Definition. Let $(X,\mu)$ be a measure space. For $f$ a measurable function on $X$, the distribution function of $f$ is the function $d_f$ defined on $[0,\infty)$ as follows: $$d_f(\alpha) := \mu\left( \left\{ x \in X : \left| f(x)\right|>\alpha\right\}\right) $$
For $f$ a $\mu$-measurable function I would like to proove the equality $$a := \inf\left\{C > 0 : \forall \alpha > 0 \>d_f(\alpha) \leqslant \frac{C^p}{\alpha^p}\right\} = \sup\left\{\gamma d_f(\gamma)^{1/p} : \gamma > 0 \right\} =: b$$
I managed to show $b \leqslant a$. For any $\alpha > 0$ we have $$\alpha d_f(\alpha)^{1/p} \leqslant C$$ and since $b$ is a least upper bound, we have $$\alpha d_f(\alpha)^{1/p} \leqslant b \leqslant C$$ However, I do not see the reversed inequality.
We can rewrite $a=\inf\left\{ C>0 \ \vert \ \forall \alpha >0 : \alpha d_f(\alpha)^{1/p}\leq C\right\}$. Let $\epsilon>0$. Then there exists $\alpha>0$ such that
$$ \alpha d_f(\alpha)^{1/p} > a - \epsilon.$$
Taking the supremum over all $\alpha>0$ yields
$$ b>a-\epsilon. $$
Now let $\epsilon \rightarrow 0$ and conclude
$$ b\geq a. $$