Let $f$ and $g$ be two positive measurable functions such that $$\mu(f\geq\epsilon)\leq\frac{1}{\epsilon}\int_{\{f\geq\epsilon\}}g\,d\mu$$ for all $\epsilon\geq0$. Show that for any $1<p<\infty$, $$\lVert f\rVert_p\leq\frac{p}{p-1}\lVert g\rVert_p$$
The first inequality looks like Markov inequality. Does anyone know how to prove it?
Let $\Omega$ denote the measure space, which we assume to be $\sigma$-finite. First, we observe that
$\displaystyle\|f\|_p^p = \int_{\Omega}f^pd\mu = \int_0^\infty x^p d(f_\ast\mu)(x) = \int_0^\infty \int_0^x pt^{p-1}~dt~d(f_\ast\mu)(x)$
$\displaystyle= \int_0^\infty\int_t^\infty pt^{p-1}~d(f_\ast\mu)(x)~dt=\int_0^\infty pt^{p-1}\int_t^\infty d(f_\ast\mu)(x) dt = \int_0^\infty pt^{p-1}\mu(\{f>t\})~dt$.
Using this and the given inequality, we obtain $\displaystyle\|f\|_p^p \leq \int_0^\infty\int_{\{f\geq t\}} pt^{p-2}g~d\mu~dt = \int_{\Omega}g(\omega)\int_0^{f(\omega)}pt^{p-2}~dt~d\mu(\omega)$
$\displaystyle = \int_{\Omega}\frac{p}{p-1}f(\omega)^{p-1}g(\omega)d\mu(\omega)$.
Now Hölder yields the claim.
Edit: As pointed out in the comments by John Dawkins, for this to work, we need that $f\in L^p$ (at least in the interesting case that $g\in L^p$). Otherwise $\Omega=[1,\infty)$ with Lebesgue measure, $p=2,~f\equiv 1, g(x)=\frac 1x$ provides a counterexample.