Outer Measure: Theorem) for an interval $I$, $|I|_e=v(I)$?

Question

I do not understand a theorem. So can someone explain for me?

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Definition

1. closed n-dimensional intervals are $$I=\{\mathbf{x}:a_j\le x_j\le b_j,~~j=1, \cdots, n\}$$
2. their volumes are $$v(I)=\prod_{j=1}^n(b_j-a_j)$$
3. A countable collection $S$ of intervals $I_k$ is $$S=\{I_1, I_2, I_3, \cdots\}$$

4. Then, $S$ is called a covering of $E$ if $$\cup_{I_j\in S}\supset E$$

5. If $\sigma(S)=\sum_{I_k\in S} v(I_k)$, then the Lebesgue outer measure of $E$ is$$ |E|_e=\inf\sigma(S)$$

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Theorem

For an interval $I$, $|I|_e=v(I)$.

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Proof

1. Since $I$ covers itself, $|I|_e \le v(I)$.
2. Let $I_k^*$ be an interval containing $I_k$ in its interior such that $$v(I_k^*)\le(1+\varepsilon)v(I_k)$$
3. Then, $$I \subset \cup_k(I_k^*)$$
4. Since $I$ is compact, there exists $N$ such that $$I\subset \cup_{k=1}^N I_k^*$$ by the Heine-Borel theorem.
5. Clearly, $$v(I)\le \sum_{k=1}^N v(I_k^*)$$
6. Hence, $$v(I)\le (1+\varepsilon)\sum_{k=1}^Nv(I_k)\le(1+\varepsilon)\sigma(S)$$
7. Since $\varepsilon$ can be chosen arbitrarily small, it follows $$v(I)\le\sigma(S) \quad \text{ and } \quad v(I)\le|I|_e$$
8. By step 1 and 7, $$v(I)\le|I|_e\le v(I)\\\therefore v(I)=|I|_e$$

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My problem

I do not understand step 1 and step 4.

Can someone explain about why $|I|_e \le v(I)$ and why $I$ is compact

Answer 1

$I$ is compact since it is a $k$-cell in $R^n$. This result is often called the Heine-Borel theorem (as mentioned in your post); you may find a proof of this in many online lecture notes or textbooks.

The definition of outer measure says this: if you can find a covering of the set $E$ with some measure $m$, the outer measure of $E$ is necessarily $\leq m$. This is because it is defined as the infimum of all the measures of the coverings.