If $f$ is non-negative and $f$ is in $L^{p}(\mathbb{R^{n}})$ then $\sum_{k=-\infty}^{k=\infty}2^{kp}\omega(2^{k})<\infty$

Question

If $f$ is non-negative and $f$ is in $L^{p}(\mathbb{R^{n}})$ then $$\sum_{k=-\infty}^{k=\infty}2^{kp}\omega(2^{k})<\infty$$

,where $\omega(\alpha)$ is the Lebesgue measure of $\{x:f(x)>\alpha\}$ for all real $\alpha$.

Here is my attempt :

\begin{align} \sum_{k=-\infty}^{k=\infty}2^{kp}\omega(2^{k})&=\sum_{k=-\infty}^{k=\infty}\int_{\{x\,:f(x)>2^{k}\,\}}2^{kp}\,dx\\ &\le\sum_{k=-\infty}^{k=\infty}\int_{\{x\,:f(x)>2^{k}\,\}}f^{p}(x)\,dx\\ &=\sum_{k=-\infty}^{k=\infty}\int_{\{x\,:f(x)>2^{k}\,\}}\int_{0}^{f(x)}p\alpha^{p-1}d\alpha dx \\ &=\sum_{k=-\infty}^{k=\infty}\int_{0}^{f(x)}\int_{\{x\,:f(x)>2^{k}\,\}}p\alpha^{p-1}dxd\alpha\\ &=\sum_{k=-\infty}^{k=\infty}\int_{0}^{f(x)}\omega(2^{k})p\alpha^{p-1}d\alpha\\ &=\sum_{k=-\infty}^{k=\infty}\omega(2^k)\int_{0}^{f(x)}p\alpha^{p-1}d\alpha\\ &=|E\,|\int_{0}^{f(x)}p\alpha^{p-1}d\alpha\\ &=\int_{E}\int_{0}^{f(x)}p\alpha^{p-1}d\alpha dx\\ &=\int_{E} f^{p}(x)\,dx=\lVert f\rVert_{L^{p}(\mathbb{R^{n}}\,)}^{p}<\infty \end{align}

,where the third equality holds by non-negative version of Fubini theorem and $|E\,|$ is the Lebesgue measure of $E$.

If have the time, please stop for a moment and check my working for validity. any valuable suggestion will be greatest appreciation.Lots of thanks.

Answer 1

\begin{align*} \sum_{k=-\infty}^{\infty}2^{kp}\omega(2^{k})&=\sum_{k=-\infty}^{\infty}2^{kp}\sum_{m=k}^{\infty}|\{2^{m}<f(x)\leq 2^{m+1}\}|\\ &=\sum_{m=-\infty}^{\infty}\sum_{k=-\infty}^{m}2^{kp}|\{2^{m}<f(x)\leq 2^{m+1}\}|\\ &=\sum_{m=-\infty}^{\infty}|\{2^{m}<f(x)\leq 2^{m+1}\}|\dfrac{2^{mp}}{1-2^{-p}}\\ &=\dfrac{1}{1-2^{-p}}\sum_{m=-\infty}^{\infty}\int_{\{2^{m}<f(x)\leq 2^{m+1}\}}2^{mp}dx\\ &\leq\dfrac{1}{1-2^{-p}}\sum_{m=-\infty}^{\infty}\int_{\{2^{m}<f(x)\leq 2^{m+1}\}}f(x)^{p}dx\\ &=\dfrac{1}{1-2^{-p}}\int_{{\bf{R}}^{n}}f(x)^{p}dx\\ &=\dfrac{1}{1-2^{-p}}\|f\|_{L^{p}({\bf{R}}^{n})}^{p}\\ &<\infty. \end{align*}