$\int^1_0 g(x) ^p-f(x)^p \,d\mu \ge 0 \Rightarrow g(x) >f(x)$?

Question

Let $\left([0,1],\mathcal{L},\mu\right)$ be a Lebesgue measure space, and $f,g$ are positive measurable function.

If $\int^1_0 g ^p-f^p \,d\mu \ge 0$ for all $p\ge 1$, then $ g(x) \ge f(x),\,\mu$ - a.e

Is this statement correct?

Answer 1

No. Try $f(x) = 1$ while $g(x) = 0$ on $[0,1/2)$, $2$ on $[1/2, 1]$. If you don't want to allow values of $0$, replace $0$ by $\epsilon$ for some $\epsilon \in (0,1)$.