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?
 A: Note first that $\|f(\cdotp+h)\|=\|f(\cdotp)\|$.
Prove it first for continuous functions with compact support (this space is dense in $L^p(\mathbb{R}^n)$): let $g$ be a such function, let $K=supp(g)$ and let $U$ be an open set of $\mathbb{R}^n$ containing $K$ such that $\mu(U)<\infty$. Since $g$ is uniformly continuous, given $\varepsilon>0$ there exists $\delta>0$ with $|g(x)-g(y)|<\varepsilon/\mu(U)^{1/p}$ if $\|x-y\|<\delta$; you may suppose that $\delta<d(K,\mathbb{R}^n\setminus U)$. This forces $$\|g(\cdotp+h)-g(\cdotp)\|=(\int_U|g(x+h)-g(x)|^pdx)^{1/p}\leq(\int_U(\varepsilon^p/\mu(U)))^{1/p}=\varepsilon.$$
I leave you the details.
A: 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).
