How to show that if $\int_a^b|{f(x)}|^p\,d\alpha = 0$ then $\int_a^b|{f(x)}|\,d\alpha = 0$ How can I prove that if $\int_a^b|{f(x)}|^p\,d\alpha = 0$ then $\int_a^b|{f(x)}|\,d\alpha = 0$? (Assuming that $\alpha$ is increasing and $p > 1$). Intuitively this is clear but I'm having trouble proving it rigorously.

Edit: Here is another proof without using Holder's inequality. 
First we prove that if 
$$
\int_A|{g(x)}|\:d\mu=0
$$
then $g(x)=0$ almost everywhere on $A$. 
Let $B_n=\{x:x\in A, \:|g(x)|\geqslant 1/n\}$. Then by Chebychev's inequality
$$
\mu(B_n)\leqslant n\int_A|{g(x)}|\:d\mu=0
$$
Hence
$$
\mu(\{x:x\in A, \:|g(x)|\ne0\})=\mu\left(\bigcup_{n=1}^{\infty}B_n\right)\leqslant \sum_{n=1}^{\infty}\mu(B_n)=0
$$
i.e. $g(x)=0$ almost everywhere on $A$. 
Now since
$$
\int_a^b|{f(x)}|^p\,d\alpha=0
$$
$f^p(x)=0$ a.e. on $[a,b]$ and so $f(x)=0$ a.e. on $[a,b]$. Thus 
$$
\int_a^b|{f(x)}|\,d\alpha=0\tag*{$\blacksquare$}
$$
This is a hint that use Holder's inequality. 
$$
\int_a^b|{f(x)}|\cdot 1\,d\alpha\leqslant \left(\int_a^b|{f(x)}|^p\,d\alpha\right)^{1/p}\left(\int_a^b1\,d\alpha\right)^{1-1/p}=0
$$
