An unbounded function with integrability for all p>0 How to construct a function $f\geq0$ such that i) $f\in L^p([0,1])\forall p>0$ and ii) $esssup_I f=+\infty$ for any interval $I\subset[0,1]$ ?
Thank you! 

How about 
$$
f(x)=|\log x|?
$$
The above is a wrong answer, since it does not fulfill 3. So let's try again.
If $0<a<1$, then
$$
\int_0^1|\log |x-a||\,dx=1 - (1 - a) \log(1 - a)- a \log a<2.
$$
Order the rationals in $[0,1]$ in a sequence $\{a_n\}$, let $f_n(x)=\bigl|\,\log |x-a_n|\,\bigr|$ and
$$
f(x)=\sum_{n=1}^\infty2^{-n}f_n(x)\ .
$$
Clearly $f$ is positive. Moreover
$$
\int_0^1f(x)\,dx=\sum_{n=1}^\infty2^{-n}\int_0^1f_n(x)\,dx\le2 \sum_{n=1}^\infty2^{-n}<\infty,
$$
so that $f$ is finite almost everywhere and in $L^1$. We have to check that $f\in L^p$ for $p>1$. Since 
$$
\|f\|_p\le\sum_{n=1}^\infty2^{-n}\|f_n\|_p,
$$
it is enough to show that $\|f_n\|_p\le C_p$ for some constant $C_p$ depending only on $p$. The function $|x-a|^{1/(2p)}\bigl|\,\log |x-a|\bigr|$ is continuous, and in particular bounded, as a function of $(x,a)$ on $[0,1]\times[0,1]$. It follows that there exists a constant $C_p$ such that
$$
\bigl|\,\log |x-a|^p\bigr|\le\frac{C_p}{\sqrt{|x-a|}},\quad0\le x\le1,\quad0\le a\le1,
$$
and the claim follows.
