showing $(ii) \Rightarrow (i)$.
Given that, for every $\epsilon>0$, one can find an open set $U$ such that $m^*(U \Delta E) \leq \epsilon$, how can I find an open set $V$, containing $E$ with $m^*(V \setminus E) \leq \epsilon ~??$
Any help would be much appreciated!
 A: First take an open set $U$ such that $m^*(U\Delta E)\leq \epsilon/4$.
This tells you that $m^*(U-E)< \epsilon/4$ and $m^*(E-U)<\epsilon/4$.
Now, since $m^*(E-U)<\epsilon/4$ we can find a countable union of intervals $B$ such that $m^*(B)<2\epsilon/4$ and $(E-U)\subseteq B$. We can change each interval into an open interval by enlarging slightly (use the geometric series) to get an open set $D$ such that $m^*(D)<3\epsilon/4$ and $(E\setminus U)\subseteq D$.
We propose the set $U\cup D$ as the required open set. Clearly it contains $E$, as $D$ contains $E\setminus U$.
Now, what about $\mu^*( (U\cup D) \setminus E)$?
We have $\mu^*( (U\cup D) \setminus E)\leq \mu^*(U-E)+ \mu^*(D-E)\leq \epsilon$
A: To show (ii)$\implies$ (i), assume $\neg$ (i) and prove $\neg$ (ii).
Let $r>0$ such  that $m^*(U$ \ $E)>r$ whenever $U$ is open and $U\supset E.$  Suppose by contradiction  there is open $V$ with $m^*(V \Delta E  )<r/4.$
Let $W$ be open with $W\supset E$ \ $V$ and $m^*(W)<m^*(E$ \ $V)+ r/4.$
Then $W\cup V \supset E$ so $$r<m^*((W\cup V) \backslash E)=$$    $$=m^*((W \backslash E)\cup (V \backslash E))\leq$$    $$\leq  m^*(W \backslash E)+m^*(V \backslash E)\leq$$    $$\leq  m^*(W )+m^*(V \Delta E)<$$    $$<  [m^*(E \backslash V)+r/4] +[r/4]\leq$$    $$\leq m^*( E \Delta V)+r/2 <$$ $$<r/4+r/2<r $$ implying $r<r.$
