Outer measure proof, assuming measurable set exists Let $A\subset X $ be a null set (so $m^*(A)=0$). Assume that $X:=[a,b]$ is a fixed interval in $\mathbb R$ and let $m^*$ be the outer measure of $X$. Show that $A\subset X$ is a measurable set if and only if for every $\epsilon>0$, there exists a measurable set $E\subset A$ such that $m^*(A\setminus E)\leq\epsilon$.
I'm struggling to prove this problem from either direction. Thanks for help!
 A: I'll assume you don't want the first sentence of your post there...

First a bit of background:
Recall that a set $A$ is measurable if and only if for any $T\subset X$ one has
$$
\mu^*(T)=\mu^*(T\cap A)+\mu^*(T\cap A^c).
$$
But,
from the subadditivity of an outer measure, to show that a set $A$ is measurable, it suffices to show that for all $T\subset X$, one has
$$\tag{1}
\mu^*(T)\ge\mu^*(T\cap A)+\mu^*(T\cap A^c).
$$

Now assume $A$ is such that for each $\epsilon>0$, there is a measurable set $E\subset A$ with $\mu^*(A\setminus E)<\epsilon$.  We will show $A$ satisfies $(1)$ for any $T\subset X$, and is  thus  measurable.
Towards this end, let $\epsilon>0$. Choose $E$ as above and let $T\subset X$.  Then the following properties hold:
$\ \ 1)$ Since $E\subset A$, we have $\mu^*(T\cap A^c)\le \mu^*(T\cap E^c)$.
$\ \ 2)$ Since  $A=E\cup(A\setminus E)$ and $\mu^*(A\setminus E)<\epsilon$, we have $\mu^*(T\cap E) \ge \mu^*(A\cap T)-\epsilon$.  
Now, from  the measurability of $E$, property $1)$, and property $2)$, we have
$$
\mu^*(T)
=\mu^*(T\cap E)+\mu^*(T\cap E^c)
\ge\mu^*(T\cap A)-\epsilon+\mu^*(T\cap A^c).
$$
As $\epsilon$ was arbitrary, it follows that 
$
\mu^*(T)\ge\mu^*(T\cap A)+\mu^*(T\cap A^c)
$, as desired.

The other direction of your proposition is trivial (take $E=A$).
A: A set, $M$, is measurable if for each $\varepsilon > 0$, there exists and open set $\mathcal{O} \supset M$, such that $m^*(\mathcal{O} \setminus M) \leq \varepsilon$.
Also, recall that a set is measurable if and only if its compliment is measurable.
$(\Leftarrow)$
Let $\varepsilon > 0$ be fixed.
Suppose there exists a measurable set $E \subset A$ such that $m^* (A \setminus E) \leq \frac{\varepsilon}{2}$. Then, $A^{\mathcal{c}} \subset E^{\mathcal{c}}$ and $A \setminus E = E^{\mathcal{c}} \setminus A^{\mathcal{c}}$. (This step only works when the set is bounded, this wouldn't make sense if the measure is infinite.)
Since $E^{\mathcal{c}}$ is measurable, there exists an open set $\mathcal{O}\supset E^{\mathcal{c}}$ such that $m^*(\mathcal{O} \setminus E^{\mathcal{c}}) \leq \frac{\varepsilon}{2}$.
By sub-additivity of outer measure, $m^*(\mathcal{O} \setminus A^{\mathcal{c}}) \leq m^*(\mathcal{O} \setminus E^{\mathcal{c}}) + M^*(E^{\mathcal{c}} \setminus A^{\mathcal{c}}) \leq \frac{\varepsilon}{2}+\frac{\varepsilon}{2} = \varepsilon$
Therefore, we have that $A^{\mathcal{c}}$ is measurable and so $A$ is measurable.
