Integral over $\mathbb R$ may be expressed as the sum of integrals over disjoint measurable subsets.

Question

I'm not sure how to prove the following:

If $E$ and $F$ are disjoint measurable subsets of $\mathbb{R}$ such that $\mathbb{R} = E \cup F$

Show that $\displaystyle \int_{\mathbb{R}} f(x) \, dm(x) = \int_E f(x) \, dm(x) + \int_F f(x) \, dm(x)$

I'm trying some solutions using the decomposition $f=f^+ - f^-$, or the fact that $\chi_{\mathbb{R}} = \chi_E + \chi_F$ but am really not sure how to go about this question.

Help is much appreciated, thanks.

Answer 1

$\displaystyle\int_{E}f(x)dm(x)=\int_{\bf{R}}f(x)\chi_{E}(x)dm(x)$, similar for the $F$. Now $\displaystyle\int_{\bf{R}}f(x)dm(x)=\displaystyle\int_{\bf{R}}f(x)(\chi_{E}+\chi_{F})dm(x)=\displaystyle\int_{\bf{R}}f(x)\chi_{E}dm(x)+\displaystyle\int_{\bf{R}}f(x)\chi_{F}dm(x)=\displaystyle\int_{E}f(x)dm(x)+\displaystyle\int_{F}f(x)dm(x)$.