Prove that if $E\cap F=\emptyset$ and $X=E\cup F$ then $\int_Xfd\mu=\int_Efd\mu+\int_Ffd\mu$

Question

I did the proof though I didn't use the hypothesis of $X=E\cup F$. Also I didn't see the why of $f$ to be in $\overline{\mathbb R}.$ Could this be any $Y$ space?

Let $f:X\to\overline{\mathbb R}$ be measurable and nonnegative. Prove that if $E,F\in\mathcal A$ are such that $E\cap F=\emptyset$ and $X=E\cup F$ then $$\int_Xfd\mu=\int_Efd\mu+\int_Ffd\mu$$

Proof

$$\int_Efd\mu+\int_Ffd\mu$$ $$=\int_Xf\mathcal X_E d\mu+\int_Xf\mathcal X_Fd\mu$$ $$=\int_X(f)*(\mathcal X_E+\mathcal X_F)d\mu$$ $$=\int_Xfd\mu$$

where the first equality is by definition and the last equality because $E\cap F=\emptyset$ thus when $\mathcal X_E=1,\mathcal X_F=0$ and vice versa.

Is the proof correct?

Answer 1

The result is valid. Your proof is correct. The last equality to be true you need $E\cap F=\emptyset$ and $X=E\cup F$. Writing the last step in details:

$$\int_X(f)(\mathcal X_E+\mathcal X_F)d\mu$$ $$=\int_X(f)(\mathcal X_{E \cup F})d\mu $$ $$=\int_Xfd\mu$$

For the first equality you need $E\cap F=\emptyset$ and for the second one, $X=E\cup F$.