$L^p$ integral on every measurable subset of $\Bbb R$

Suppose $f:\Bbb R \to \Bbb R$ is in $L^p$ for some $p>1$ and also in $L^1$. Prove there exist constants $c>0$ an $\alpha \in (0,1)$ such that

$\int_A|f(x)|dx\le cm(A)^{\alpha}$, for every Borel measurable set $A\subset \Bbb R$, where $m$ Lebesgue measure.

Actually, I didn't get any concrete idea how to start. I tried Holders, splitting $f(x)$ into two parts with conjugate powers. It took me nowhere.. Can you just give me idea so that i can start it..

Hint: Holder's inequality with $f$ and $g(x)=1$.
• More of hint: $\alpha=1/q$, $c=||f||_p$. – abnry Jul 19 '14 at 19:39
• Ya, I did same but my confusion was what would happen if measure of A is $\infty ?$ – Toeplitz Jul 19 '14 at 19:51
• If the measure of $A$ is $\infty$, $||f||_1 \leq \infty$. – abnry Jul 19 '14 at 19:51
• The inequality will still be true in that case. More of a concern is when $m(A)<\infty$ so we need hypothesis that $f\in L^{1}$. – user71352 Jul 19 '14 at 19:53