I am reading Real Analysis by Yeh and have a question about the following result
(I cut off the case where $\mu_L(E)= \infty$ since I have no question about it).
Here are Yeh's notations: $\mathfrak{M}_L$ is the $\sigma$-algebra of Lebesgue measurable subsets of $\mathbb{R}$, $\mu_L$ denotes Lebesgue measure. If we're not sure $E \in \mathfrak{M}_L$, then $\mu_L^*(E)$ denotes the Lebesgue outer measure of E.
Question: Where is the fact that $E \in \mathfrak{M}_L$ used? In other words, does the above result hold for arbitrary $E \subseteq \mathbb{R}$ (and using outer measure $\mu_L^*$ instead of $\mu_L$ where appropriate, of course)? Here are my observations:
- Note equation (1) holds for arbitrary $E \subseteq \mathbb{R}$ (Lemma 3.21 found earlier in the book does not assume $E \in \mathfrak{M}_L$)
- $E \in \mathfrak{M}_L$ is seemingly used to show $\mu_L(E)= \sum_{n \in \mathbb{N}} \mu_L(E \cap I_n)$ right before equation (2), but even if $E$ were not measurable, the fact that outer measure is still countably subadditive gives us $\mu_L^*(E) \leq \sum_{n \in \mathbb{N}} \mu_L^*(E \cap I_n)$, and this is actually all that is needed to derive equation (2) (again, replacing $\mu_L$ with $\mu_L^*$).
Thanks a lot for your consideration
You are correct. However, if we no longer assume $E$ is measurable then we have to use the outer measure (as you pointed out) and in general we much prefer to deal with measures. The outer measure is a useful tool for constructing measures, but once we have succeeded in doing that we tend to discard it.