TITLE: $\delta$-fined division and Lebesgue measure $\mu$
QUESTION [5 upvotes]: Let $E$ be a measurable subset of $[0,1]$. Then we know that we can choose an open set $G$ and a closed set $F$ such that $F\subset E \subset G \subset [0,1]$. For each $t\in [0,1]$, define
$$\delta(t) = 
 \begin{cases}
\text{dist}(t,G^c),&\text{if $t\in F$}\\
\text{min}\{\text{dist}(t,b(G)),\text{dist}(t,F)\},&\text{if $t\in G\smallsetminus F$}\\
\text{dist}(t,F),&\text{if $t\in [0,1]\smallsetminus G$.}  
 \end{cases} 
$$
Here we use the notation $b(G)$ to mean the boundary of $G$. Because the sets $G^c$, $b(G)$, and $F$ are closed, it follows that $\delta(t)>0$ for each $t\in [0,1].$ This defines a positive function $\delta$ defined on $[0,1].$ Cousin's Lemma therefore assures that a division $D=\{(t_i,I_i)\}_{i=1}^{n}$ exists such that for each $i=1,\dots,n$ we have $t_i\in [0,1]$ and
$$I_i\subset (t_i-\delta(t_i),t_i+\delta(t_i)).$$
Some authors call $D$ as a $\delta$-fined free tagged division of $[0,1].$
My question:
How do we show (if it is true) that
$$(D)\sum_{t_i\in E}[\mu(I_i)-\mu(E\cap I_i)] \leq \mu(G\smallsetminus F)$$ and
$$(D)\sum_{t_i\notin E}\mu(E\cap I_i)] \leq \mu(G\smallsetminus F)$$
where $\mu$ denotes the Lebesgue measure.
Any tips?Thanks in advance.

REPLY [2 votes]: The choice of $\delta$ implies the following two properties of the tagged partition $(t_i,I_i)$: 
$$t_i\in G\implies I_i\subseteq G \tag{1}$$
$$t_i\notin F\implies I_i\cap F=\varnothing \tag{2}$$
Since the intervals form a partition of $[0,1]$, property (1) implies
$$\sum_{t_i\in G} \mu(I_i) \le \mu(G)\tag{3}$$
and property (2) implies
$$\sum_{t_i\in G} \mu(F\cap I_i) = \mu(F)\tag{4}$$
Therefore, 
$$\sum_{t_i\in G} (\mu(I_i) - \mu(F\cap I_i)) \le \mu(G\setminus F)\tag{5}$$
which is stronger than the your first inequality. 
If $t_i\notin F$, then by (2) $E\cap I_i\subseteq G\setminus F$. Therefore,
$$\sum_{t_i\notin F} \mu(E\cap I_i)\le \mu(G\setminus F)\tag{6}$$ 
which is stronger than your second inequality.