Why outer measure is defined by $m_n^*(A)=\inf \bigg\{ \sum_{i=1}^\infty l(I_i):\mathbb{F}=\{I_1,I_2,… \}\mbox{ is Lebesgue cover of } A \bigg\}$?

My questions is:

• Why the outer measure $m_n^*(A)$ is defined by $$m_n^*(A)=\inf \left\{ \sum_{i=1}^{\infty} l(I_i): \mathbb{F}= \{I_1,I_2,\cdots \} \mbox{ is Lebesgue cover of } A \right\}$$ for $A \subset \mathbb{R}^n$ ?
• Is it true that $m_n^*(A) \leq \sum_{i=1}^{\infty}l(I_i) \leq l(I)$? For example can you compute $m_1^*(A)$, where $A=\{x \in \mathbb{R}: x \in [a,b] \}$?

I think you should know that whether $A=I$. Then at least $m_n^*(A) = l(I)$ But what is $I$? I'm aware about these many questions and apologize for that.

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$I$ is n-interval such that $I=I_1 \times I_2 \times \cdot$, where $I_i \in \mathbb{R}$. But notice that $I_i \subset \mathbb{R}^n$ in $m_n^*(A)$ is not same as $I_i \subset \mathbb{R}$( definition of $I$). –  laovultai Jan 23 '13 at 16:41