Without measure, need $\int_a^b |f|^pd\alpha =0\Rightarrow \int_a^b |f|d\alpha=0$

Question

As in the title, I want to show $\int_a^b |f|^pd\alpha =0\Rightarrow \int_a^b |f|d\alpha=0$

With measure theory, this implication is obvious. But using the definition, as a Riemann Integral, seems impossible to me right now. This material was so long ago for me. Could someone get me started?

EDIT: sorry, important details: $f$ is continuous and $p>1$.

Answer 1

This is a consequence of the more general result:

If $f\colon[a,b]\to\Bbb R$ is continuous, $f\ge0$ and $\int_a^bf=0$, then $f(x)=0$ for all $x\in[a,b]$.

Proof. Suppose that $f(\xi)\ne0$ for some $\xi\in[a,b]$ and show (using the fact that $f$ is continuous at $\xi$) that then $\int_a^bf>0$.