Suppose \(X\) is a topological space, \(\mathcal{B}_X\) its Borel-\(\sigma\)-algebra, \(\mu\) some measure on \(\mathcal{B}_X\) and \(f\colon (X,\mathcal{B}_X,\mu)\to \mathbb{C}\) is a continuous function.
Assume there are constants \(C>0\) and \(\alpha>0\) such that for every measurable \(S\subseteq X\) with nonzero measure we have
\[ C \mu(S)^{\alpha} \le \sup_{x\in S} \lvert f(x)\rvert. \]Then for every measurable \(S\subseteq X\) with \(\mu(S)>0\) a similar \(L^p\) estimate hold, i.e. for \(p\ge 1\) we have
\[ C\biggl(\frac{\mu(S)}{2}\biggr)^{\alpha+\frac{1}{p}}\le \lVert f\rVert_{L^p(S)}. \]Consider \(M=\Bigl\{x\in X \mid \lvert f(x)\rvert < C\Bigl(\frac{\mu(S)}{2}\Bigr)^\alpha\Bigr\}\). Since \(f\) is continuous , \(M\) is open and therefore measurable. If it is non empty, we are allowed to use the assumption and thus
\[ C\mu(M)^\alpha\le \sup_{x\in M} \lvert f(x)\rvert \le C\biggl(\frac{\mu(S)}{2}\biggr)^\alpha.\]This yields \(\mu(M)\le \frac{\mu(S)}{2}\) since \(C\) and \(\alpha\) are positive. If \(\mu(M)=\emptyset\), this estimate is still true. For the complement we then obtain
\[ \mu(M^c\cap S)=\biggl\{x\in S\mid \lvert f(x)\rvert\ge C\biggl(\frac{\mu(S)}{2}\biggr)^\alpha\biggr\}\ge \frac{\mu(S)}{2}. \]Finally, this yields
\[ \int_{S} \lvert f\rvert^p \ge \int_{X} 𝟙_{M^c\cap S}\lvert f\rvert^p \ge \mu(M^c\cap S) C^p\biggl(\frac{\mu(S)}{2}\biggr)^{\alpha p} \ge C^p \biggl(\frac{\mu(S)}{2}\biggr)^{\alpha p + 1}. \]