Showing that $\lim_{x \to x_0} \|f_x - f_{x_0}\|_p = 0$ under certain circumstances 
For $f: \mathbb{R} \to \mathbb{R}$ and $x\in \mathbb{R}$ define the
  function $f_x: \mathbb{R} \to \mathbb{R}$ by $f_x(t) := f(x - t)$.
  Suppose $1 \leq p < \infty$ and $f \in
 \mathcal{L}^p(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$. Then show
  that $$\lim_{x \to x_0} \|f_x - f_{x_0}\|_p = 0$$ for any $x_0 \in
 \mathbb{R}$.

A hint is to show this first for a step function with compact support. So let $$f(t) = \sum_{i = 1}^na_i \chi_{A_i}(t)$$ where $\operatorname{supp} f$ compact. Let $x \in \mathbb{R}$. Then $$f_x(t) = \sum_{i = 1}^na_i \chi_{A_i}(x - t) = \sum_{i = 1}^na_i \chi_{x - A_i}(t)$$ where $-A$ is defined as usual. So for $x_0 \in \mathbb{R}$ $$\begin{align}\|f_x - f_{x_0}\|_p^p &= \int_{\mathbb{R}}\left|\sum_{i = 1}^na_i (\chi_{x - A_i}(t) - \chi_{x_0 - A_i}(t))\right|^pdt\\
&\leq \int_{\mathbb{R}} \left(\sum_{i = 1}^n |a_i||\chi_{x - A_i}(t) - \chi_{x_0 - A_i}(t)|\right)^pdt\\
&= \int_{\mathbb{R}} \left(\sum_{i = 1}^n |a_i|\chi_{(x - A_i)^c \cap (x_0 - A_i)^c}(t)|\right)^pdt\end{align}$$ which does not bring me further. Any help would be appreciated.
 A: I am not sure this will help but I think it is related to your problem.
Problem:
Suppose that $f\in L^p(\mathbb{R})$, where $1\leq p <\infty$. For $h\in\mathbb{R}$, write $\tau_hf(x)=f(x-h)$. Show that
$$\lim_{h\to 0}\|\tau_hf - f\|_{L^p}=0$$
Note that $\tau_h f(x) = f(x − h)$.
Proof:
Let $\mu$ be the Lebesgue measure on $\mathbb{R}$. Since $f\in L^p(\mathbb{R})$, we have
$$ \int |f|^p d\mu <\infty $$
For each $n\in \mathbb{N}$, we have that $|f|^p\chi_{[-n,n]}$ pointwise montonically to $|f|^p$. So using the Monotone Convergence Theorem,  we have that
$$ \int |f|^p \chi_{[-n,n]} d\mu \nearrow  \int |f|^p d\mu $$
Since $ \int |f|^p d\mu <\infty$, we have that, given $\epsilon >0$, there is $k \in \mathbb{N}$, $k>1$ such that
$$ \int_{\mathbb{R}\setminus [-k+1,k-1]} |f|^p d\mu = \int |f|^p d\mu - \int |f|^p \chi_{[-k+1,k-1]} d\mu <\frac{\epsilon^p}{5^p}$$
Then, for any $h\in \mathbb{R}$, $|h|<1$, we have $[-k+1,k-1] \subset [-k+h,k+h]$ and
\begin{align*}
\int |\tau_h f\chi_{\mathbb{R}\setminus[-k,k] }|^pd\mu & = \int |f(x-h)\chi_{\mathbb{R}\setminus [-k,k]} (x-h)|^p dx= \\
& =\int_{\mathbb{R} \setminus [-k+h,k+h]} |f|^p d\mu  \leq \int_{\mathbb{R}\setminus [-k+1,k-1]} |f|^p d\mu <\frac{\epsilon^p}{5^p}
\end{align*}
So we have , for any $h\in \mathbb{R}$, $|h|<1$,
\begin{equation}
\|\tau_h f\chi_{\mathbb{R}\setminus[-k,k] }\|_p < \frac{\epsilon}{5} \ \ \ (1)
\end{equation}
In particular, we have (taking $h=0$)
\begin{equation}
 \| f\chi_{\mathbb{R}\setminus[-k,k] }\|_p < \frac{\epsilon}{5} \ \ \ \  \  \  \ (2)
\end{equation}
Now, note that $f \chi_{[-k,k]}\in L^p([-k,k])$ and, since the simple functions are dense in $L^p$, we have that there is $g$ a simple function defined on $[-k,k]$, such that
\begin{equation}
\|f \chi_{[-k,k]} -g\|_p < \frac{\epsilon}{5} \ \ \ \ \ \ (3)
\end{equation}
It easy to see that, for any $h\in \mathbb{R}$,
\begin{equation}
\|\tau_hf \chi_{[-k,k]} -\tau_hg\|_p =\|f \chi_{[-k,k]} -g\|_p < \frac{\epsilon}{5} \ \ \ \ \ \ (4)
\end{equation}
Now, let $\{h_i\}_i$ be any sequence of real numbers converging to $0$. We can, without loss of generality, assume that $|h_i|\leq 1$, for all $i$. Also, note that there is $M>0$ such that $|g(x)| <M$, for all $x\in\mathbb{R}$.
So we have that $\tau_{h_i}g$ converges pointwise to $g$, $|g|<M\chi_{[-k-1,k+1]} $ and, for all $i$, $|\tau_{h_i}g|<M\chi_{[-k-1,k+1]} $. So, $|\tau_{h_i}g-g|^p$ converges pointwise to $0$ and for all $i$, $$|\tau_{h_i}g-g|^p\leq2^p(|\tau_{h_i}g|^p+|g|^p)<2^pM^p\chi_{[-k-1,k+1]} $$
Since $2^pM^p\chi_{[-k-1,k+1]}$ is integrable, we can apply the Dominated Convergence Theorem and conclude that
$$ \int |\tau_{h_i}g-g|^p d\mu \to 0  \quad \textrm{ as } i \to \infty $$
So,
$$ \|\tau_{h_i}g-g\|_p \to 0  \quad \textrm{ as } i \to \infty $$
So, there is $N\in \mathbb{N}$, such that if $i\geq N$,
\begin{equation}
 \|\tau_{h_i}g-g\|_p <\frac{\epsilon}{5} \ \ \ \ \ \ (5)
\end{equation}
So, using $(1)$, $(2)$, $(3)$ and $(4)$, we have
\begin{align*}
\|\tau_{h_i}f-f\|_p &= \|\tau_{h_i}(f\chi_{\mathbb{R}\setminus[-k,k] }+ f\chi_{[-k,k] })-(f\chi_{\mathbb{R}\setminus[-k,k] }+ f\chi_{[-k,k] })\|_p = \\
&= \|(\tau_{h_i}f\chi_{\mathbb{R}\setminus[-k,k] }-f\chi_{\mathbb{R}\setminus[-k,k] })+ (\tau_{h_i}f\chi_{[-k,k] }- f\chi_{[-k,k] })\|_p \leq\\
& \leq \|\tau_{h_i}f\chi_{\mathbb{R}\setminus[-k,k] }-f\chi_{\mathbb{R}\setminus[-k,k] }\|_p + \|\tau_{h_i}f\chi_{[-k,k] }- f\chi_{[-k,k] }\|_p \leq\\
& \leq \|\tau_{h_i}f\chi_{\mathbb{R}\setminus[-k,k] }\|_p + \|f\chi_{\mathbb{R}\setminus[-k,k] }\|_p + \|\tau_{h_i}f\chi_{[-k,k] }- f\chi_{[-k,k] }\|_p <\\
& < \frac{\epsilon}{5} + \frac{\epsilon}{5} + \|\tau_{h_i}f\chi_{[-k,k] }- f\chi_{[-k,k] }\|_p =\\
&  = \frac{2\epsilon}{5} +  \|(\tau_{h_i}f\chi_{[-k,k] }- \tau_{h_i}g)- (f\chi_{[-k,k] }-g)+(\tau_{h_i}g-g)\|_p\leq \\
& \leq \frac{2\epsilon}{5} + \|\tau_{h_i}f\chi_{[-k,k] }- \tau_{h_i}g\|_p + \|f\chi_{[-k,k] }-g\|_p+ \|\tau_{h_i}g-g\|_p\leq\\
& \leq \frac{2\epsilon}{5} +  \frac{\epsilon}{5} + \frac{\epsilon}{5} + \|\tau_{h_i}g-g\|_p= \\
& =  \frac{4\epsilon}{5} +  \|\tau_{h_i}g-g\|_p
\end{align*}
So, using $(5)$, we have that  there is $N\in \mathbb{N}$, such that if $i\geq N$,
$$ \|\tau_{h_i}f-f\|_p <  \frac{4\epsilon}{5} +  \|\tau_{h_i}g-g\|_p = \frac{4\epsilon}{5}+ \frac{\epsilon}{5}= \epsilon$$
Remark:
The above mentioned solution can be make shorter by using the fact that the continuous function with compact support (or the simple functions with compact support) are dense in $L^p(\mathbb{R})$. However, in Folland's book sections $6.1$ and $6.2$ and in their exercises, it  is not mentioned that continuous (resp. simple) function with compact support are dense in $L^p(\mathbb{R})$.
Shorter Proof:
Let $\mu$ be the Lebesgue measure on $\mathbb{R}$. Since $f\in L^p(\mathbb{R})$, and the continuous (resp. simple ) functions with compact support are dense in $L^p(\mathbb{R})$, let $g$ be a continuous (resp. simple ) function with compact support, such that
\begin{equation}
\|f  -g\|_p < \frac{\epsilon}{3} \ \ \ \ (6)
\end{equation}
It easy to see that, for any $h\in \mathbb{R}$,
\begin{equation}
\|\tau_hf  -\tau_hg\|_p =\|f -g\|_p < \frac{\epsilon}{3} \ \ \ \ \ (7)
\end{equation}
Now, since $g$ has compact support, there is $k\in \mathbb{N}$ such that $\textrm{supp} g \subseteq [-k,k]$. Let $\{h_i\}_i$ be any sequence of real numbers converging to $0$. We can, without loss of generality, assume that $|h_i|\leq 1$, for all $i$. Also, note that there is $M>0$ such that $|g(x)| <M$, for all $x\in\mathbb{R}$.
So we have that $\tau_{h_i}g$ converges pointwise to $g$, $|g|<M\chi_{[-k-1,k+1]} $ and, for all $i$, $|\tau_{h_i}g|<M\chi_{[-k-1,k+1]} $. So, $|\tau_{h_i}g-g|^p$ converges pointwise to $0$ and for all $i$, $$|\tau_{h_i}g-g|^p\leq2^p(|\tau_{h_i}g|^p+|g|^p)<2^pM^p\chi_{[-k-1,k+1]} $$
Since $2^pM^p\chi_{[-k-1,k+1]}$ is integrable, we can apply the Dominated Convergence Theorem and conclude that
$$ \int |\tau_{h_i}g-g|^p d\mu \to 0  \quad \textrm{ as } i \to \infty $$
So,
$$ \|\tau_{h_i}g-g\|_p \to 0  \quad \textrm{ as } i \to \infty $$
So, there is $N\in \mathbb{N}$, such that if $i\geq N$,
\begin{equation}
 \|\tau_{h_i}g-g\|_p <\frac{\epsilon}{3} \ \ \ \ \ (8)
\end{equation}
So, using $(6)$ and $(7)$, we have
\begin{align*}
\|\tau_{h_i}f-f\|_p &=  \|(\tau_{h_i}f- \tau_{h_i}g)- (f-g)+(\tau_{h_i}g-g)\|_p\leq \\
& \leq \|\tau_{h_i}f- \tau_{h_i}g\|_p + \|f-g\|_p+ \|\tau_{h_i}g-g\|_p\leq\\
& \leq \frac{\epsilon}{3} + \frac{\epsilon}{3} + \|(\tau_{h_i}g-g)\|_p= \\
& =  \frac{2\epsilon}{3} +  \|(\tau_{h_i}g-g)\|_p
\end{align*}
So, using $(8)$, we have that  there is $N\in \mathbb{N}$, such that if $i\geq N$,
$$ \|\tau_{h_i}f-f\|_p <  \frac{2\epsilon}{3} +  \|(\tau_{h_i}g-g)\|_p = \frac{2\epsilon}{3}+ \frac{\epsilon}{3}= \epsilon$$
So, for any $\{h_i\}_i$ sequence of real numbers converging to $0$, we have
$$\lim_{i \to \infty}\|\tau_{h_i}f - f\|_{L^p}=0$$
So, we have
$$\lim_{h\to 0}\|\tau_hf - f\|_{L^p}=0$$
So, for any $\{h_i\}_i$ sequence of real numbers converging to $0$, we have
$$\lim_{i \to \infty}\|\tau_{h_i}f - f\|_{L^p}=0$$
So, we have
$$\lim_{h\to 0}\|\tau_hf - f\|_{L^p}=0$$
Hope this helps
A: Let's start by considering a single step function. If $f=\chi_{A}$, $A=(a,b)$, then when $x$ and $x_{0}$ are close (and supposing that $x<x_{0},$ the other case is similar): $$(f_{x}-f_{x_{0}})(t)=\begin{cases}1&\text{if }x-b\leq t\leq x_{0}-b\\-1&\text{if }x-a\leq t\leq x_{0}-a\\0&\text{otherwise.}\end{cases}$$ Thus (once $x$ and $x_{0}$ are close enough), $\|f_{x}-f_{x_{0}}\|_{p}=2|x-x_{0}|$, and clearly this tends to 0 as $x\rightarrow x_{0}.$
For the step function $f=\sum_{i=1}^{n}a_{i}\chi_{A_{i}}$, we have $$\|f_{x}-f_{x_{0}}\|_{p}\leq\sum_{i=1}^{n}|a_{i}|\|(\chi_{A_{i}})_{x}-(\chi_{A_{i}})_{x_{0}}\|_{p}\rightarrow0\text{ as }x\rightarrow x_{0},$$ since this is true for each $\chi_{A_{i}}$. 
Now for $f\in L^{p}$, given $\varepsilon>0$, there is a step function $g$ with compact support such that $\|f-g\|_{p}<\varepsilon$. Clearly, this also means for any $x$ that $\|f_{x}-g_{x}\|_{p}<\varepsilon$. Now we have: $$\|f_{x}-f_{x_{0}}\|_{p}\leq \|f_{x}-g_{x}\|_{p}+\|g_{x}-g_{x_{0}}\|_{p}+\|f_{x_{0}}-g_{x_{0}}\|_{p}<\|g_{x}-g_{x_{0}}\|_{p}+2\varepsilon.$$ This means that for every $\varepsilon>0$, $$\lim\sup_{x\rightarrow x_{0}}\|f_{x}-f_{x_{0}}\|_{p}\leq 2\varepsilon,$$ which completes the proof.
