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$E$ is measurable (lebesgue and on Real line) if for each $\epsilon >0$ ,there is a closed set $F$ and open set $O$ for which $F\subseteq E \subseteq O $ and $ m^{*}(O-F)<\epsilon $ ?

Edit:- Definition of Measurable set:- A set $E$ is measurable if for each set $A$ of Real number , $m^{*}(A)=m^{*}(A\cap E)+m^{*}(A\cap E^{c})$.

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    $\begingroup$ You only gave one definition of measurable here. If you want to ask if it is equivalent, or why it is equivalent, to another definition, you should give that definition as well. $\endgroup$ – Asaf Karagila Nov 5 '13 at 19:09
  • $\begingroup$ You are right... I felt most of the book when give definition of measurable function will give based on outer measure , so I felt its not necessary ... $\endgroup$ – henry Nov 5 '13 at 19:16
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Yes they are equivalent. Consider $E \subseteq \mathbb{R}$.

1.(Definition due to Carathéodory) $E$ is measurable if for each set $A \subset \mathbb{R}$, $m^{*}(A)=m^{*}(A\cap E)+m^{*}(A\cap E^{c})$.

The following are equivalent:

i. E is measurable (according to definition 1)

ii. For each $\epsilon >0$ , there is a closed set $F$ and open set $O$ for which $F\subseteq E \subseteq O $ and $ m^{*}(O-F)<\epsilon $

Let us also have the definition of Lebesgue outer measure $m^{*}: \mathscr{P}(\mathbb{R}) \rightarrow \bar{\mathbb{R}}^{+}$ given by,

$ m^{*}(E) = \inf \{ \sum l(I_{n}) : (I_n)_{n \in \mathbb{N}}$ is a collection of open intervals such that $E \subset \cup I_{n} \}$

When $m^{*}$ is defined on $\mathscr{P}(\mathbb{R})$ it is only countably sub-additive. But if we restrict $m^{*}$ to $\mathscr{M}\subset \mathscr{P}(\mathbb{R})$, where $\mathscr{M}$ is the collection of all sets that satisfies 1, then it has been proved that $m^{*}|_{\mathscr{M}}$ is countably additive and that $\mathscr{M}$ is a $\sigma$-algebra (containing the intervals).

Proof: i. $\implies$ ii. (Case $m^{*}(E) < +\infty$ )

Given an $\epsilon > 0$ there is a collection $\{I_{n} \}$ of open intervals such that $m^{*}(E) < l(\cup \{I_{n} \}) < m^{*}(E) + \epsilon $, this is so because $m^{∗}(E)$ is an infimum. Define $O = \bigcup_{n =1}^{\infty} \{I_{n} \}$. Note that $O$ is an open set, $O \in \mathscr{M}$, $E \subset O$, and $m^{*}(O) = l(O)$. Now we have,

$ m^{*}(O) < m^{*}(E) + \epsilon \implies m^{*}(O\backslash E) = m^{*}(O) - m^{*}(E) < \epsilon $

Let us find now that closed $F$.

Notice that $ E \subset \mathscr{M} \implies E^{c} \subset \mathscr{M}$. By the first part of this proof there is an open set $O \in \mathscr{M}$ such that $E^{c} \subset O$ and $ m^{*}(O\backslash E^{c}) < \epsilon$.

Take $ F = O^{c}$ closed. Then, $ F \subset E$ and,

$m^{*}(E \backslash F) = m^{*}((E^{c})^{c} \cap F^{c}) = m^{*}((E^{c})^{c}\cap O) = m^{*}(O \backslash E^{c}) < \epsilon$

Finally we observe that,

$(O \backslash F) = (O \backslash E) \bigcup (E \backslash F)$ (disjoint)

hence,

$m^{*}(O \backslash F) = m^{*}(O \backslash E) + m^{*}(E \backslash F) < \epsilon + \epsilon = 2 \epsilon$.

To prove ii. $ \implies$ i. first prove that ii. implies that exists a $G \in \mathscr{G}_{\delta}$ such that $m^{*}(G \backslash E) = 0$. Then use it to prove i.

Once this has been done, we have to think on the case $m^{*}(E) = \infty$.

I hope this could help a bit. I recommend reading Chapter 3 from Royden's book, the overall structure of the proof I tried to sketch here are suggested there.

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