Suppose that $f$ is integrable and for each Borel $A$ there is an $\alpha \in (0,1)$ and $c$ such that $\int_A|f| \leq c m(A)^{\alpha}$, with $m$ Lebesgue measure. Then $f \in L^p(\Bbb R)$ for some $p > 1$.
The converse of this theorem is an easy application of Holder's inequality, so I tried to show that $f \in L^{\frac{1}{1-\alpha}}$, but can't seem to get anywhere. Any tips are appreciated.
For a fixed $p>1$ (to be specified later), start by writing $$ \int_{\mathbb{R}}|f|^p=\int_{|f|\leq 1}|f|^p+\sum_{k=1}^{\infty}\int_{2^{k-1}<|f|\leq 2^k}|f|^p\leq \int_{|f|\leq 1}|f|+\sum_{k=1}^{\infty}2^{k(p-1)}\int_{2^{k-1}<|f|\leq 2^k}|f| $$ Then applying the given condition to $A=\{2^{k-1}<|f|\leq 2^k\}$, we get $$ \int_{\mathbb{R}}|f|^p\leq ||f||_1+c\sum_{k=1}^{\infty}2^{k(p-1)}m(2^{k-1}<|f|\leq 2^k)^{\alpha} $$ Next, we have $$ m(2^{k-1}<|f|\leq 2^k)\leq m(|f|>2^{k-1})\leq \frac{||f||_1}{2^{k-1}}=\frac{2||f||_1}{2^k} $$ by Chebyshev's inequality, hence $$\int_{\mathbb{R}}|f|^p\leq||f||_1+c2^{\alpha}||f||_1^{\alpha}\sum_{k=1}^{\infty}2^{k(p-1-\alpha)}$$
The sum will be finite as long as $p-1-\alpha<0$, i.e. $p<1+\alpha$, and since $\alpha>0$ we can choose $p>1$ satisfying this condition.