$f \in L^p \iff \sum_{k\in\mathbb{Z}}2^{kp}\lambda_f(2^k)<\infty$

Question

The problem is,

$$f \in L^p \iff \sum_{k\in\mathbb{Z}}2^{kp}\lambda_f(2^k)<\infty$$

I have the solution for $\Rightarrow$, but I don't understand a determined step.

Let $f\in L^p$,

\begin{align*} \sum_{\mathbb{Z}} \lambda_f(2^k)2^{kp}&=\sum_{\mathbb{Z}}\lambda_f(2^k)\frac{p}{1-2^{-p}}\int_{2^{k-1}}^{2^k} \alpha^{p-1}\,d\alpha\\ &\le \sum_{\mathbb{Z}}\frac{p}{1-2^{-p}}\int_{2^{k-1}}^{2^k}\lambda_f(\alpha) \alpha^{p-1}\,d\alpha \quad(*)\\ &=\frac{p}{1-2^{-p}}\int_{0}^{\infty}\lambda_f(\alpha) \alpha^{p-1}\,d\alpha\\ &=\frac{p}{1-2^{-p}}\int|f|^p < \infty \end{align*}

I couldn't understand the step $(*)$. I probably can't understand it because I'm having a little difficulty understanding exactly what it would be

$$\int\lambda_f(\alpha)\,d\alpha$$

Recalling that $\lambda_f(\alpha)$ is the distribution function, given by $\lambda_f(\alpha)=\mu(\{x:|f(x)|>\alpha\})$.

Answer 1

Hint

If $\alpha \in [2^{k-1},2^k]$, then

$$|f(x)|>2^k\implies |f(x)|>\alpha ,$$ and thus $$ \lambda _{f}(2^k)\leq \lambda _f(\alpha ).$$

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Edit

Therefore $$\alpha ^{p-1}\lambda_f(2^k)\leq \alpha ^{p-1}\lambda _f(\alpha ),$$ for all $\alpha \in [2^{k-1},2^k]$, and thus $$\lambda _f(2^k)\int_{2^{k-1}}^{2^k}\alpha ^{p-1}\,\mathrm d \alpha \leq \int_{2^{k-1}}^{2^k}\lambda _f(\alpha )\alpha ^{p-1}\,\mathrm d \alpha .$$