Show that$\,\,\int_{|f|\leq s}|f\,|^{q}\,d\mu\leq\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p\,\,,$ where $0<p<q<\infty$

Let $(X,\mu)$ be a measure space and let $s>0$

Let $f$ be a measurable function on $X$ .Show that $0<p<q<\infty$ we have $$\int_{|f|\leq s}|f\,|^{q}\,d\mu\leq\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p$$

Here's my done :

If $\lVert f\,\rVert_{L^{p,\infty\,}}=\infty$ then there is nothing to prove.So assume $\lVert f\,\rVert_{L^{p,\infty\,}}<\infty$ Then we have

\begin{align} \int_{|f|\leq s}|f\,|^{q}\,d\mu&=q\int_{0}^{s} \alpha^{q-1}\omega(\alpha)\,d\alpha\\ &\leq q\int_{0}^{s} \alpha^{q-1-p}\lVert f\,\rVert_{L^{p,\infty}}^p\,\,\,d\alpha\\ &=\frac{q}{q-p}s|_{0}^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p\\ &=\frac{q}{q-p}s^{q-p}\lVert f\,\rVert_{L^{p,\infty}}^p \end{align}

where $\omega(\alpha)$ is the distribution function of $f$.

If you have the time,please check my proof for validity with me. Any advice would appreciate.Thanks for considering my request.

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I take note below this line that anyone can ignore it or give your valuable suggestion.

\begin{align} \int_{|f|\leq s}|f(x)|^{q}\,d\mu(x)&= q\int_{0}^{\infty}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha \,+ q\int_{s}^{\infty}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\chi_{\{\alpha\,:\, 0<\alpha\leq s\}}\omega(\alpha)\,d\alpha \,+ 0\\ &=q\int_{0}^{s}\alpha^{q-1}\omega(\alpha)\,d\alpha \end{align}

Okay, or if we let $E=\{x\in X: |f(x)|\leq s\}$, then \begin{align*} \int_{|f|\leq s}|f|^{q}d\mu&=\int_{E}|f|^{q}d\mu\\ &=q\int_{0}^{\infty}\alpha^{q-1}\omega_{E}(\alpha)d\alpha\\ &=q\int_{0}^{\infty}\alpha^{q-1}\chi_{0\leq\alpha<s}\omega_{E}(\alpha)d\alpha\\ &=q\int_{0}^{s}\alpha^{q-1}\omega(\alpha)d\alpha, \end{align*} where $\omega_{E}(\alpha)=\{x\in E:|f(x)|>\alpha\}$, if $\alpha\geq s$, then $\omega_{E}(\alpha)=0$, if $\alpha<s$, then $\omega_{E}(\alpha)=\omega(\alpha)$.