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

alt text http://i43.tinypic.com/144ck1w.png

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): measurableMeasurable 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!

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

alt text http://i43.tinypic.com/144ck1w.png

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!

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

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.