# Why for measurable sets $\inf\{\sum_{i=1}^\infty |E_i|: A\subset \bigcup_{i=1}^\infty E_i\}=\sup\{\sum_{i=1}^\infty |E_i|...\}$

Let $$E$$ being a measurable set. We say that $$E$$ is Jordan measurable if $$\sup_{S\subset E}J(S)=\inf_{S\supset E}J(S),$$ where $$S$$ is a simple set (finite union of interval of the form $$[a,b]$$) and $$J$$ the Jordan measure. Could we say that $$E$$ is Lebesgue measurable if $$\inf\{\sum_{i=1}^\infty (b_n-a_n)\mid E\subset \bigcup_{i=1}^\infty [a_n,b_n]\}=\sup\{\sum_{i=1}^\infty (b_n-a_n)\mid E\supset \bigcup_{i=1}^\infty [a_n,b_n]\},$$ or both can be equal without $$E$$ being Lebesgue measurable ?

You need to be a little careful in the description of the sup. For example, if $$a_n=0$$ and $$b_n=1-1/n$$, then $$\bigcup_{n=1^\infty}[a_n,b_n]\subseteq[0,1)$$ but $$\sum_{n=1}^\infty(b_n-a_n)=\infty$$.
To fix that, consider the statement \begin{align*} &\inf\left\{\sum_{n=1}^\infty(b_n-a_n):E\subseteq\bigcup_{n=1}^\infty[a_n,b_n]\right\}\\ &=\sup\left\{\sum_{n=1}^\infty(b_n-a_n):\bigcup_{n=1}^\infty[a_n,b_n]\subseteq E,\ [a_n,b_n]\text{ pairwise disjoint}\right\}\tag{*} \end{align*}
If $$(*)$$ holds and the value above is finite, then we can approximate $$E$$ by Borel sets $$A_n\subseteq E\subseteq B_n$$ (namely countable unions of appropriate intervals) such that $$\mu(B_n\setminus A_n)<1/n$$. Then letting $$A=\bigcup_n A_n$$ and $$B=\bigcap B_n$$ we have $$A\subseteq E\subseteq B$$ and $$\mu(B\setminus A)=0$$, so since the Lebesgue $$\sigma$$-algebra is complete (wrt Lebesgue measure) then $$E$$ is Lebesgue measurable.
If the value above is infinite then $$E$$ might not be Lebesgue measurable even if $$(*)$$ holds. For example, if $$X\subseteq[0,1]$$ is not Lebesgue measurable just take $$E=X\cup [2,\infty)$$. Then $$(*)$$ holds but $$E$$ is not Lebesgue measurable.
In the other direction, $$E$$ may be Lebesgue measurable but $$(*)$$ does not hold: A fat Cantor set has positive measure, so the supremum in $$(*)$$ is $$>0$$, but it also has empty interior, so the infimum in $$(*)$$ is $$0$$.