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The definition of Lebesgue measure is

$E$ is Lebesgue measurable if $\forall \epsilon > 0,$ there exists an open set $O$ such that $E \subset O$ and $m_*(O-E) < \epsilon$.

Caratheorody measurable is

$E$ is Caratheodory measurable, if $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ for all $A \in \Omega$.

I want to show that the top definition implies the bottom definition. Since $m_*(A) \le m_*(A \cap E) + m_*(A \cap E^c)$ is trivial by subadditivity, we only need to show $m_*(A) \ge m_*(A \cap E) + m_*(A \cap E^c)$. In order to use the Lebesgue measurability, let $E$ be Lebesgue measurable, and let $O$ be some open set such that $A \subset O$. Then \begin{align*} m_*(A \cap E) + m_*(A \cap E^c) \le m_*(O \cap E) + m_*(O \cap E^c) \end{align*} But I'm not sure where I go from there. I want to use $\epsilon$ somewhere but I'm not sure how. Any hint would be much appreciated.

MoneyBall
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  • Are you using $m_$ to mean the outer* measure? Usually the outer mesure is represented as $m^*$. – Ramiro Dec 13 '20 at 04:47
  • @Ramiro Yes, I'm using $m_*$ to denote exterior measure on $\mathbb{R}^d$. This is the notation used in my textbok. – MoneyBall Dec 13 '20 at 05:17
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  • See https://people.math.gatech.edu/~heil/6337/spring11/section1.5.pdf for a simpler solution (note that the Lebesgue outer measure of $E$ is denoted by $|E|_e$ in the document). – ho boon suan Dec 13 '21 at 10:16

1 Answers1

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Attention: We are going to use the terminology used in the question, including $m_*$ to denote Lebesgue outer measure:

$E$ is Lebesgue measurable if $\forall \epsilon > 0$, there exists an open set $O$ such that $E \subset O$ and $m_*(O-E) < \epsilon$.

and

$E$ is Caratheodory measurable, if $m_*(A) = m_*(A \cap E) + m_*(A \cap E^c)$ for all $A \subseteq \Omega$.

Let us prove that if $E$ is Lebesgue measurable, then $E$ is Caratheodory measurable.

Suppose $E$ is is Lebesgue measurable. Given any $\epsilon >0$, let $O_\epsilon$ be an open set such that $E \subset O_\epsilon$ and $m_*(O_\epsilon-E) < \epsilon$.

Given any $A \subseteq \Omega$, we have, taking any any $\epsilon >0$ \begin{align*} \mu_*(A) &\leqslant \mu_*(A \cap E) + \mu(A\cap E^c) \\ & \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*((A\cap O_\epsilon ^c) \cup (A\cap O_\epsilon \cap E^c))\\ & \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*(A\cap O_\epsilon ^c) +\mu_*(A\cap O_\epsilon \cap E^c)\\ & \leqslant \mu_*(A \cap O_\epsilon ) + \mu_*(A\cap O_\epsilon ^c) +\mu_*(O_\epsilon \cap E^c)\\ & \leqslant \mu_*(A) + \epsilon \end{align*} where in the last step we used the fact that open sets are Caratheodory measurable.

So we have proved that, for any $\epsilon >0$, $$ \mu_*(A) \leqslant \mu_*(A \cap E) + \mu(A\cap E^c) \leqslant \mu_*(A) + \epsilon $$ So $$ \mu_*(A) = \mu_*(A \cap E) + \mu(A\cap E^c) $$ So $E$ is Caratheodory measurable.

Ramiro
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  • I think you meant $E$ is Caratheodory measurable at the end. Otherwise, thank you so much for the detailed answer. It all makes sense now! – MoneyBall Dec 13 '20 at 12:54
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    @MoneyBall , Yes, you are right. In the last line I meant: "So $E$ is Caratheodory measurable.". I have fixed it, thanks. You are welcome. I am happy I could help you. – Ramiro Dec 13 '20 at 14:59