How do I show that a distribution is locally p-integrable? 
Let $\Omega \subseteq \mathbb{R}^n$ (open) and $u \in \mathcal{D}'$ 
  be a distribution, that has a distributional derivative
  which is in $L^p(\Omega)$ (for some $p \geq 1$). 
Show that $u \in L^p_{loc}(\Omega)$

...where $L^p(\Omega)$ is defined as in
https://en.wikipedia.org/wiki/Lp_space#Lp_spaces
and $L^p_{loc}(\Omega)$ is defined as in: 
https://en.wikipedia.org/wiki/Locally_integrable_function#Generalization:_locally_p-integrable_functions 

My thoughts: 
1) For notational simplicity let's call $Du = D_i u$ and $\phi' = \partial_{x_i}\phi$ for some $i \in \lbrace 1,2 \dots n\rbrace$
Then by definition of distributional derivatives, we have that: 
$\int_{\Omega} (Du) \phi dx=-\int_{\Omega} u \phi' dx, \forall \phi \in C^{\infty}(\Omega)$ with compact support
2) Let also $K \subset \Omega$ (compact). We need to show that $\int_K \vert u \vert^p dx < \infty$
3) A simple density argument of the test functions should help us go from $K$ to $\Omega$ so this shouldn't be a big deal
4) We can also use something like Holder's inequality (and the fact that $Du \in L^p$) to show something among the lines of: 
$\vert \int_{\Omega} (Du) \phi  dx \vert \leq \int_{\Omega} \vert (Du) \phi \vert dx \leq (\int_{\Omega} \vert Du \vert^p)^{\frac{1}{p}} (\int_{\Omega} \vert \phi \vert^q)^{\frac{1}{q}}< \infty$
5) Then from the definition of $Du$ we also have that: 
$\vert \int_{\Omega} u \phi'  dx \vert = \vert \int_{\Omega} (Du) \phi  dx \vert$
6) But how do we find an inequality that involves $\int \vert u \vert ^p$ and $\int\vert u \vert$ (in this order) to fill in the chain of inequalities? (Assuming of course, that such a "chain" is the way to go in this problem)  
7) Another idea I am toying with (see update below), is to use some sort of a density argument to go from - say - $\int\vert u \vert ^p$ to $\int\vert u \vert \phi_n$ for an appropriate sequence of $\phi_n$. Not sure if this is possible though. 
Update:

Question: What density arguments are available for locally integrable
  functions? For instance, since $u \in L^1_{loc}$, is it possible to
  find a sequence $\phi_n$ in $C^{\infty}_0(\Omega)$ or even in
  $C^{\infty}(\Omega)$ such that: 
$$\lim\limits_{n \rightarrow \infty}\phi_n = \vert u \vert^{p-1}$$
I think that would complete the chain of inequalities and thus the proof.

 A: I will outline the solution for the case $n=1$ and $p=2$, and if you can not generalize it to $p\ge1$ and $n\ge 1$, ask in comments, we will dive into the murky details. Without further ado:
Let $K$ be a non-trivial compact set, $\phi$ be a test function with support in $K$ and $u'\in L^2(K)\cap D'(\Bbb R)$.
We can say that by Cauchy-Bunyakovskiy-Schwarz inequality
$$|\langle u',\phi\rangle| = \left|\int_K u'(x)\phi(x)dx\right|\le \|u',L^2(K)\|\|\phi,L^2(K)\|$$
Then again, by Poincaré's inequality, for a constant $C$ that depends only on $K$, $$\|\phi,L^2(K)\|\le C \|\phi',L^2(K)\|.$$
This gives us 
$$|\langle u,\phi'\rangle| \le C\|u',L^2(K)\| \|\phi',L^2(K)\| .$$
Therefore, we can already say that $u$ is a continuous linear operator over the set $L^2(K)\cap \{f: \int_Kf=0\}$. We can extend it to the whole $L^2(K)$ by noticing that if $\psi$ is a test function supported in $K$ with $\int_k\psi=1$, then for any $\phi$ as above we have
$$\langle u,\phi\rangle = \left\langle u,\phi-\psi\int_K\phi\right\rangle + \left(\int_K\phi\right)\langle u,\psi\rangle.$$
The test function $\phi-\psi\int_K\phi$ is a derivative of a test function (easy exercise), hence you can apply the estimations above; and the term $\left(\int_K\phi\right)\langle u,\psi\rangle$ is bounded by $C_1 \|\phi,L^2(K)\|$ (essentially, $C_1$ acts like an arbitrary constant when you find an antiderivative). Thus, $\langle u,\phi\rangle$ is bounded by $$C\|u',L^2(K)\|  \left\| \phi-\psi\int_K\phi,L^2(K)\right\| + C_1 \|\phi,L^2(K)\|.$$
It easy to see that $u$ now can be seen as continuous linear functional over $L^2(K)$. Therefore, $u$ itself belongs to $L^2(K)$.
Notes for the  generalizations: the Poincaré's inequality works for $p\in [1,\infty)$ and $n\ge 1$, so that part is ok. The representation theorem definitely works for $p>1$, some tinkering might be necessary to cover the case $p=1$.
