# $\delta$-fined division and Lebesgue measure $\mu$

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 $[0,1].$

And here is 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.

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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.