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