Prove that $g^+= max(0,g(x))$ is integrable on $I$

Question

Let $I$ be a bounded and closed rectangle in $\mathbb{R}^3$. $g$ is integrable function on $I$.

Prove that $g^+= max(0,g(x))$ is integrable on $I$ and $|g|$ are integrable on $I$.

Answer 1

Divide $I$ into two regions: $I = I_+ \cup I_{-}$, where

$$I_+ = \{x \in I ~|~ g(x) \geq 0\} ~~~ \mathrm{and} ~~~ I_{-} = \{x \in I ~|~ g(x)<0\}$$

Since $g$ is integrable on $I$, it is integrable on both $I_+$ and $I_{-}$. And you can see that $\int_I g^+ \, dx = \int_{I_+} g \, dx$, because $g^+ = 0$ on $I_{-}$.

Also, $|g| = g^+ + (-g)^+$, so the fact that $|g|$ is integrable follows from above.