Uniform convergence of $L^1$ norms Let $f \in L^1(\mathbb{R}^d)$. How can one show that
\begin{equation}
\sup_{|y| \leq \frac{1}{n}} \int_{\mathbb{R}^d} |f(x-y)-f(x)|dx \longrightarrow 0
\end{equation}
as $n \longrightarrow \infty$? 
Edit: Is there another way to prove this result? (preferably without using a density argument)
 A: Consider $\varphi \in C_c(\mathbb{R}^d)$. The support of $\varphi$ being compact, we deduce that $\varphi$ is uniformly continous, i.e.
\begin{equation}
\forall \varepsilon > 0, \exists \delta > 0 : \forall x, y \in supp(\varphi), |x-y| \leq \delta \Rightarrow |\varphi(x)-\varphi(y)| \leq \varepsilon.
\end{equation}
The first step is to prove that for any $\varepsilon >0$, there is $N \in \mathbb{N}$ such that for all $n \geq N$ we have 
\begin{equation}
\int_{\mathbb{R}^d} |\varphi(x-y)-\varphi(x)|dx \leq \varepsilon, \quad \forall |y| \leq \frac{1}{n}.
\end{equation}
We observe (if needed I can write it down) that the domain of integration can be reduced to the support of $\varphi$, $supp(\varphi)$. This allows us to use the uniform continuity of $\varphi$ to obtain
\begin{equation}
\int_{\mathbb{R}^d}|\varphi(x-y)-\varphi(x)|dx = \int_{supp(\varphi)}|\varphi(x-y)-\varphi(x)|dx \leq \varepsilon \text{mes}(supp(\varphi)),
\end{equation}
whenever $|y| \leq \delta$. We choose $N = [\frac{1}{\delta}]+1$ and so the result holds for $\varphi \in C_c(\mathbb{R}^d)$. 
Now let $f \in L^1(\mathbb{R}^d)$. By density of $C_c(\mathbb{R}^d)$, we may choose $\varphi \in C_c(\mathbb{R}^d)$ such that $\|f - \varphi\|_{L^1(\mathbb{R}^d)} \leq \varepsilon$. Thus, by using the triangle inequality, we obtain
\begin{align}
\int_{\mathbb{R}^d} |f(x-y)-f(x)|dx \leq \int_{\mathbb{R}^d} |f(x-y)-\varphi(x-y)|dx &+ \int_{\mathbb{R}^d} |\varphi(x-y)-\varphi(x)|dx \\
&+ \| \varphi - f \|_{L^1(\mathbb{R}^d)}, 
\end{align} 
and we may conclude by using the established estimates, which hold uniformly for any $|y| \leq \frac{1}{n}$.
Remark: the estimates could be sharpened to obtain a clear $\leq \varepsilon$ in the end.
