Integrable function with given condition is in $L^p$

Question

Suppose $f:\Bbb R \to \Bbb R$ is integrable and there exist constant $c\gt 0$ and $\alpha \in (0,1)$ such $$\int_A |f(x)|dx\le cm(A)^\alpha$$

for every Borel measurable set $A\subset \Bbb R,$ where $m$ is lebesgue measure. Prove that there exists $p\gt1$ such that $f\in L^p$.

I tried to use holders on $\int_A |f(x)|^pdx$ taking $p=\frac1p$, $q$ as conjugate of $\frac1p$ and $A$ as some finite measure set. But, when $p\lt 1$ we get reversed holder and i got stuck.

Please could anyone help on this.

Answer 1

Since everything is non-negative and measurable, we can write

$$\begin{align} \int_\mathbb{R} \lvert f(x)\rvert^p\,dx &= \int_\mathbb{R} \int_0^{\lvert f(x)\rvert^p}\,dt\,dx\\ &= \iint_{0 \leqslant t \leqslant \lvert f(x)\rvert^p}\,dt\,dx\\ &= \int_0^\infty m\left(\{ x : \lvert f(x)\rvert \geqslant t^{1/p}\}\right)\,dt \end{align}$$

for any $p > 1$. Use the given estimate to find a range for $p$ such that

$$\int_1^\infty m\left(\{ x : \lvert f(x)\rvert \geqslant t^{1/p}\}\right)\,dt < \infty.$$

Argue that then also

$$\int_0^\infty m\left(\{ x : \lvert f(x)\rvert \geqslant t^{1/p}\}\right)\,dt < \infty,$$

since $f$ is integrable.

Answer 2

Let $A(k) = \{x: |f(x)| \geq k\}$

$ km(A(k))\leq \int_{A(k)}|f(x)|dx \leq cm(A(k))^\alpha$ gives $m(A(k))^{1-\alpha}\leq \frac{c}{k}$ i.e $m(A(k)) = O(k^{-\frac{1}{1-\alpha}})$

Then remark that \begin{align} \int_R |f(d)|^pdx &=\sum_{k=0}^{+\infty} \int_{A(k)-A(k+1)}|f(x)|^pdx \\ &\leq \sum(k+1)^p(m(A(k))- m(A(k+1))) \\ &= \sum ((k+1)^p - k^p)m(A(k)) \end{align}

and $((k+1)^p-k^p) m(A(k)) = O(k^{p-1} k ^{-\frac{1}{1-\alpha}})$, it suffices to have $p-1-\frac{1}{1-\alpha}< -1$ so that the above sum will converge.

Therefore take $p$ such that $p<\frac{1}{1-\alpha}$