If $X\geq 0$, and $\mathbb{E}(X^\alpha)<\infty$ for $0<\alpha<1$ show that $\mathbb{E}(\min (X,t))$ is $o(t^{1-\alpha})$.
We know that that $$\mathbb{E}(\min(X,t))=\int_{X\leq t}Xd\mathbb{P}+\int_{X>t}td\mathbb{P}=\int_{X\leq t}Xd\mathbb{P}+t\mathbb{P}(X>t)$$
Using Chebychev's inequality for $x^\alpha$, I get the second term is bounded by $$t\mathbb{P}(X>t)\leq t^{1-\alpha}\mathbb{E}(X^\alpha)$$
But that doesn't decay faster than $t^{1-\alpha}$ and I have no idea how to get a bound on the first term other that bounding $X$ by $t$, which isn't very helpful. I am just looking for a hint on how to start.
The asymptotics are when $t\to\infty$.
The first step is to rephrase everything in terms of $G(t)=P(X\geqslant t)$, namely, $$E(X^a)=\int_0^\infty at^{a-1}G(t)\mathrm dt,$$ and $E(\min(X,t))=J(t)$, where $$J(t)=\int_0^tG(s)\mathrm ds.$$ Assume that $E(X^a)$ is finite. Then, when $t\to\infty$, $$\int_{t}^\infty as^{a-1}G(s)\mathrm ds\to0,$$ and, for every $t$, $$\int_{t}^\infty as^{a-1}G(s)\mathrm ds\geqslant\int_{t}^{2t}as^{a-1}G(s)\mathrm ds\geqslant \int_t^{2t}as^{a-1}G(2t)\mathrm ds=(2^a-1)t^aG(2t),$$ hence, when $t\to\infty$, $$t^aG(t)\to0.$$ (This property is slightly weaker than the finiteness of $E(X^a)$ but it suffices to conclude.)
Fix $\varepsilon\gt0$ and choose some $t_\varepsilon$ such that $t^aG(t)\leqslant\varepsilon(1-a)$ for every $t\geqslant t_\varepsilon$. For every $t\geqslant t_\varepsilon$, $$J(t)=J(t_\varepsilon)+\int_{t_\varepsilon}^tG(s)\mathrm ds\leqslant J(t_\varepsilon)+\varepsilon\int_{t_\varepsilon}^t(1-a)s^{-a}\mathrm ds\leqslant J(t_\varepsilon)+\varepsilon t^{1-a},$$ in particular, $$\limsup_{t\to\infty}t^{a-1}J(t)\leqslant\varepsilon.$$ This holds for every $\varepsilon\gt0$ hence $t^{a-1}J(t)\to0$ when $t\to\infty$, QED.