$\newcommand{\c}{\mathfrak{c}}$
Let $\c$ denote the cardinal of the continuum and wellorder the Borel subsets of $[0,1]$ as $(B_\alpha)_{\alpha < \c}$. We build by transfinite recursion a sequence $(x_\alpha)_{\alpha < \c}$ of elements of $[0,1]$ such that:
(a) $x_\alpha$ is Vitali inequivalent to $x_\beta$ for all $\beta < \alpha$, and
(b) $x_\alpha \in [0,1] \setminus B_\alpha$ if $[0,1] \setminus B_\alpha$ is uncountable.
Note that this process can't get stuck, since if the complement of $B_\alpha$ is uncountable then it has cardinality $\c$, and thus it meets an unused Vitali equivalence class (since at most $|\alpha| < \c$ have been used so far). Then by setting $X = \{x_\alpha : \alpha < c\}$ we obtain a set such that whenever $B$ is a Borel set with $X \subseteq B$, then $B$ has countable complement (and in particular has measure $1$). So $X$ has outer measure $1$ as desired.
Lemma: Let $A \subseteq \Omega$ and $E, F \in M$ such that $E \cap F=\emptyset$, then
$$\mu^*(A \cap (E \cup F)) = \mu^*(A \cap E) + \mu^*(A \cap F)$$
Proof: Since $E \in M$ and $E \cap F=\emptyset$, we have
\begin{align}
\mu^*(A \cap (E \cup F)) &= \mu^*(A \cap (E \cup F) \cap E) + \mu^*(A \cap (E \cup F) \cap E^c) =\\
&= \mu^*(A \cap E) + \mu^*(A \cap F)
\end{align}
$\square$
Now for the main result:
Since for all $n$, $A \cap E_n \subseteq \lim_k A \cap E_k$, we have
$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)$.
Now, let $E= \lim_n E_n = \bigcup_n E_n$. Let us define $F_0 = E_0$ and, for each $n$, $F_{n+1} =E_{n+1} \setminus \bigcup_{k=0}^n E_k$. Then the sets $F_n$ are disjoint set and for all $n$, $F_n \in M$. Moreover $E_n= \bigcup_{k=0}^n F_k$ and $E= \lim_n E_n = \bigcup_n E_n= \bigcup_n F_n$.
So we have
$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)= \mu^* \left (\bigcup_n ( A \cap F_n) \right ) \leqslant \sum_n\mu^*(A\cap F_n)$$
where the last step is $\sigma$-sub-additivity of $\mu^*$. So we have:
$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) \tag{1}$$
Using the Lemma and induction, we have that for all $k$,
$$\sum_{n=0}^k\mu^*(A\cap F_n)= \mu^*\left (A \cap \bigcup_{n=0}^k F_n \right ) = \mu^*(A \cap E_k) \tag{2}$$
Combining $(1)$ and $(2)$:
$$\lim_n \mu^*( A \cap E_n) \leqslant \mu^*(\lim_n A \cap E_n)\leqslant \lim_k\sum_{n=0}^k\mu^*(A\cap F_n) = \lim_k\mu^*(A \cap E_k) $$
So $\lim_n \mu^*( A \cap E_n) = \mu^*(\lim_n A \cap E_n)$
Best Answer
Since I have the copyright on an explanation that might help I can repost it here.
The problem that most students encounter for this one is that they have aleady see the Lebesgue definition using outer and inner measures. But that approach does not work in general situations. The Greek mathematician Constantin Carathéodory (1873–1950) was the one who developed the correct approach. But why does it work and why doesn't the simpler inner/outer measure approach work?
This example from the reference [1] was meant to illustrate.
Example 2.29 Let $X = \{1,2,3\}$. Let $μ^∗( \emptyset) = 0$, $μ^∗(X) = 2$, and $μ^∗(A) = 1$ for every other set $A⊂X$. It is a routine matter to verify that $μ^∗$ is an outer measure.
Suppose now that we wish to mimic the procedure that worked so well for the Peano–Jordan content and the Lebesgue measure. We could take our cue from the formula in assertion 2.4 and define a version of the inner measure for this example as $$μ_∗(A) = μ^∗(X)−μ^∗(X \setminus A) = 2−μ^∗(X \setminus A).$$
If we then call $A$ measurable provided that $μ^∗(A) = μ_∗(A)$, and let $μ(A) = μ^∗(A)$. for such sets, our process is complete. We find that all eight subsets of X are measurable by this definition, but $μ$ is clearly not additive on $2^X$ . The classical inner–outer measure procedure completely fails to work in this simple example!
A bit of reflection pinpoints the problem. The inner–outer measure approach puts a set $A$ to the following test stated solely in terms of $μ^∗$:
Is it true that $μ^∗(A)+μ^∗(X \setminus A) = μ^∗(X)$?
[Here] every $A ⊂ X$ passed this test. But, for $A = \{1\}$ and $E = \{1,2\}$, we see that $$μ^∗(A)+μ^∗(E \setminus A) = 2 > 1 = μ^∗(E).$$
Thus, while $μ^∗$ is additive with respect to $A$ and its complement in $X$, it is not with respect to $A$ and its complement in $E$.
REFERENCE:
[1] http://classicalrealanalysis.info/documents/BBT-AlllChapters-Landscape.pdf