TITLE: Class of functions so that $\int_0^t|f(x)|dx\leq C\cdot t^\alpha$.
QUESTION [5 upvotes]: I have done something with Lebesgue integrals and It leads me to the following class of functions: Let $\alpha$ be a real number. We consider all Lebesgue measurable function $f$ on $[0;1]$ such that there exists a constant $C>0$ and $\int_0^t|f(x)|dx\leq C\cdot t^\alpha$ for almost everywhere $t\in(0;1]$.
Does this class has a name? I think it look simple and usual. Could we construct some non-trivial functions which belong to this class? (I may assume further that $\alpha>0$).

REPLY [3 votes]: If $\alpha \le 0$ then any integrable function $f$ belongs to the class. 
If $0 < \alpha < 1$ you may use Holder's inequality to write
$$
\int_0^t |f(s)| \, ds \le \left(\int_0^t \, ds\right)^{\alpha} \left( \int_0^t |f(s)|^{\frac{1}{1-\alpha}}\, ds \right)^{1-\alpha} = t^\alpha \left( \int_0^t |f(s)|^{\frac{1}{1-\alpha}}\, ds \right)^{1-\alpha}$$ so that any function $f \in L^{1/(1-\alpha)}([0,1])$ belongs to the class.
If $\alpha \ge 1$, then any integrable function satisfying $$\limsup_{\epsilon\to 0^+} \epsilon^{-\alpha} \int_0^\epsilon |f(s)| \, ds < \infty$$ belongs to the class.