Show a set must be measurable Below is a problem I found, however, after many attemps I can not seem to get a solution.
Problem: Let $E \subset [0,1]$. Show that if $m^*(E) + m^*([0,1] \setminus E) = 1$, then $E$ is measurable.
(my attempted) Solution: Notice, the following is true $[0,1] \setminus E = [0,1] \cap E^c$. Therefore, we can rewrite the given equations as
$$m^*(E) + m^*([0,1] \cap E^c) = 1. \quad (i)$$
Also, notice that $E = [0,1] \cap E$, hence, we can rewrite $(i)$ as the following
$$m^*([0,1] \cap E) + m^*([0,1] \cap E^c) = 1. \quad (ii)$$
Since $m^*([0,1]) = 1$ we can, again, rewrite $(ii)$ as follows
$$m^*([0,1] \cap E) + m^*([0,1] \cap E^c) = m^*([0,1]). \quad (iii)$$
For this path, this is where I get stuck. In other words, obviously if $[0,1]$ could be replaced by any set $A \subseteq \mathbb{R}$ then, sure, $E$ is measurable. Therefore, I do not think this is the "correct path" to take.
I feel like one way is, possibly, to show that for any $\epsilon > 0$, there exists a closed set F, such that, $E \subseteq F$ and $m^*(F \setminus E) < \epsilon$. Taking $F = [0,1]$, we have $E \subset F$ and
$m^*(F \setminus E) < 1 - m^*(E)$. Hence, if I take $\epsilon = 1 - m^*(E)$ can I conclude the proof?
 A: Let $A$ be any Lebesgue measurable subset of $[0,1]$. So, Lebesgue measurability of $A$ gives $$m^*(E)=m^*(A\cap E)+m^*(A^c\cap E)$$ and $$m^*(E^c)=m^*(A\cap E^c)+m^*(A^c\cap E^c).$$
Adding these two we have, $$m^*([0,1])=m^*(E)+m^*(E^c)$$$$=\bigg[m^*(A\cap E)+m^*(A\cap E^c)\bigg]+\bigg[m^*(A^c\cap E)+m^*(A^c\cap E^c)\bigg]$$$$\geq m^*(A)+m^*(A^c)\geq m^*([0,1]).$$ Both inequalitities are due to the fact $m^*$ is sub-additive. Hence, $$m^*(A)+m^*(A^c)=m^*(A\cap E)+m^*(A\cap E^c)+m^*(A^c\cap E)+m^*(A^c\cap E^c).$$ But, $$m^*(A^c)\leq m^*(E\cap A^c)+m^*(E^c\cap A^c)$$$$\implies m^*(A)\geq m^*(E\cap A)+m^*(A\cap E^c).$$
Hence, using sub-additivity of $m^*$ we have, $m^*(A)\geq m^*(E\cap A)+m^*(A\cap E^c)$, so $E$ is measurable.

Note that we are using the following equivalent conditions:
$(1)$ $E$ is measurable
$(2)$ For every $Y\subseteq [0,1]$ we have $m^*(Y)\geq m^*(Y\cap
 E)+m^*(E^c\cap Y)$.
$(3)$ For every measurable set $A\subseteq [0,1]$ we have $m^*(A)\geq m^*(A\cap
 E)+m^*(E^c\cap A)$.

Let us prove the non-trivial part, namely $(3)\implies (2)$. Let $Y\subseteq [0,1]$ and for a given $\varepsilon>0$ choose Lebesgue measurable subsets $A_1,A_2,...$ of $[0,1]$ so that $Y\subseteq \bigcup_{n=1}^\infty A_n$ with $m^*(Y)+\varepsilon>\sum_{n=1}^\infty m^*(A_i)$.
Next, since we are assuming $(3)$ we have  $m(A_n)\geq m^*(A_n\cap E^c)+m^*(A_n\cap E)$. So, we have $$m^*(Y)+\varepsilon>\sum_{n=1}^\infty m^*(A_i)$$$$\geq \sum_{n=1}^\infty m^*(A_n\cap E^c)+\sum_{n=1}^\infty m^*(A_n\cap E)$$$$\geq m^*(Y\cap E)+m^*(Y\cap E^c).$$ Finally, letting $\varepsilon\to 0+$ we are done.
A: $(2)\implies(1)$
Proof:
Assume that $\;E\;$ is a subset of $\;[0,1]\;$ such that (2) holds.
Let $\;Y\;$ be a $\;G_\delta\;$ set ($Y$ is Lebesgue measurable) such that $\;E\subseteq Y\subseteq [0,1]\;$ and $\;m(Y)=m^*(E)\;$.
From $\;(2)\;$ it follows that
$m^*(E)=m(Y)=m^*(Y)\ge m^*(Y\cap E)+m^*(Y\cap E^c)=$
$=m^*(E)+m^*(Y\cap E^c)\;,\;$ hence
$m^*(Y\cap E^c)=0\;$ and so $\;Z=Y\cap E^c\;$ is Lebesgue measurable.
Since $\;E=Y-Z\;,\;$ it is Lebesgue measurable as well.
Addendum:
Now we are going to prove that there exists a $\;G_\delta\;$ set $\;Y\;$ such that $\;E\subseteq Y\subseteq [0,1]\;$ and $\;m(Y)=m^*(E)\;$.
$m^*(E)=\inf\left\{\sum\limits_{n=1}^\infty m(I_n): \left(I_n\right)_{n\in\mathbb{N}}\text{ is a sequence of open}\\\text{ intervals such that } E\subseteq\bigcup_\limits{n=1}^\infty I_n\right\}.$
Let $\;A=\bigcup_\limits{n=1}^\infty I_n\implies A\;$ is a open set, $\;A\supseteq E\;$ and $\;m(A)\le\sum\limits_{n=1}^\infty m(I_n)\;.$
There exists a sequence $\;\left(A_k\right)_{k\in\mathbb{N}}\;$ of open sets such that $\;A_k\supseteq E\;$ and $\;m(A_k)<m^*(E)+\cfrac{1}{k}\;,\;$ for all $\;k\in\mathbb{N}\;.$
Let $\;B_k=A_k\cap\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$ for all $\;k\in\mathbb{N}\;.$
Hence,
$B_k\;$ is a open set, $\;E\subseteq B_k\subseteq\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$ and $\;m(B_k)<m^*(E)+\cfrac{1}{k}\;,\;$ for all $\;k\in\mathbb{N}\;.$
Let $\;Y=\bigcap\limits_{k=1}^\infty B_k\implies Y\;$ is a $\;G_\delta\;$ set.
It results that $\;E\subseteq Y\subseteq\left]-\frac{1}{k},1+\frac{1}{k}\right[\;$ for all $\;k\in\mathbb{N}\;,$
so $\;E\subseteq Y\subseteq[0,1]\;,\;$ moreover,
$m(Y)<m^*(E)+\cfrac{1}{k}\;$ for all $\;k\in\mathbb{N}\;,\;$ hence
$\;m(Y)\le m^*(E)\;.$
Since $\;E\subseteq Y\;,\;$ it follows that $\;m^*(E)\le m(Y)\;,\;$ hence $\;m(Y)=m^*(E)\;.$
