Suppose that we already know $A \subseteq B \implies \lambda^*(A) \le \lambda^*(B)$ for $A, B \in \mathbb{R}$.
Then, consider a closed interval $[a, b]$ with $a < b$. Then there exists some $\epsilon > 0$ such that $(a + \epsilon, b – \epsilon) \subseteq [a, b] \subseteq (a – \epsilon, b + \epsilon)$ and then $b – a – 2\epsilon = \lambda^*(a + \epsilon, b – \epsilon) \le \lambda^*([a, b]) \le \lambda^*(a – \epsilon, b + \epsilon) = b – a + 2\epsilon$ and so taking the supremum on the left and infimum on the right we should get $\lambda([a, b]) = b – a$.
I don't see why this wouldn't be valid, but the book I'm looking at warns the reader against a simple proof of this fact and then launches into a construction directly from the definition of $\lambda^*$ via the Heine-Borel theorem.
Is my proof correct or am I just blind to something obvious?
Best Answer
The proof provided is valid if it is already known that $\lambda^*(a, b) = b - a$. However, my confusion simply stemmed from assuming that given the definition of the length of an interval as $\ell(a, b) = b - a$ that it was also immediate for $\lambda^*$, which is not the case. For completeness:
$$\lambda^*(A) = \text{inf}\{\sum_{k = 1}^\infty \ell(I_k)\ |\ I_1, ... \text{are open intervals s.t.}\ A \subseteq\bigcup_{k = 1}^\infty I_k\},$$
and where $A$ is any subset of $\mathbb{R}$.