A question on weak derivative - Sobolev spaces Let $\Omega$ be an open set in $R^n$, and $f \in L^1_{loc}(\Omega)$, such that for each multiindex $\alpha\in N^n$, $|\alpha| = l$  f has weak derivative $D^\alpha f$ in $L^p(\Omega)$, with $1\leq p\leq \infty$. 
In general it is not true that $f\in L^p(\Omega)$, but it has to be true that $f\in L^p_{loc}(\Omega)$. How can this be shown?
 A: (Original answer edited to make it shorter)
It suffices to show this for $l = 1$. It also suffices to show that $f$ is locally in $L_q$ for some $q \ge p$. But this follows immediately by the Sobolev inequality.
A: Actually more is true. It suffices to assume that $f$ is a distribution on $\Omega\subset\mathbb{R}^n$ such that all its distributional derivatives of order $l$ are in $L^p(\Omega)$.
Then $f\in L^p_\mathrm{loc}(\Omega)$. A proof based on convolution and the fundamental solution of the polyharmonic operator can be found in the Theorem of Section 1.1.2 in the following book:

Vladimir G. Maz'ja, Sobolev spaces. 
  Translated from the Russian by T. O.
  Shaposhnikova. Springer Series in
  Soviet Mathematics. Springer-Verlag,
  Berlin, 1985. ISBN: 3-540-13589-8

A: Nicolo, I'm not entirely sure if I understand your question but I will attempt an answer. What you're saying seems to be trivial actually. Just take any constant function $f=C$. Then $f \in L_{loc}^1(\Omega)$ and $f$ has a weak derivative lying in every $L^p(\Omega)$. Moreover $f \notin L^p(\Omega)$ but $f \in L_{loc}^p(\Omega)$ for all $p \in [1,\infty]$.
