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.
the measurability of sets given their union are measurable
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measure-theory
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2What definition of measurability are you using? – 2012-12-29
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3This 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
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0@Clayton , I have specified and edited my question, now it seems clear. Thanks – 2012-12-29
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0@Giuseppe Negro, – 2012-12-29
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0@ Andres Caicedo – 2012-12-29
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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
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0exactly!@Clayton – 2012-12-29
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1View [this question](http://math.stackexchange.com/questions/72729/proving-sets-are-measurable) for help. It is essentially the same question. – 2012-12-29
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0oh, yeah.thanks Clayton. – 2012-12-29