The Lebesgue outer measure on $\mathbb{R}$ is defined as:

$\lambda^{*}(A)$ = $inf${$\sum_{n=1}^{\infty}(b_{n}-a_{n}): A \subset \bigcup_{n=1}^{\infty}(a_{n}, b_{n}) $}

I want to show that $\lambda^{*}([a,b]) = b-a$. Although I have proofs for this lemma, they are incomplete and jump from one statement to another without justification. So would anyone be kind enough to post a proof which is relatively simple?


Here's a sketch that leaves (straightforward) details to fill in.

First to work with closed intervals: Compactness guarantees that every covering by open intervals may be replaced by a finite covering $(a_{i}, b_{i})$, $i = 1, \dots, n$.

If necessary, re-order your intervals (possibly discarding some of them) so that $a \in (a_{1}, b_{1})$, $b_{1} \in (a_{2}, b_{2})$, ..., $b \in (a_{m}, b_{m})$. (Consequently, $a < a_{1}$, $a_{2} < b_{1}$, ..., $a_{m} < b_{m-1}$, and $b < b_{m}$.) The sum of the lengths of the covering intervals clearly exceeds $b - a$. (That is, a formal estimate is easy, thanks to telescoping sums and the inequalities in the preceding parentheses. A diagram should help organize things.)

To handle open intervals, note that $[a + \varepsilon, b - \varepsilon] \subset (a, b)$ for all $\varepsilon > 0$, so by the preceding paragraph, $b - a - 2\varepsilon \leq \lambda^{*}(a, b)$ for all $\varepsilon > 0$. (The inequality $\lambda^{*}(a, b) \leq b - a$ is obvious.)


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