[Math] the difference between outer measure and Lebesgue measure

lebesgue-measuremeasure-theoryreal-analysis

What is the difference between outer measure and Lebesgue measure?

We know that there are sets which are not Lebesgue measurable, whereas we know that outer measure is defined for any subset of $\mathbb{R}$.

Best Answer

The Lebesgue measure and Lebesgue outer measure coincide on Lebesgue measurable sets, which can be defined in several equivalent ways. Let $m$ and $m^*$ denote the Lebesgue measure and the Lebesgue outer measure respectively. These are some possible definitions of $A\subset\mathbb{R}^n$ being measurable:

  1. For all $B\subset\mathbb{R}^n$ $$ m^*(B)=m^*(B\cap A)+m^*(B\setminus A) $$
  2. For all $\epsilon>0$ there exist an open set $G$ and a closed set $F$ such that $F\subset A\subset G$ and $m^*(G\setminus F)<\epsilon$. (Note that since $G\setminus F$ is open, it is measurable, so that $m^*(G\setminus F)=m(G\setminus F)$.)
  3. $A=F\cup N$, where $F$ is an $F_\sigma$ (i.e. a countable union of closed sets) and $m(N)=0$.
  4. $A=G\setminus N$, where $G$ is a $G_\delta$ (i.e. a countable intersection of open sets) and $m(N)=0$.

The reason for the need of two different concepts is that neither of them is "perfect":

  • $m$ is a measure, but is not defined for all subsets of $\mathbb{R}^n$
  • $m^*$ is defined for all subsets of $\mathbb{R}^n$, but is not additive: here exist disjoint sets $A$ and $B$ such that $m^*(A\cup B)\ne m^*(A)+m^*(B)$.
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