1. ## measure theory intergration

The question is:
Show that if $\displaystyle g:\mathbb{R} \longrightarrow [0,\infty)$ equals 0 outside a bounded interval and $\displaystyle \int g^2 < \infty$ , then $\displaystyle \int g < \infty$

It doesn't appear so difficult, but i dont know how to progress in it. I say let I be the bounded interval, then $\displaystyle \int g^2 = \int_{\mathbb{R}} g^2 \chi_I < \infty$ how do i progress from here? i dont know how to get $\displaystyle \int g$ from this! thanks

2. Divide $\displaystyle I=A\cup B$, $\displaystyle A=\{x\in I: |f(x)|\leq 1\}$ and $\displaystyle B:=I\setminus A$. Both sets are measurable. Show that $\displaystyle f$ is integrable both in $\displaystyle A$ and $\displaystyle B$ because it is dominated by the identically 1 function in $\displaystyle A$ (which is of finite measure) and by $\displaystyle f^2$ in $\displaystyle B$, integrable by hypothesis.

3. Originally Posted by Enrique2
Divide $\displaystyle I=A\cup B$, $\displaystyle A=\{x\in I: |f(x)|\leq 1\}$ and $\displaystyle B:=I\setminus A$. Both sets are measurable. Show that $\displaystyle f$ is integrable both in $\displaystyle A$ and $\displaystyle B$ because it is dominated by the identically 1 function in $\displaystyle A$ (which is of finite measure) and by $\displaystyle f^2$ in $\displaystyle B$, integrable by hypothesis.
Thanks a lot!!!!!!! I wouldn't have ever thought of dividing I!!!