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This is Tao Analysis II Prop. 19.3.3. (b)

Let $\Omega \subseteq \mathbb R^n$ measurable and $f,g: \Omega \rightarrow \mathbb R$ absolutely integrable functions. Then $f+g$ is absolutely integrable and $$ \int_\Omega f+g = \int_\Omega f + \int_\Omega g $$

How can I prove that ?

My first idea was $f+g = f^+ + g^+ - (f^- + g^-)$. But then I get a problem with "$-$"-sign.

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well, you can show that $(f+g)^+\leq f^+ + g^+$ and the same for the negative parts - just using the properties of the $\max$ function. That gives you the integrability. –  Ilya Jan 11 '13 at 12:53
    
Yes. This was the easy part :D –  André Jan 11 '13 at 12:55
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1 Answer

up vote 1 down vote accepted

Copied from Folland, Proposition 2.21. Let $h = f+g$, then we know that $$ h^+ - h^- = f^++g^+-(f^-+g^-) $$ and by regrouping $$ h^+ + f^-+g^- = h^- + f^++g^+. $$ Since all functions are positive, $$ \int h^+ + \int f^-+\int g^- = \int h^- + \int f^++\int g^+ $$ and thus $$ \int h^+ - \int h^- = \int f^++\int g^+-\int f^--\int g^- =\int f - \int g $$

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In other words, you should first prove your result for nonnegative functions. –  GEdgar Jan 12 '13 at 13:05
    
@GEdgar: for them it follows directly from the linearity for simple functions and the definition of the integral. It seems from OP, that OP doesn't have this problem. –  Ilya Jan 12 '13 at 13:14
    
Thanks. Quite simple indeed. –  André Jan 12 '13 at 13:32
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