These questions comes from theorem 19.C, page 81-82, in Halmos' "Measure Theory", as the image below shows.

Question 1): The 4th line of the proof says "we restrict our attention to finite valued functions" and the proof is carried out for finite *f* and *g*. Why can we "restrict our attention to finite valued functions"? How to extend the conclusion from finite case to extended real valued case?

Question 2): measurable functions, by definition in page 76-77, are defined on the whole *X*. But it is possible that *f* + *g* has no definition at some point *x* of *X*, e.g. $f(x)=+\infty$ but $g(x)=-\infty$. The product *fg* has the same situation. Of course we can assume in advance that *f* + *g* must be defined on the whole *X* in order for the theorem to hold, but violation of such an assumption occurs immediately: in the equation in the last line of the proof, $fg=[(f+g)^2-(f-g)^2]/4$, even if we assume *f* + *g* is defined on the whole *X*, we cannot guarantee that *f* - *g* and $(f+g)^2-(f-g)^2$ is meaningful for all *x* in *X*; in the paragraph that follows the proof, $f^+=f\cup 0=(f+|f|)/2$, if $f(x)=-\infty$ for some *x*, we cannot apply the theorem to obtain the measurability of (*f*+|*f*|)/2 and in turn of $f^+$ because *f*+|*f*| is not definable on whole *X*. So we have to allow that *f* + *g* (and *fg*) has domain smaller than *X*, but this violates the definition of measurable function in page 76-77. What, then, does the conclusion of the theorem "so also are *f* + *g* and *fg*" mean exactly on earth? and how to apply it to $fg=[(f+g)^2-(f-g)^2]/4$ and $f^+=(f+|f|)/2$?

Thanks!