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Definition of Lebesgue Outer Measure: Given a set $E$ of $\mathbb R$, we define the Lebesgue Outer Measure of $E$ by, $$m^*(E) = \inf \left\{\sum_{n=1}^{+\infty} \ell(I_n): E \subset \bigcup_{n=1}^{+\infty}I_n \right\}$$ where $\ell(I_n)$ denotes the length of interval (bounded and nonempty interval).

Definition of measurable set: A set $E$ measurable if $$m^*(T) = m^*(T \cap E) + m^*(T \cap E^c)$$ for every subset of $T$ of $\mathbb R$.

If $E \subset \mathbb R$ with $m^*(E) = 0$ and $\exists$ finite interval $I,$ such that $E \subset I, $, then is $E$ measurable?

If $E \subset \mathbb R$ with $m^*(E) = 0$, then is $E$ measurable?

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    $\begingroup$ All measure-zero sets are Lebesgue measurable even though not all of them are Borel-measurable. ${}\qquad{}$ $\endgroup$ – Michael Hardy Jul 18 '15 at 18:46
  • $\begingroup$ @MichaelHardy: "The Lebesgue measurable sets are said to be complete because every subset of a null set is again measurable". So is there any subset of null set non-measurable?? $\endgroup$ – Bear and bunny Jul 18 '15 at 18:50
  • $\begingroup$ If ${\frak m}^\ast E_1 = 0$ and $E_2 \subseteq E_1$, then $0 \leq {\frak m}^\ast E_2 \leq {\frak m}^\ast E_1 = 0$ gives ${\frak m}^\ast E_2 = 0$, so $E_2$ is also measurable. $\endgroup$ – Ivo Terek Jul 18 '15 at 18:52
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Yes. Suppose that ${\frak m}^\ast E = 0$. On one hand, the inequality: $${\frak m}^\ast T \leq {\frak m}^\ast(T \cap E) + {\frak m}^\ast(T \cap E^c)$$always holds. On the other hand: $${\frak m}^\ast(T \cap E) \leq {\frak m}^\ast E = 0 \implies {\frak m}^\ast(T \cap E) = 0, \quad {\frak m}^\ast(T \cap E^c) \leq {\frak m}^\ast T$$gives the other inequality.

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  • $\begingroup$ Yes, you are right. $\endgroup$ – Bear and bunny Jul 18 '15 at 19:21

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