# Show that Lebesgue outer measure is the infimum of Lebesgue measure

Let $$\lambda^*(S) = \inf\left\{\sum_{j=0}^\infty (b_j - a_j)\mid a_j, b_j \in \mathbb{R},S \subseteq \bigcup_{j=0}^\infty [a_j,b_j[ \right\}$$ be the Lebesgue outer measure. Using Carathedory's method, the Lebesgue measure and $\lambda$ and Lebesgue measurable sets $\Sigma \subseteq \mathcal{P}(\mathbb{R})$ can be defined as $$\Sigma = \{S\mid \forall U \subseteq \mathbb{R}, \lambda^*(U \cap S) + \lambda^*(U \setminus S) = \lambda^*(U)\}, \lambda = \lambda^*\restriction_\Sigma$$

Theorem: For every $S \subseteq \mathbb{R}$, $$\lambda^*(S) = \inf\{\lambda(U)\mid U \in \Sigma, S \subseteq U\}$$

How to prove this theorem? The question hints to consider sets of the form $$U = \bigcup_{j=0}^\infty [a_j,b_j[$$ They are certainly measurable, but how can the infimum be attained?

• What is $\mu$? Should that be $\lambda$? Dec 22, 2015 at 4:20
• @user295959 Sorry that was a typo. It is fixed. Dec 22, 2015 at 4:21

This is almost just the definition if you've already proven that unions of half-open intervals are measurable. Given an $S$, the definition of $\lambda^*$ gives the existence of covers by half-open intervals that approximate arbitrarily well from above.
Then it just remains to show that there is no union by half-open intervals $[a_i, b_i)$ so that
$$S \subseteq \bigcup_{i = 1}^{\infty} [a_i, b_i)$$ yet $$\sum_{i = 1}^{\infty} (b_i - a_i) < \lambda^*(S)$$ This is an immediate consequence of the monotonicity of $\lambda^*$.