If $f \in L^1$ is bounded $(0 \leq f \leq 1)$ and has compact support then $f^p\in L^1$ for $p\in(0,1)$

Question

According to my textbook, the statement

If $f \in L^1$ is bounded $(0 \leq f \leq 1)$ and has compact support then $f^p\in L^1$ for $p\in(0,1)$

obviously follows from the theorem

$f \in L^1$ iff for all $\varepsilon>0$ there is $g\in C_c$ with $\| f-g \|_{L^1} < \varepsilon$

Sorry if the statement looks a little bit unmotivated, it's a part of a larger proof I'm trying to understand.

I've noticed that $\| f-f^p \|_{L^1} \leq 1$ and tried the usual things with triangle inequality yet it doesn't seem to work. Any hints and appreciated.

Answer 1

The use of the theorem is unnecessary - if f is bounded with compact support then $f^p$ is, so it's in $L^1$.