If $f \in L^p(\mathbb{R}^n)$ then $\Vert f(\cdot + h) - f(\cdot) \Vert_{L^p(\mathbb{R}^n)} \rightarrow 0 $ Let $1 \leq p < \infty$. Show that: If $f \in L^p(\mathbb{R}^n)$ then 
$$\Vert f(\cdot + h) - f(\cdot) \Vert_{L^p(\mathbb{R}^n)} \rightarrow 0 \ \ \text{ for } h \in \mathbb{R}^n \text{ with } h \rightarrow 0$$
where $f(\cdot + h)$ denotes the function $x \mapsto f(x+h)$.
I have no clue how to show this, can someone get me started and give me a hint?
Thanks in advance.
EDIT: Someone has provided an answer using the density of $C^\infty_{0} \subset L^p$. Is there a way to show it without?

The proof given by Angel is the usual one (you can use just density of step functions and use step functions instead of smooth-functions).
Another way to see the problem is as $Tg$, where $T$ is a convolution operator (or more specifically, a convolution operator minus Id).
By using the Hausdorff-Young inequality, you end up studying a suitable multiplier, which is easily analyzed (one might need to regularize the situation a bit, or localize, as Id does not have Fourier transform, if one feels happy with distributions then it is okay).
