Show that a set $A$ is Lebesgue measurable.

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Show that a set $A\subseteq \mathbb{R}$ is Lebesgue measurable iff for every $a$ and $b$ we have $|[a,b]\cap A| + |[a,b]\setminus A|=b-a$.

I started this problem by supposing the condition stated holds and I want to arrive at the conclusion that $A$ is Lebesgue measurable. I am thinking that I need to show one of the following equivalences for being a Lebesgue measurable set:

(a) There exists a Borel set $B\subset A$ such that $|A\setminus B|=0$.

(b) For each $\epsilon >0$, there exists an open set $G\subset A$ such that $|G\setminus A|<\epsilon$.

(c) There exist open sets $G_{1}, G_{2},...$ containing $A$ such that $(\bigcap_{k=1}^{\infty}G_{k})\setminus A|=0$

(d) There exists a Borel set $A\subset B$ such that $|B\setminus A|=0$.

My idea was to break up the proof in cases:

Case 1: $A$ and $[a,b]$ are disjoint. Then \begin{equation} \begin{split} |[a,b]\cap A|+|[a,b]\setminus A|&=0+b-a\\ &=b-a \end{split} \end{equation} I was thinking that we can use (b) in this case but I am not sure of the role that $[a,b]$ has, or why we even need the condition.

The second case that $A$ and $[a,b]$ have a nonempty intersection can be broken up into three parts:

(I.) $A\subset [a,b]$

(II.)$[a,b]\subset A$

(III.) $[a,b]\cap A=B$

I'm not entirely sure where to start or if I should even be looking at those particular equivalences. Any advice would be helpful. Thank you.