If $\Omega$ bounded, is it true that $f \in L^p(\Omega) \Rightarrow f \in L^{p+\epsilon}(\Omega)$ for small $\epsilon$? Let $\Omega$ be a bounded subset of $\mathbb{R^n}$, and $f \in L^p(\Omega)$. Prove or disprove that there exits $\epsilon > 0$ such that $f \in L^{p+\epsilon}(\Omega)$. 
It seems to me that there is no direct proof, but I can't find a counterexample. Any comments and ideas would be appreciated. Thanks!
 A: Let $p \in [1, \infty)$. Take $f(x) := 1 / (x \log^2 x)^{1/p}$ on $\Omega := (0, \theta)$, where $\theta \in (0, 1)$, e.g. $\theta = 1/2$. Then $$\int_\Omega |f(x)|^p dx = \int_\Omega \frac{1}{x} \frac{1}{(\log x)^2} = \left. -\frac{1}{\log{x}} \right|_{x = 0}^{x = \theta} = -\frac{1}{\log \theta}$$ is finite. But $$\int_\Omega |f(x)|^{p + p \epsilon} dx = \int_\Omega \frac{1}{x^{1+\epsilon}} \frac{1}{\log^{2(1+\epsilon)} x} dx$$ does not exist for any $\epsilon > 0$.
PS. Of course, the term $\left. \frac{-1}{\log x}\right|_{x = 0}^{x = \theta}$ is to be understood with care. One justifiable way is to view it as 
$$
\left. \frac{-1}{\log x}\right|_{x = 0}^{x = \theta}
=
\lim_{\epsilon \searrow 0}
\left. \frac{-1}{\log x}\right|_{x = \epsilon}^{x = \theta}
=
\frac{-1}{\log \theta}
+
\lim_{\epsilon \searrow 0} \frac{1}{\log \epsilon}
=
\frac{-1}{\log \theta}
,
$$
where $\frac{1}{\log \theta}$ is ok for $\theta \in (0, 1)$.
A: Hint: Take $\Omega= [0,1]$, and $f$  constant value $a_n>0$ on each interval $[\frac{1}{2^{n-1}}, \frac{1}{2^{n}}]$. You are interested in exponents $\theta$ so that 
$$\sum_{n\ge 1} \frac{a_n^{\theta}}{2^{n}} < \infty$$
To find an example of a sequence $(a_n)$ for which the set of exponents $\theta$ is the closed interval $[0,1]$ study a similar problem : find a sequence $b_n >0$ such that $\sum b_n^{\theta}$ converges precisely when $\theta\in [1, \infty)$. 
$\bf{Added:}$ Take $b_n = \frac{1}{n (\log n)^2}\ \ $, you will get 
$$a_n = \frac{2^n}{n (\log n)^2}$$
A: Let $x_0$ be a point of Lebesgue density $1$ of $\Omega$. Since $\Omega$ is bounded, the integral $$\int_\Omega \frac{1}{|x-x_0|^\alpha} \, dx$$ converges if and only if $\alpha < n$.  For each $k \ge 1$ define $\displaystyle \alpha_k = \frac{n}{p} \left( 1 - \frac{1}{k} \right)$ so that $\alpha_k p = n(1-1/k)$ and thus $$\int_\Omega \frac{1}{|x - x_0|^{\alpha_k p}} \, dx < \infty.$$ Set
$$A_k = \left( \int_\Omega \frac{1}{|x - x_0|^{\alpha_k p}} \, dx \right)^{1/p}.$$
Let $$ f(x) = \sum_{k=1}^\infty \frac{1}{A_k 2^k} \frac{1}{|x - x_0|^{\alpha_k}}.$$ Then $\|f\|_p \le 1$, but for any $\epsilon > 0$ you have  $$\int_\Omega |f(x)|^{p + \epsilon} \, dx \ge \int_\Omega \frac{1}{A_k 2^k} \frac{1}{|x - x_0|^{\alpha_k(p + \epsilon)}} \, dx = \infty$$ provided that $\alpha_k(p + \epsilon) \ge n$, which holds for all $k \ge \dfrac{p + \epsilon}{\epsilon}$.
