I am trying to prove the equivalence of the following statements:
(a) $\forall\epsilon,$ $\exists$ an open set $U\supset E$ s.t. m$^*$(U\E)$\lt\epsilon$.
(b) $\forall\epsilon,$ $\exists$ an open set $U$ s.t. m*(U$\Delta$E)$\lt$$\epsilon$.
(c) $\forall\epsilon,$ $\exists$ a Lebesgue measurable set $E_\epsilon$ s.t. m*($E_\epsilon$$\Delta$E)$\lt$$\epsilon$.
Now, I have no trouble proving (a) to (b) or (b) to (c). Here is my approach for (c) to (a):
Suppose (c) holds. Since $E_\epsilon$ is Lebesgue measurable, $\exists$ an open set $U_\epsilon$ that contains E s.t. m$^*$($U_\epsilon$$\backslash$$E_\epsilon$)$\lt\epsilon$.
Then, $U_\epsilon$\E $\subset$ $U_\epsilon$$\backslash$$E_\epsilon$ $\bigcup$ $E_\epsilon$\E $\subset$ $U_\epsilon$$\backslash$$E_\epsilon$ $\bigcup$ $E_\epsilon$$\Delta$$E$.
This implies m$^*$($U_\epsilon$$\backslash$$E$)$\leq$ m$^*$($U_\epsilon$$\backslash$$E_\epsilon$)$\bigcup$m$^*$($E_\epsilon$$\Delta$E) $=2\epsilon$
However, I am having trouble showing that $U_\epsilon$ contains the set E. As a side note, I apply a similar strategy for (b) to (a) (i.e., I cover the symmetrical difference with an open set $E_\epsilon$) and am having the same issue there as well. Intuitively, I am thinking because the symmetrical difference between $E_\epsilon$ and E, meaning $E_\epsilon$ approximates E and vice versa with error so arbitrary small, if there exists an open set $U_\epsilon$ that contains $E_\epsilon$, with difference in measure also arbitrarily small, then this open set must contain E as well. I am just having a hard time formalizing this as part of the proof (tried argument by contradiction but met a dead end). Any help or hints would be appreciated!