Proving convergence in $L^p$ Suppose $f \in L^p(\mathbb{R}) , 1 \leq p < \infty$. Prove that $$\lim_{h\rightarrow 0} \int_{\Bbb R} |f(x+h)-f(x)|^p = 0$$ 
I was thinking about showing that the sequence $\{f(x+\frac{1}{n}) - f(x)\}_n$ converges to zero in $L^p(\mathbb{R})$ so that I could pass the limit under the integral and conclude that the limit is zero, but I am having problems showing this convergence. 
I'm also not sure what other approach to take/what other convergence theorems I could use.  

Let $f=1_[a,b]$ where $A \subseteq \mathbb{R}$.
$\forall h \in (0,1)$ we have that $(\int_{\mathbb{R}}|1_{[a,b]}(x+h)-1_{[a,b]}(x)|^p)^{1/p}=(\int_{\mathbb{R}}|1_{[a,b]-h}(x)-1_{[a,b]}(x)|^p)^{1/p}=(\int_{([a,b]-h) \triangle [a,b]})^{1/p}=m(([a,b]-h) \triangle [a.b])^{1/p} \leqslant m([a-h,a) \cup(b-h,b])^{1/p} \leqslant (2h)^{1/p} \rightarrow 0 $
In the same way you can prove the above $\forall h \in (-1,0)$
Now if $f$ is a step function with compact support then its easy to see that $\int_{\mathbb{R}}|f(x+h)-f(x)|^p \rightarrow 0$ from the fist step.
Now we know that the set of step functions with compact support is dence in $L^p(\mathbb{R})$.
Let $\epsilon >0$ and  $f \in L^p(\mathbb{R})$  
Exist a step  function $h$ with compact support  such that $||f(x)-s(x)||_p \leqslant \epsilon$
$\forall h \in \mathbb{R}$ we have that $$\||f(x+h)-f(x)||_p \leqslant$$ $$\||f(x+h)-s(x+h)||_p+||s(x+h)-s(x)||_p+||f(x)-s(x)||_p$$ $$=2||f(x)-s(x)||_p+||s(x+h)-s(x)||_p$$
Thus $$\limsup_{h \rightarrow 0}||f(x+h)-f(x)||_p \leqslant 2 \epsilon +0=2 \epsilon$$
Therefore $\int_{\mathbb{R}}|f(x+h)-f(x)|^p \rightarrow 0$ as $h \rightarrow 0$
