# Thread: measure theory intergration

1. ## measure theory intergration

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

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

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

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