# Sum of Lebesgue integral (absolutely integrable functions)

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 – Epsilon Jan 11 '13 at 12:55

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 $$\int h^+ - \int h^- = \int f^++\int g^+-\int f^--\int g^- =\int f - \int g$$