1
$\begingroup$

Given a set $E$ is Lebesgue measurable, and $E$=$E_1$$\cup$$E_2$, if $m(E)$ is less than infinity, $m(E)$ = $m^{*}(E_1) + m^{*}(E_2)$,where $m(E)$ is the Lebesgue measure of $E$,and $m^{*}(E_1)$ and $m^{*}(E_2)$ are the outer measure of $E_1$ and $E_2$ how to prove that $E_1$ and $E_2$ are measurable.

  • 2
    What definition of measurability are you using?2012-12-29
  • 3
    This is surely false without additional conditions (take Lebesgue measure on $E=\mathbb R$, a non-measurable set of infinite outer measure, and its complement). Is $E$ assumed to have finite measure? What else are you assuming?2012-12-29
  • 0
    @Clayton , I have specified and edited my question, now it seems clear. Thanks2012-12-29
  • 0
    @Giuseppe Negro,2012-12-29
  • 0
    @ Andres Caicedo2012-12-29
  • 0
    @Alex: I have seen multiple characterizations of what it means to be Lebesgue measurable; is there a definition you prefer? For example, there is Caratheodory's characterization, or the one that I used in my Real Analysis class which is for every $\varepsilon>0$, there exists an open set $G$ such that $m^*(G-E)<\varepsilon$.2012-12-29
  • 0
    exactly!@Clayton2012-12-29
  • 1
    View [this question](http://math.stackexchange.com/questions/72729/proving-sets-are-measurable) for help. It is essentially the same question.2012-12-29
  • 0
    oh, yeah.thanks Clayton.2012-12-29

0 Answers 0