Let $E \subset \mathbb{R}^{d}$. Suppose that, for any $\epsilon > 0$, there is a measurable $F \subset \mathbb{R}^{d}$ such that $$m_{\ast}(E \triangle F) < \epsilon.$$ Show that $E$ is measurable.
I need to show that:
for any $\epsilon > 0$, there is an open $O$ with $E \subset O$ such that $m_{*}(O - E) \leq \epsilon$.
This question doesn't seem to require specific properties of measure theory. It seems viable to try to show that it is worth the definition. But I couldn't find a good way, maybe some specific property of symmetric differences would be useful.
I don't like the answer of the question, I just wanted a hint to try. Thanks for the advance!
You know that for any $\varepsilon >0$ there exists an open set $O \subset \mathbb{R}^d$ satisfying $$m_{*}(O \setminus F) < \varepsilon,$$ because $F$ is measurable. Now we can use this in order to show that we also have $m_{*}( O \setminus E) < \varepsilon$.
In fact, note that $$O \setminus E \subset E \Delta F \cup O \setminus F.$$ This can be easily checked. Using the $\sigma$-subadditivity of the outer measure $m_{*}$ shows already that $m_{*}( O \setminus E) < \varepsilon$, as claimed.
However, we don't need to use the special properties of the Lebesgue-measure. The above statement is in general true for the "Carathéodory-$\sigma$-algebra" defined for an outer-measure. Let $\mu_{*}$ be any outer-measure on a set $X$. We need to prove that for an $Q \subset X$ the equality $$\tag{1}\mu_{+}(Q) = \mu_{*}(E \cap Q) + \mu_{*}(Q \cap E^c).$$ holds. Since (1) is true for $F$ instead of $E$, it is enough to verify that $\mu_{*}(F \cap Q) = \mu_{*}(E \cap Q)$, resp. $\mu_{*}(F^c \cap Q) = \mu_{*}(E^c \cap Q)$.
As above, we have $F \cap Q \subset E \cap Q \subset E \Delta F$ and also the same for $E$ replaced by $F$. Thus $\mu_{*}(F \cap Q) = \mu_{*}(E \cap Q)$. Similiar we can show that $\mu_{*}(F^c \cap Q) = \mu_{*}(E^c \cap Q)$.