Possible typo in Bogachev's Measure Theory I've a problem with the definition of a measure space in Bogachev's Measure Theory. The author (Definition 1.3.2, p. 9) assumes a measure to be a countably additive set function defined on an algebra of sets (Definition 1.2.1, p. 3), and let a countable additive set function be a real-valued set function (for which the usual conditions hold). But (see the top of p. 9) a real-valued function means, in the terminology of Bogachev, a function with values in the open interval $(-\infty, +\infty)$, so in particular a measure is not allowed to take the value $+\infty$. This looks like a typo to me, so I'd like to know (i) if I am missing something obvious and (ii) if there is an errata corrige for the book somewhere.
Edit. I'm all the more convinced that it is a typo because later on, in Example 4.7.89 (p. 311), the author mentions the counting measure, which he defines, as usual, as the cardinality of the measured set. But I'd really like to hear from somebody else.
Thanks.
 A: Sorry, I'm going to reply my own question, because I've just realized that Bogachev distinguishes between finite measures, which he refers to simply as measures, and infinite measures, which are introduced later in Section 1.6. So, there is no typo at all in his definition(s) of (a) measure(s).
A: Your suspicions are correct. It's supposed to be extended real-valued function (i.e. from sets to $\mathbb{R} \cup \{-\infty\} \cup \{+\infty\}$). Here is how a finitely additive set function is defined in the textbook I learned from (Dudley, Real Analysis and Probability):

A function $\mu$ from $\mathcal{C}$ into $[-\infty,\infty]$ is said to be finitely additive iff $\mu(\emptyset)=0$ and whenever $A_i$ are disjoint, $A_i \in \mathcal{C}$ for $i=1,\ldots,n$ and...

There is a theorem about countably additive set functions that requires the function to be finite-valued though. Maybe you can think of a counterexample to illustrate why:

Let $\mu$ be a finitely additive, real-valued function on an algebra $\mathcal{A}$. Then $\mu$ is countably additive if and only if $\mu$ is continuous at $\emptyset$.

