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 ?


1 Answer 1


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 $0$.


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