On page 29 of Folland's real analysis book one can read the following proposition:
Proposition 1.10. Let $\mathcal{E} \subset \mathcal{P}(X)$ and $\rho: \mathcal{E} \mapsto [0,\infty]$ be such that $\emptyset \in \mathcal{E}, X \in \mathcal{E}$, and $\rho(\emptyset)=0$.
For any $A \subset X$, define $$\mu^*(A)=\inf \bigg\lbrace \sum_{i=1}^{\infty} \rho(E_i):\ \{E_i\}_{i=1}^\infty\subset\mathcal{E}, \,\, A \subset \bigcup_{i=1}^{\infty}E_i \bigg\rbrace$$ Then $\mu^*$ is an outer measure.
I don't see why the condition $X\in\mathcal{E}$ is necessary.
Folland writes: "for any $A \subset X$ there exists $\{E_i\}_{i=1}^\infty\subset\mathcal{E}$ such that $A \subset \bigcup_{i=1}^{\infty}E_i$ (Take $E_i=X$ for all $i$) so the definition of $\mu^*$ makes sense."
But if there is no such cover then by definition $\mu^*(A)=\infty.$
Also it seems to me that the properties of an outer measure remain satisfied without this condition. For example in proving countable subadditivity $\mu^*(\bigcup_{i=1}^{\infty}A_i)\leq \sum_{i=1}^\infty \mu^*(A_i)$ we only need to consider the case where $\mu^*(A_i)<\infty$ for each $i$, and then the same argument as in the book goes through.
Am I missing something?
Best Answer
I believe this boils down to the author's preferences. I imagine Folland includes the condition $X \in \mathcal{E}$ to avoid having to explicitly state the convention that $\inf \emptyset := \infty$. Here is some brief justification for this thinking. For clarity, we will say Proposition 1.10$\star$ to refer to Proposition 1.10 with the condition $X \in \mathcal{E}$ replaced by the convention $\inf \emptyset := \infty$ (as you suggest).
Suppose $\mathcal{E}, \rho$ satisfy all the conditions for Proposition 1.10$\star$ where $X \notin \mathcal{E}$ and $\rho(X)$ is not defined. The outer measure obtained by invoking Proposition 1.10$\star$ on $\mathcal{E}, \rho$ coincides with the outer measure obtained by invoking Proposition 1.10 on $\mathcal{E'} := \mathcal{E} \cup \{X\}$ and $$\rho'(A) := \begin{cases} \infty & \text{if } A =X \\ \rho(A) & \text{if } A \in \mathcal{E}. \end{cases}$$
Conversely, suppose $\mathcal{E}, \rho$ satisfy all the conditions for Proposition 1.10 meaning $X \in \mathcal{E}$ and $\rho(X)$ is defined. The outer measure obtained by invoking Proposition 1.10$\star$ on $\mathcal{E}, \rho$ clearly coincides with the outer measure obtained by invoking Proposition 1.10 on $\mathcal{E}, \rho$.
So, ultimately, the author had a stylistic choice to make and decided to go with Proposition 1.10 as you have stated it, requiring $X \in \mathcal{E}$.