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)|
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)|
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
