Let $p: [a, b] \longrightarrow \mathbb{R}$ integrable and $p(x) \geq 0$ for all $x \in [a, b]$. Prove that if $\displaystyle \int\limits_{a}^{b}p(x)dx = 0$, then the set of points $x \in [a,b]$ such that $p(x) = 0$ is dense in $[a, b]$.
I proved in a previous question that $``\varphi: [a,b] \longrightarrow \mathbb{R}$ is a integrable and positive function, then $\int\limits_{a}^{b}\varphi(x)dx > 0"$.
$\textbf{My idea:}$ Let $X$ the set of points $x \in [a,b]$ such that $p(x) = 0$. If $X$ not is dense in $[a, b]$, for all partitions of $[a, b]$, there exist a interval $[a_{P}, b_{P}]$ such that $p(x) > 0$ for all $x \in [a_{P}, b_{P}]$. By the previous question, $\displaystyle \int\limits_{a_{P}}^{b_{P}}p(x)dx > 0$. Therefore, $\displaystyle \int\limits_{a}^{b}p(x)dx > 0$. Contradiction!
Is a correct ideia?