Suppose that $f$ is non-negative and measurable on $E$ and that $\omega$ is finite on $(0,\infty)$.
If $\int_{0}^{\infty}\alpha^{p-1}\omega(\alpha)d\alpha$ is finite , show that $\lim_{a\rightarrow0+}a^{p}\omega(a)=\lim_{b\rightarrow\infty}b^{p}\omega(b)=0$,
where $p\in(0,\infty)$,and $\omega(\alpha)=\mu\{x\in E :f(x)>\alpha\},\mu$ is the Lebesgue measure.
Here's my working :
By assumption, $\int_{0}^{\infty}\alpha^{p-1}\omega(\alpha)d\alpha$ is finite.
For each $\epsilon>0$ there exists a small enough $a\in\mathbb{R_{>0}}$ such that $\int_{0}^{a}\alpha^{p-1}\omega(\alpha)d\alpha<\epsilon$ .
But,$$\int_{a/2}^{a}\alpha^{p-1}\omega(\color{red}a)d\alpha\leq\int_{a/2}^{a}\alpha^{p-1}\omega(\alpha)d\alpha$$
and$$\int_{a/2}^{a}\alpha^{p-1}\omega(a)d\alpha=\omega(a)\int_{a/2}^{a}\alpha^{p-1}d\alpha=\omega(a)\bigg(p^{-1}\alpha^p|_{a/2}^{a} \bigg)=\dfrac{2^p-1}{p2^p}a^p\omega(a)$$
For then,$$a^p\omega(a)\leq\dfrac{p2^p}{2^p-1}\int_{0}^{a}\alpha^{p-1}\omega(\alpha)d\alpha\leq\dfrac{p2^p}{2^p-1}\epsilon$$
Whence,$$lim_{a\rightarrow0+}a^{p}\omega(a)\leq\dfrac{p2^p}{2^p-1}\epsilon$$
By the arbitrariness of $\epsilon>0$,one has $lim_{a\rightarrow0+}a^{p}\omega(a)=0$ since $\dfrac{p2^p}{2^p-1}$ is a finite number
I think the other part is similar to the way , i.e. $lim_{b\rightarrow\infty}b^{p}\omega(b)=0$
Anyone can help me to check this? Maybe I missed something.Thanks a lot.
Generally we don't have $\alpha^{p-1}\geq(\alpha/2)^{p-1}$ because $p$ could be strictly less than $1$. My proof goes like this:
\begin{align*} b^{p}\omega(b)&\leq\int_{f(x)>b}f^{p}(x)d\mu(x)\\ &=\int\chi_{f(x)>b}(x)\int_{0}^{\infty}\chi_{t<f(x)}(t)pt^{p-1}dtd\mu(x)\\ &=p\int_{b}^{\infty}\int\chi_{f(x)>t}(x)d\mu(x)t^{p-1}dt\\ &=p\left(\int_{0}^{\infty}\omega(t)t^{p-1}dt-\int_{0}^{b}\omega(t)t^{p-1}dt\right)\\ &\rightarrow 0 \end{align*} as $b\rightarrow\infty$ by Monotone Convergence Theorem.
Meanwhile,
\begin{align*} a^{p}\omega(a)&=p\int_{0}^{a}t^{p-1}\omega(a)dt\\ &\leq p\int_{0}^{a}t^{p-1}\omega(t)dt\\ &\rightarrow 0 \end{align*} as $a\rightarrow 0^{+}$ by Lebesgue Dominated Convergence Theorem.