The $L^1$ convergence of $f(x - a_n) \to f(x)$

I'm trying to solve something, and I'm stuck.

Suppose you have a function $$f \in L^1(\mathbf R) $$ and a sequence of real numbers that converges to zero: $$ a_n \rightarrow 0 $$ define a sequence of functions by $$ f_n (x)=f(x-a_n)$$ does this sequence converges to $f(x)$ in the norm?

So what you are trying to do is prove that

$$\int |f(x) - f_n(x)| \, \text{d}x \to 0.$$

Note that $|f(x) - f(x - a_n)| \to 0$ pointwise if $f$ is continuous with compact support. Furthermore, note that $f$ is uniformly continuous because it has compact support (and hence we can concatenate the zero function with a function on a bounded interval). So, given $\epsilon > 0$ we have $|f(x) - f_n(x)| < \epsilon$ for all $x$ and $n$ large enough. This means that for $n$ large enough we can find a compact support for the function $f - f_n$ so, we can bound this by an integrable function. So we can apply the Lebesgue Dominated Convergence Theorem to obtain the result for continuous $f$ with compact support.

Now we know that the compactly supported continuous functions are dense in $L^1(\mathbf R)$. So, let $f$ be a general $L^1$ function and let given $\epsilon > 0$, $g$ be a compactly supported continuous function such that $\|f - g\| < \epsilon$.

So, we get $\|f - f_n\|_1 \leq \|f - g\| + \|g - f_n\|$.

This requires us to show that $\|g - f_n\|$ can be made arbitrary small for large enough $n$. So given $\epsilon > 0$, let $g_n(x) = g(x - a_n)$. So,

$$\|g - f_n\| \leq \|g - g_n\| + \|g_n - f_n\|$$

The first term on the RHS goes to $0$ and the second one can be made arbitrarily small by density.

So, to solve these kinds of problems, first pick a class of functions that is dense in the required set where the problem becomes much easier.