If $f \in L^p$ then $\lim_{x\rightarrow \infty} x R_p(x) =0$

Question

I am stuck on the following problem:

Let $p \in [0, \infty)$ and let $f$ be a real valued measurable function. Define $R_p(x) = x^{p-1}\mu(\{|f|>x\})$, $x>0$. Prove if $f \in L^p(\mu)$ then $\lim_{x\rightarrow \infty} x R_p(x) =0$.

I know by Tchebyshev's inequality that $x^p \mu(\{|f|>x\}) \leq \int|f|^p <\infty$ (where the last inequality follows since $f\in L^p(\mu)$) but I am not sure where to go from here. Any pointers in the right direction is appreciated.

Answer 1

By Markov-Chebyshev's inequality:

$$\lambda R_p(\lambda)=\lambda^p\mu(|f|>\lambda)\leq\int_{\{|f|>\lambda\}}|f|^p\xrightarrow{\lambda\rightarrow\infty}0$$ where the last conclusion follows by dominated convergence.