I am trying to prove that: $ \subseteq \mathbb{R}$ is Lebesgue measurable if and only if for every $a$ and $b$ we have $|[a,b] \cap A|+|[,]\backslash |=b-a$.
I just have to prove the backwards direction now. I have noted that it suffices to show that $A$ intersect $[n,n+1]$ is Lebesgue measurable for every $n$, so I can assume that $A$ is a subset of some closed interval $[a,b]$. This, along with our hypothesis, gives $|A| + | [a,b] \backslash A| = b-a$. I am stuck on where to go from here. I am trying to leverage the above statement to prove that a closed set $F$ exists which is a subset of $A$ with $|A\backslash F|< \epsilon$.
Things we know:
Closed sets are measurable
A set $E$ is measurable if for all $\epsilon > 0$ I can find $F \subseteq E$ closed so that $|E \setminus F| < \epsilon.$
Things I'm assuming: I'm assuming outer measure is defined as $$ |A| = \inf\{|U| : A \subseteq U, U \text{ is open}\}.$$
Step 1: Intervals are measurable.
If $I$ is a closed interval, we're done.
If $I$ is a half open interval, say $I = (a,b]$, then $[a-\epsilon, b] \subseteq I$ is a closed interval and we have $|I \setminus [a-\epsilon,b]| < \epsilon.$ We can do this for all $\epsilon$, so we're done. Same argument works the other way.
If $I$ is an open interval, do the same kind of trick. Fix $\epsilon > 0$. The closed interval $[a+\epsilon/2, b-\epsilon/2] \subseteq (a,b)$ and satisfies $|U \setminus [a+\epsilon/2, b-\epsilon/2]| = |(a, a+\epsilon/2) \sqcup (b-\epsilon/2, b)| = \epsilon$. I can make this arbitrarily small.
Step 2: A countable union of measurable sets is measurable.
Assume $\{E_n\}_{n=1}^\infty$ are closed. We want to show that $E = \bigcup_{n=1}^\infty E_n$ is measurable. A finite union of closed sets is closed, and $F_n = \bigcup_{n=1}^N E_n$ is a sequence of closed sets such that $F_n \nearrow E$, so we can choose $n$ large enough so that $|E \setminus F_n| < \epsilon.$ Can do this for all $\epsilon > 0$.
Now examine a set $E$. If for all $\epsilon > 0$, I can find a measurable set $F \subseteq E$ so that $|E \setminus F| < \epsilon$, then I claim that $E$ is measurable. Fix $\epsilon > 0$. Find $F \subseteq E$ so that $|E \setminus F| < \epsilon/2$. Since $F$ is measurable, I can find $G \subseteq F$ closed so that $|F \setminus G| < \epsilon/2$. Now $|E \setminus G| \leq |E \setminus F| + |F \setminus G| < \epsilon.$ The choice of $\epsilon > 0$ arbitrary, so we have the result.
Now let $\{E_n\}_{n=1}^\infty$ be a collection of measurable sets, $E = \bigcup_{n=1}^\infty E_n$. Fix $\epsilon > 0$. For each $n$, I can find $F_n \subseteq E_n$ closed so that $|E_n \setminus F_n| < \epsilon 2^{-n}.$ Notice by the first bullet point $F =\bigcup_{n=1}^\infty F_n$ is measurable, and $|E \setminus F| \leq \epsilon$. Since $\epsilon > 0$ arbitrary, I get that $E$ is measurable.
Step 3: Open sets are measurable.
Step 3: Complements of measurable sets are measurable.
Using Steps 1-3, we get that $G_\delta$ and $F_\sigma$ sets are measurable. Recall $G_\delta$ sets are countable unions of open sets, and $F_\sigma$ sets are countable unions of closed sets.
Step 4: On measurable sets, we have countable additivity. That is, if $\{A_n\}$ a sequence of disjoint measurable sets, then $$ \left|\bigcup A_n\right| = \sum |A_n|.$$
For nonoverlapping closed intervals, we have $$\left| \bigcup_1^N I_k \right| = \sum_1^N |I_k|.$$ This is not really that trivial. Use compactness and cover with open intervals. (I leave as an exercise).
Two sets $A$ and $B$ are separated if $d(A,B) = \inf\{d(x,y) : x \in A, y \in B\} > 0$. For compact separated sets, $|A \cup B| = |A| + |B|$. (I leave as an exercise. Again, use intervals.)
Assume that the $A_n$ are bounded. Use compact sets $F_k \subseteq A_k$ which approximate well. Use the prior step. (I leave as an exercise.)
For unbounded, examine $I_k = [-k,k]$, $F_k = I_k \setminus I_{k-1}$. Then $$ A_k = \bigcup_{k=1}^\infty (A_k \cap F_k).$$ Now use prior results (I leave as an exercise.)
Step 5: For every set $A$, there exists a $G_\delta$ set (i.e. measurable) $H$ so that $A \subseteq H$ and $|H| = |A|$.
Step 6: A set $E$ is measurable if and only if for all $A$, we have $$ |A| = |A \cap E| + |A \setminus E|.$$
$(\implies):$ Notice $|A| \leq |A \cap E| + |A \setminus E|.$ Use Step 5 to find $H$ with $|H| = |A|.$ Notice $H = (H \cap E) \sqcup (H \setminus E).$ Using prior steps, this is a disjoint union of measurable sets, so $$ |A| = |H| = |H \cap E| + |H \setminus E| \geq |A \cap E| + |A \setminus E|.$$
$(\impliedby):$ We can assume $E$ is bounded (if not, take $E \cap I_k$ as above and do a similar trick). Let $E \subseteq H$, $H$ measurable, and $|H| = |E|$. Then $$ |E| = |H| = |H \cap E| + |H \setminus E| = |E| + |H \setminus E|.$$ Conclude that $|H \setminus E|$ has measure zero since $E$ is bounded. (I leave as an exercise.)
Call this condition the C condition (for Caratheodory).
Step 7: A set $E$ satisfies the C condition if and only if it satisfies the interval condition in the problem.
$(\implies):$ This is trivial (if it satisfies the C condition, it works for every set).
$(\impliedby):$ Let's show if it holds for closed intervals, it holds for half open intervals. Let $[a,b)$ be a half open interval. Now
$$ |[a,b)| \leq |[a,b) \cap E| + |[a,b) \setminus E| \leq |[a,b] \cap E| + |[a,b] \setminus E| = b-a = |[a,b)|.$$
So the result holds.
Similarly, we get it holds for open intervals.
Suppose we have a union of disjoint open intervals $I = \bigcup I_k$. Then $$ |I| \leq |I \cap E| + |I \setminus E| \leq \sum_k |I_k \cap E| + |I_k \setminus E| = \sum_k |I_k| = |I|.$$ So it holds.
We can write open sets as unions of disjoint open intervals, so it holds for open sets.
Now we have it for open sets. Let $A$ be any set. We can find $U$ open so that $A \subseteq U$ (claim). Now $A \cap E \subseteq U \cap E$ and $A \setminus E \subseteq U \setminus E$. Now $$ |A| \leq |A \cap E| + |A \setminus E| \leq |U \cap E| + |U \setminus E| = |U|.$$
By the supremum property of outer measure, for every $\epsilon > 0$ I can find open $U$ so that $A \subseteq U$ and $|U| \leq |A| + \epsilon$. So we have
$$|A| \leq |A \cap E| + |A \setminus E| \leq |U \cap E| + |U \setminus E| = |U| \leq |A| + \epsilon.$$
But this is for all $\epsilon > 0$. Take $\epsilon \rightarrow 0$ to get the result.