I am trying to get a grasp of the following
Lemma. Let $(X,\mathcal{A},\mu)$ be a measure space and let $u:X \to \mathbb{R}$ be a measurable function on $X$. If $s > 1$ and $\mu(\{|u| > t\}) \leq C_0 t^{-s}$, then $\forall q \in [1;s)$ we have $$\|u\|_q \leq \left(\frac{s}{s-q}\right)^{1/q} C_0^{1/s} \mu(X)^{(s-q)/sq}.$$
What I have managed to do is the following:
(1) Since $u$ is measurable, so is $|u|:X \to [0;\infty)$, and hence $\{|u|>t\} \in \mathcal{A}$, since it is the preimage of the borel set $(t;\infty) \in \mathcal{B}(\mathbb{R})$ under the map $|u|$.
(2.1) If $\mu(X) = 0$, $\mu$ is the zero measure on $X$ and the claim holds trivially true.
(2.2) If $\mu(X) = \infty$, the claim reduces to $\|u\|_q \leq \infty$, which is also always true.
(3) WLOG one may therefore assume $0 < \mu(X) < \infty$.
(4) Trying to estimate the $q$-norm, I find (defining $A:= \{|u|>t\}$ and $B:=\{|u|\leq t\}$) that $$ \begin{align*} \|u\|_q^q &= \int_X |u(x)|^q \mathrm{d}\mu(x) \\ &= \int_{A \sqcup B} |u(x)|^q \mathrm{d}\mu(x) \\ &= \int_{A} |u(x)|^q \mathrm{d}\mu(x) + \int_{B} |u(x)|^q \mathrm{d}\mu(x) \\ &\leq \int_{A} |u(x)|^q \mathrm{d}\mu(x) + \int_{B} t^q \mathrm{d}\mu(x) \\ &= \int_{A} |u(x)|^q \mathrm{d}\mu(x) + t^q \mu(B) \\ &= \int_{A} |u(x)|^q \mathrm{d}\mu(x) + t^q \mu(X \setminus A) \\ &= \int_{A} |u(x)|^q \mathrm{d}\mu(x) + t^q \mu(X) - t^q \mu(A) \\ &\leq \||u|^q\|_{\infty,A} \mu(A) + t^q \mu(X) - t^q \mu(A) \\ &\leq \||u|^q\|_{\infty,A} C_0 t^{-s} + t^q \mu(X) - t^q \mu(A) \end{align*} $$
but here I either went in the wrong direction or am not seeing how to proceed...
First, i'll mention a useful way of computing $p$-th moments. Since $$|x|^p=\int_0^\infty 1\{|x|^p\geq t\}dt$$ we get by Fubini's theorem that $$\Vert u\Vert_q=\int_0^\infty \mu(\{|u|^q\geq t\})dt.$$
Now you can perform a substitution $s\leftarrow t^{1/q}$, rewrite the expression given above and use the inequality you are given to show the result.