Bounded set has finite outer measure

The definition of outer measure is as follows: $$m^*(E) = \inf \left\{ \sum\limits_{k=1}^{\infty} l(I_k) : A \subset \bigcup\limits_{k=1}^{\infty} I_k \right\}$$

If a set is bounded then for some $p\in E$ every $e \in E$, $d(e, p) \leq M$.

In my mind I interpret this as saying that every point has a finite distance less than or equal to M. However, I'm not sure how to translate that to the definition of outer measure. My first idea was that the intervals $I_k$ are the lengths between a fixed $p$ and $e$. Since they would all clearly be less than $M$, then they are all finite. Thus the outer measure must be finite.

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Do you mean $$m^*(E) = \inf\left\{\sum\limits_{k=1}^{\infty} l(I_k) : A \subset \bigcup\limits_{k=1}^{\infty} I_k\right\}\;?$$ –  Brian M. Scott Sep 30 '12 at 21:50
$E$ is covered by the interval $[p-M,p+M]$, so $m^*(E)\le 2M$. –  Brian M. Scott Sep 30 '12 at 21:53

Let $p\in E$ be arbitrary, $I_1=[p-M,p+M]$. Then $E\subseteq I_1$, hence $m^*(E)\le l(I_1)$.