The problem from my real analysis coursework is:
Let $0<p<\infty$ prove there exists constants $c_1, c_2$ independent of $f$ such that $$ c_1\left(\sum_{k=-\infty}^\infty 2^{kp}\omega_f(2^k)\right)^{\frac{1}{p}}\leq \Vert f\Vert_p \leq c_2 \left(\sum_{k=-\infty}^\infty 2^{kp}\omega_f(2^k)\right)^{\frac{1}{p}}$$ Where $\omega_f(\alpha)=\mu(\left\{|f|>\alpha\right\})$
My attempt
I've read this other problem which is very much the same (except for the second bound and the fact we don't know $f\in L^p$), where the answer proves the first part if $f\in L^p$. However, if we don't know $f\in L^p$ then it isn't really working anymore because after summation by parts.
$$\int |f|^p\geq \limsup_{n\to\infty\\m\to-\infty}\;\; 2^{mp}\omega(2^m)-2^{np}\omega(2^{n+1})+2^{-p}\sum_{k=m+1}^n 2^{kp}\omega(2^k)$$ $$\geq \limsup_{n\to\infty\\m\to-\infty}\;\; -2^{np}\omega(2^{n+1})+2^{-p}\sum_{k=m+1}^n 2^{kp}\omega(2^k)$$ I was only able to prove that $2^{np}\omega(2^{n+1}) \leq \int_{|f|>2^{n+1}} |f|^p$ which goes to zero only if $f\in L^p$.
As for the upper bound, I tried doing the same trick $$ \int_{2^n\leq|f|<2^{n+1}} |f|^p \leq 2^{n+1}\cdot(\omega(2^n)-\omega(2^{n+1}))$$ but now the problem lies, that I know nothing about the other term $2^{(m+1)p}\omega(2^m)$.
I also tried using Cauchy's condensation test for series with $\int p\alpha^{p-1}\omega(\alpha)d\alpha$ but $g(\alpha)=p\alpha^{p-1}\omega(\alpha)$ has to be non-decreasing which is something I don't necessarily know (because $\alpha^p$ is increasing and $\omega(\alpha)$ is decreasing). Any help would be very much appreciated.
Hint:
$$\begin{align} \int|f|^p\,d\mu&=\int^\infty_0pt^{p-1}\mu(|f|>t)\,dt\\ &=p\Big(\int^1_0t^{p-1}\mu(|f|>t)\,dt+\int^\infty_1t^{p-1}\mu(|f|>t)\,dt\Big)\end{align}$$
Splitting the integrals $$\begin{align} \int^1_0&=\sum^\infty_{k=0}\int^{2^{-k}}_{2^{-k-1}}\\ \int^\infty_1 &=\sum^\infty_{k=0}\int^{2^{k+1}}_{2^k} \end{align}$$
and using the monotonicity of $t\mapsto\mu(|f|>t)$ will gives what you are looking for.