I'm reading the book Real Analysis from Elias.M.Stein and Rami Shakarchi.
In the chapter about abstract integration, when the book starts to construct the integral, it limits itself to $\sigma$-finite spaces. I've been told that I can define the integral in general space without any problem but I am struggling to get a result that is obvious if one has a $\sigma$-finite space, maybe here is not true.
This book starts defining the integral for simple functions, meaning lineal combinations of characteristic of measurable sets of finite measure, putting $$\int \sum_{1}^nc_i\chi_{E_i} \ d\mu = \sum_1^nc_i\mu(E_i)$$
After that it basically define the integral of a positive function as $$\int f \ d\mu = sup\{\int g, \ 0\leq g\leq f\}$$ with $g$ simple. At this point one can observe that the first step in the definition, i.e, the definition for simple functions does not tell someone the value of $$\int \chi_E \ d\mu$$ when $\mu(E) = \infty$. However, once one has the second defintion, if the space is $\sigma$-finite, the next reasoning proves that the integral above has the value $\infty$ as one may expect; As the space is $\sigma$-finite, $X = \cup_{i=1}^{\infty}E_i$ with $E_i$ of finite measure, then $E = \cup_{i=1}^{\infty}(E\cap E_i)$. Let me call $B_n = \cup_{i=1}^{n}(E\cap E_i)$. Now by the second definition, or by monotony, $$\int \chi_E \ d\mu \geq \int \chi_{B_n} \ d\mu = \mu(B_n)$$ where the last equal follows from definition for simple functions, as $B_n$ has finite measure. To finish if one takes $n \to \infty$, by the continuity of the measure one has $\mu(B_n) \to \mu(E) = \infty$ and then $$\int \chi_E \ d\mu = \infty$$
This seems really clear but I am getting confused to get the same result if one doesn't have a $\sigma$-finite spaces. It is clear that I can't use the same reasoning.
One important observation is that I have seen books where admit sets of infinite measure in the first definition of integral, but this is not the case in this book. (Maybe because they restrict to $\sigma$-finite spaces).