If $f\in L^1$ has a compact support and $0 \leq p \leq1$ then $|f|^p\in L^1$ My text proved that

If $f\in L^1$ is bounded and $p \geq1$ then $|f|^p\in L^1$

I wanted to prove the seemingly very similar statement:

If $f\in L^1$ has a compact support and $0 \leq p \leq1$ then $|f|^p\in L^1$

I was able to mimic the proof from the text yet I guess one can use the first statement to prove the second. Any idea how?
 A: Let $B$ be any bounded set that contains the support of $f$. By Jensen's Inequality,
$$
\frac1{|B|}\int_B|f(x)|^p\mathrm{d}x\le\left(\frac1{|B|}\int_B|f(x)|\mathrm{d}x\right)^{\large p}
$$
since $|x|^p$ is concave for $0\le p\le1$.
A: They are quite different; the first I imagine you used that $|f|^p \leq |f| M^{1-p}$ where $M$ is the bound for $f$. 
You'll want to bound the $L^1$ norm of $|f|^p$ by the $L^1$ norm of $f$ multiplied by something to do with the measure of it's support; the following inequality may come in useful: for $1\leq p,q\leq \infty$ with $1/p + 1/q=1$, $\int fg \leq (\int |f|^p)^{1/p}(\int |g|^q)^{1/q}$. (Notice if you choose $g$ to be the characteristic function of the support of $f$, the second term on the RHS will be the measure of the support of $f$ to a power)
A: Let
\begin{align*}
A\equiv&\,\{x\in \mathbb R^n\,|\,0<|f(x)|\leq 1\},\\
B\equiv&\,\{x\in \mathbb R^n\,|\,|f(x)|> 1\}.
\end{align*}
Note that both sets are contained in the support of $f$ and since $f$ is compactly supported, it follows that $A$ and $B$ are bounded sets. In particular, $m(A)<\infty$ and $m(B)<\infty$, where $m$ is the Lebesgue measure on $\mathbb R^n$.
Now, let $p\in[0,1]$. Then,
\begin{align*}
\int_{\mathbb R^n}|f(x)|^p\,\mathrm d x=&\,\int_A|f(x)|^p\,\mathrm d x+\int_B|f(x)|^p\,\mathrm d x\\
\leq&\,\int_A|f(x)|^p\,\mathrm d x+\int_{B}|f(x)|\,\mathrm d x\\
\leq&\,\int_A|f(x)|^p\,\mathrm d x+\int_{\mathbb R^n}|f(x)|\,\mathrm d x\\
\leq&\,\int_A\,\mathrm d x+\int_{\mathbb R^n}|f(x)|\,\mathrm d x\\
=&\,m(A)+\int_{\mathbb R^n}|f(x)|\,\mathrm d x<\infty,
\end{align*}
given that $f\in L^1$.
